Chapter 7 Algebraic Expression, Identities & Factorisation

Like terms are terms that have the same variables raised to the same powers. The numerical coefficient can be different.

The given term is $24a^2bc$. The variable part is $a^2bc$ (which means $a^2b^1c^1$).

Let's simplify each option:

(a) $13 \times 8a \times 2b \times c \times a = (13 \times 8 \times 2) \times (a \times a) \times b \times c = 208 \times a^{1+1} \times b \times c = 208a^2bc$.

The variable part is $a^2bc$.

(b) $8 \times 3 \times a \times b \times c = (8 \times 3) \times a \times b \times c = 24abc$.

The variable part is $abc$.

(c) $3 \times 8 \times a \times b \times c \times c = (3 \times 8) \times a \times b \times (c \times c) = 24abc^2$.

The variable part is $abc^2$.

(d) $3 \times 8 \times a \times b \times b \times c = (3 \times 8) \times a \times (b \times b) \times c = 24ab^2c$.

The variable part is $ab^2c$.

Comparing the variable parts of the options with the given term $24a^2bc$, only option (a) has the variable part $a^2bc$.

Therefore, $13 \times 8a \times 2b \times c \times a$ is a like term as $24a^2bc$.

The correct answer is (a).

An identity is an equation that is true for all possible values of the variables involved.

We need to check each given option to see which equation holds true for any values of $p$ and $q$.

(a) $(p + q)^2 = p^2 + q^2$

Let's expand the left side: $(p + q)^2 = (p+q)(p+q) = p^2 + pq + qp + q^2 = p^2 + 2pq + q^2$.

So the equation is $p^2 + 2pq + q^2 = p^2 + q^2$.

Subtracting $p^2 + q^2$ from both sides, we get $2pq = 0$. This is only true if $p=0$ or $q=0$, not for all values of $p$ and $q$. Thus, it is not an identity.

(b) $p^2 – q^2 = (p – q)^2$

Let's expand the right side: $(p – q)^2 = (p-q)(p-q) = p^2 - pq - qp + q^2 = p^2 - 2pq + q^2$.

So the equation is $p^2 – q^2 = p^2 - 2pq + q^2$.

Subtracting $p^2$ from both sides, we get $-q^2 = -2pq + q^2$.

Adding $q^2$ and $2pq$ to both sides, we get $2pq = 2q^2$, or $2q(p-q)=0$. This is only true if $q=0$ or $p=q$, not for all values of $p$ and $q$. Thus, it is not an identity.

(c) $p^2 – q^2 = p^2 + 2pq – q^2$

Subtracting $p^2$ and adding $q^2$ to both sides, we get $0 = 2pq$. This is only true if $p=0$ or $q=0$, not for all values of $p$ and $q$. Thus, it is not an identity.

(d) $(p + q)^2 = p^2 + 2pq + q^2$

As expanded in option (a), the left side is $(p + q)^2 = p^2 + 2pq + q^2$.

The right side is also $p^2 + 2pq + q^2$.

Since the left side is equal to the right side for all values of $p$ and $q$, this equation is true for all values of the variables. Thus, it is an identity.

The correct answer is (d).

To find the irreducible factorization of $3a^3 + 6a$, we need to find the greatest common factor (GCF) of the terms $3a^3$ and $6a$.

The term $3a^3$ can be written as $3 \times a \times a \times a$.

The term $6a$ can be written as $2 \times 3 \times a$.

The common factors are $3$ and $a$.

The GCF of $3a^3$ and $6a$ is $3a$.

Now, we factor out the GCF from the expression:

The factors are $3a$ and $(a^2 + 2)$. The term $3a$ is irreducible because it is a product of a prime number and a single variable.

The term $a^2 + 2$ is irreducible over the real numbers because $a^2$ is always non-negative, so $a^2 + 2$ is always positive and cannot be factored into simpler linear factors with real coefficients.

Comparing this factorization with the given options:

(a) $3a (a^2 + 2)$ - This matches our result.

(b) $3 (a^3 + 2)$ - This is a factorization, but $a^3 + 2$ is not irreducible in general algebraic factoring.

(c) $a (3a^2 + 6)$ - This is a factorization, but $3a^2 + 6$ is not irreducible as it can be factored further: $3a^2 + 6 = 3(a^2 + 2)$. So this factorization is $a \times 3(a^2 + 2) = 3a(a^2+2)$, but the intermediate factor $a(3a^2+6)$ is not the final irreducible form.

(d) $3 \times a \times a \times a + 2 \times 3 \times a$ - This is the expanded form of the original expression, not a factorization.

The irreducible factorization of $3a^3 + 6a$ is $3a(a^2 + 2)$.

The correct answer is (a).

The given equation is $a(b + c) = ab + ac$.

Let's consider the definition of the properties listed:

The commutative property states that the order of operands does not change the result for addition and multiplication (e.g., $x + y = y + x$ and $x \times y = y \times x$).

The associative property states that the grouping of operands does not change the result for addition and multiplication (e.g., $(x + y) + z = x + (y + z)$ and $(x \times y) \times z = x \times (y \times z)$).

The closure property states that the set is 'closed' under an operation if performing the operation on elements of the set always produces an element of the same set.

The distributive property of multiplication over addition states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products. It is expressed as $a \times (b + c) = (a \times b) + (a \times c)$, or simply $a(b+c) = ab + ac$.

Comparing the given equation $a(b + c) = ab + ac$ with the definitions, we see that it directly matches the form of the distributive property.

The correct answer is (b).

The process of writing an algebraic expression as a product of two or more expressions is called factorization.

The representation of an expression as the product of its factors is called factorisation.

To find the missing term, we need to expand the expression $(x + a)(x + b)$.

Using the distributive property, we multiply each term in the first parenthesis by each term in the second parenthesis:

Apply the distributive property again:

Group the terms with $x$:

Comparing this expanded form with the given expression $x^2 + (a + b)x + \text{________}$, the missing term is $ab$.

So the completed statement is:

$(x + a)(x + b) = x^2 + (a + b)x + \mathbf{ab}$.

An identity is, by definition, an equation that holds true for every possible value that the variables can take. This is what distinguishes it from a conditional equation, which is only true for specific values of the variables.

Therefore, the statement "An identity is true for all values of its variables" is correct.

The statement is True (T).

We need to find the common factors of the terms $x^2y$ and $-xy^2$.

Let's write the factors of each term:

$x^2y = x \times x \times y$

$-xy^2 = -1 \times x \times y \times y$

The factors common to both terms are $x$ and $y$. The product of these common factors is $x \times y = xy$. While $-1$ is a factor of the second term, it is not a common factor of both terms (unless we consider numerical factors generally, in which case 1 and -1 are common numerical factors, leading to common factors $xy$ and $-xy$). However, when asked for "the common factor", the term with positive numerical coefficient is usually implied if variables are involved.

The greatest common factor (GCF) in terms of variables is $xy$.

The statement says the common factor is $xy$. This is correct as $xy$ divides both $x^2y$ and $-xy^2$.

The statement is True (T).

We need to evaluate the left side of the given equation and compare it with the right side.

The left side is $(3x + 3x^2) \div 3x$.

We can perform this division by dividing each term in the numerator by $3x$:

Simplify each term:

So, the left side of the equation simplifies to $1 + x$.

The given equation is $1 + x = 3x^2$.

This equation is only true for specific values of $x$ (solutions to $3x^2 - x - 1 = 0$), not for all values of $x$. Therefore, it is not an identity, and the statement is not always true.

The statement "(3x + 3x²) ÷ 3x = 3x²" is false.

The statement is False (F).

To simplify the expressions, we perform the indicated multiplications.

(i) Simplify $- pqr (p^2 + q^2 + r^2)$

We distribute the monomial $-pqr$ to each term inside the parenthesis using the distributive property $a(b+c+d) = ab + ac + ad$.

Now, we multiply the terms, adding the exponents of the same variables ($x^m \times x^n = x^{m+n}$).

(ii) Simplify $(px + qy) (ax – by)$

We multiply these two binomials. We can use the distributive property twice, multiplying each term of the first binomial by each term of the second binomial (often remembered by the acronym FOIL: First, Outer, Inner, Last).

Now distribute $px$ and $qy$ into the second parenthesis:

Perform the multiplications:

The $xy$ terms can be combined by factoring out $xy$, although they are not like terms if $qa \neq -pb$. We can write it as:

Both forms are considered simplified.

Solution:

(i) (3x + 7y) (3x – 7y)

We can use the algebraic identity: $(a+b)(a-b) = a^2 - b^2$.

Here, $a = 3x$ and $b = 7y$.

Substituting these values into the identity:

$(3x + 7y)(3x - 7y) = (3x)^2 - (7y)^2$

$(3x)^2 = 3^2 x^2 = 9x^2$

$(7y)^2 = 7^2 y^2 = 49y^2$

Therefore, the expansion is:

$(3x + 7y)(3x - 7y) = 9x^2 - 49y^2$

(ii) $\left( \frac{4x}{5}+\frac{y}{4} \right)\left( \frac{4x}{5}+\frac{3y}{4} \right)$

We can use the algebraic identity: $(u+a)(u+b) = u^2 + (a+b)u + ab$.

Here, $u = \frac{4x}{5}$, $a = \frac{y}{4}$, and $b = \frac{3y}{4}$.

Substituting these values into the identity:

$\left( \frac{4x}{5}+\frac{y}{4} \right)\left( \frac{4x}{5}+\frac{3y}{4} \right) = \left(\frac{4x}{5}\right)^2 + \left(\frac{y}{4} + \frac{3y}{4}\right)\left(\frac{4x}{5}\right) + \left(\frac{y}{4}\right)\left(\frac{3y}{4}\right)$

Let's calculate each term:

First term: $\left(\frac{4x}{5}\right)^2 = \frac{(4x)^2}{5^2} = \frac{16x^2}{25}$

Second term: $\left(\frac{y}{4} + \frac{3y}{4}\right)\left(\frac{4x}{5}\right) = \left(\frac{y+3y}{4}\right)\left(\frac{4x}{5}\right) = \left(\frac{4y}{4}\right)\left(\frac{4x}{5}\right) = (y)\left(\frac{4x}{5}\right) = \frac{4xy}{5}$

Third term: $\left(\frac{y}{4}\right)\left(\frac{3y}{4}\right) = \frac{y \times 3y}{4 \times 4} = \frac{3y^2}{16}$

Combining the terms, the expansion is:

$\left( \frac{4x}{5}+\frac{y}{4} \right)\left( \frac{4x}{5}+\frac{3y}{4} \right) = \frac{16x^2}{25} + \frac{4xy}{5} + \frac{3y^2}{16}$

Solution:

(i) Factorise 21x²y³ + 27x³y²

We find the greatest common factor (GCF) of the two terms.

The GCF of the coefficients 21 and 27 is 3.

The lowest power of $x$ is $x^2$.

The lowest power of $y$ is $y^2$.

So, the GCF of $21x^2y^3$ and $27x^3y^2$ is $3x^2y^2$.

Factor out the GCF:

$21x^2y^3 + 27x^3y^2 = 3x^2y^2 \left(\frac{21x^2y^3}{3x^2y^2} + \frac{27x^3y^2}{3x^2y^2}\right)$

$21x^2y^3 + 27x^3y^2 = 3x^2y^2(7y + 9x)$

(ii) Factorise a³ – 4a² + 12 – 3a

We rearrange the terms and group them:

$a^3 - 4a^2 - 3a + 12$

Group the first two terms and the last two terms:

$(a^3 - 4a^2) + (-3a + 12)$

Factor out the common factor from each group:

$a^2(a - 4) - 3(a - 4)$

Now, we have a common binomial factor $(a - 4)$. Factor it out:

$(a - 4)(a^2 - 3)$

(iii) Factorise 4x² – 20x + 25

This expression is a trinomial. We check if it is a perfect square trinomial of the form $(A - B)^2 = A^2 - 2AB + B^2$.

Here, $4x^2 = (2x)^2$ and $25 = 5^2$.

Let $A = 2x$ and $B = 5$.

Check the middle term: $-2AB = -2(2x)(5) = -20x$.

This matches the given middle term.

So, the expression is a perfect square trinomial.

$4x^2 - 20x + 25 = (2x)^2 - 2(2x)(5) + 5^2 = (2x - 5)^2$

(iv) Factorise $\frac{y^2}{9}$ - 9

This expression is in the form of a difference of squares, $A^2 - B^2$, which factors as $(A - B)(A + B)$.

Here, $A^2 = \frac{y^2}{9} = \left(\frac{y}{3}\right)^2$, so $A = \frac{y}{3}$.

And $B^2 = 9 = 3^2$, so $B = 3$.

Using the identity $A^2 - B^2 = (A - B)(A + B)$:

$\frac{y^2}{9} - 9 = \left(\frac{y}{3}\right)^2 - 3^2 = \left(\frac{y}{3} - 3\right)\left(\frac{y}{3} + 3\right)$

(v) Factorise x⁴ – 256

This expression is a difference of squares: $x^4 = (x^2)^2$ and $256 = 16^2$.

Using the identity $A^2 - B^2 = (A - B)(A + B)$ with $A = x^2$ and $B = 16$:

$x^4 - 256 = (x^2)^2 - 16^2 = (x^2 - 16)(x^2 + 16)$

Now, notice that the first factor $(x^2 - 16)$ is also a difference of squares: $x^2 = x^2$ and $16 = 4^2$.

Using the identity again with $A = x$ and $B = 4$ for $(x^2 - 16)$:

$x^2 - 16 = x^2 - 4^2 = (x - 4)(x + 4)$

The second factor $(x^2 + 16)$ cannot be factorised further into real linear factors.

Combining the factors, the complete factorisation is:

$x^4 - 256 = (x - 4)(x + 4)(x^2 + 16)$

Solution:

(i) Evaluate (48)²

We can write $48$ as $50 - 2$.

So, $(48)^2 = (50 - 2)^2$.

We use the identity $(a - b)^2 = a^2 - 2ab + b^2$, where $a = 50$ and $b = 2$.

$(50 - 2)^2 = (50)^2 - 2(50)(2) + (2)^2$

$(50)^2 = 2500$

$2(50)(2) = 200$

$(2)^2 = 4$

$(50 - 2)^2 = 2500 - 200 + 4$

$2500 - 200 + 4 = 2300 + 4 = 2304$

Thus, $(48)^2 = 2304$.

(ii) Evaluate 181² – 19²

This is in the form of a difference of squares, $a^2 - b^2$.

We use the identity $a^2 - b^2 = (a - b)(a + b)$, where $a = 181$ and $b = 19$.

$181^2 - 19^2 = (181 - 19)(181 + 19)$

Calculate the terms inside the parentheses:

$181 - 19 = 162$

$181 + 19 = 200$

$181^2 - 19^2 = (162)(200)$

$162 \times 200 = 32400$

Thus, $181^2 - 19^2 = 32400$.

(iii) Evaluate 497 × 505

We can write $497$ as $500 - 3$ and $505$ as $500 + 5$.

So, $497 \times 505 = (500 - 3)(500 + 5)$.

We use the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$, where $x = 500$, $a = -3$, and $b = 5$.

$(500 - 3)(500 + 5) = (500)^2 + (-3 + 5)(500) + (-3)(5)$

$(500)^2 = 250000$

$(-3 + 5)(500) = (2)(500) = 1000$

$(-3)(5) = -15$

$(500 - 3)(500 + 5) = 250000 + 1000 - 15$

$250000 + 1000 - 15 = 251000 - 15 = 250985$

Thus, $497 \times 505 = 250985$.

(iv) Evaluate 2.07 × 1.93

We can write $2.07$ as $2 + 0.07$ and $1.93$ as $2 - 0.07$.

So, $2.07 \times 1.93 = (2 + 0.07)(2 - 0.07)$.

This is in the form of a difference of squares, $(a + b)(a - b)$.

We use the identity $(a + b)(a - b) = a^2 - b^2$, where $a = 2$ and $b = 0.07$.

$(2 + 0.07)(2 - 0.07) = (2)^2 - (0.07)^2$

$(2)^2 = 4$

$(0.07)^2 = 0.07 \times 0.07 = 0.0049$

$(2 + 0.07)(2 - 0.07) = 4 - 0.0049$

$4 - 0.0049 = 3.9951$

Thus, $2.07 \times 1.93 = 3.9951$.

Solution:

We need to verify that $(3x + 5y)^2 – 30xy = 9x^2 + 25y^2$.

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS = $(3x + 5y)^2 – 30xy$

First, expand the term $(3x + 5y)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$.

Here, $a = 3x$ and $b = 5y$.

So, $(3x + 5y)^2 = (3x)^2 + 2(3x)(5y) + (5y)^2$

$(3x)^2 = 9x^2$

$2(3x)(5y) = 30xy$

$(5y)^2 = 25y^2$

Substituting these back into the expansion:

$(3x + 5y)^2 = 9x^2 + 30xy + 25y^2$

Now, substitute this back into the LHS expression:

LHS = $(9x^2 + 30xy + 25y^2) – 30xy$

Combine the like terms ($30xy$ and $-30xy$):

LHS = $9x^2 + (30xy - 30xy) + 25y^2$

LHS = $9x^2 + 0 + 25y^2$

LHS = $9x^2 + 25y^2$

This is the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is verified.

Therefore, $(3x + 5y)^2 – 30xy = 9x^2 + 25y^2$ is true.

Solution:

We need to verify that $(11pq + 4q)^2 – (11pq – 4q)^2 = 176pq^2$.

We will start with the Left Hand Side (LHS) of the equation and simplify it.

LHS = $(11pq + 4q)^2 – (11pq – 4q)^2$

This expression is in the form of a difference of squares, $A^2 - B^2$, where $A = (11pq + 4q)$ and $B = (11pq - 4q)$.

We use the identity $A^2 - B^2 = (A - B)(A + B)$.

Let's find $(A - B)$:

$A - B = (11pq + 4q) - (11pq - 4q)$

$A - B = 11pq + 4q - 11pq + 4q$

$A - B = (11pq - 11pq) + (4q + 4q)$

$A - B = 0 + 8q = 8q$

Now, let's find $(A + B)$:

$A + B = (11pq + 4q) + (11pq - 4q)$

$A + B = 11pq + 4q + 11pq - 4q$

$A + B = (11pq + 11pq) + (4q - 4q)$

$A + B = 22pq + 0 = 22pq$

Now, substitute these results back into the difference of squares identity:

LHS = $(A - B)(A + B) = (8q)(22pq)$

Multiply the terms:

LHS = $8 \times 22 \times q \times pq$

LHS = $176 \times p \times q^2$

LHS = $176pq^2$

This is the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is verified.

Therefore, $(11pq + 4q)^2 – (11pq – 4q)^2 = 176pq^2$ is true.

Alternate Solution:

We can also expand each term separately.

Using $(a+b)^2 = a^2 + 2ab + b^2$ for the first term, with $a=11pq$ and $b=4q$:

$(11pq + 4q)^2 = (11pq)^2 + 2(11pq)(4q) + (4q)^2$

$(11pq + 4q)^2 = 121p^2q^2 + 88pq^2 + 16q^2$

Using $(a-b)^2 = a^2 - 2ab + b^2$ for the second term, with $a=11pq$ and $b=4q$:

$(11pq - 4q)^2 = (11pq)^2 - 2(11pq)(4q) + (4q)^2$

$(11pq - 4q)^2 = 121p^2q^2 - 88pq^2 + 16q^2$

Now, subtract the second expansion from the first:

LHS = $(121p^2q^2 + 88pq^2 + 16q^2) - (121p^2q^2 - 88pq^2 + 16q^2)$

LHS = $121p^2q^2 + 88pq^2 + 16q^2 - 121p^2q^2 + 88pq^2 - 16q^2$

Combine like terms:

LHS = $(121p^2q^2 - 121p^2q^2) + (88pq^2 + 88pq^2) + (16q^2 - 16q^2)$

LHS = $0 + 176pq^2 + 0$

LHS = $176pq^2$

This is the Right Hand Side (RHS) of the given equation.

Since LHS = RHS, the identity is verified.

Given:

Area of the rectangle = $x^2 + 12xy + 27y^2$

Length of the rectangle = $(x + 9y)$

To Find:

The breadth of the rectangle.

Solution:

The formula for the area of a rectangle is:

Area = Length $\times$ Breadth

To find the breadth, we can rearrange the formula:

Breadth = $\frac{\text{Area}}{\text{Length}}$

Substitute the given expressions for Area and Length:

Breadth = $\frac{x^2 + 12xy + 27y^2}{x + 9y}$

Now, we need to simplify this expression. We can factorise the numerator, which is a quadratic expression in terms of $x$ and $y$. We look for two terms that multiply to $27y^2$ and add up to $12y$ (when considering the coefficient of $x$). These terms are $3y$ and $9y$.

So, we can factorise the numerator as:

$x^2 + 12xy + 27y^2 = x^2 + 3xy + 9xy + 27y^2$

$= x(x + 3y) + 9y(x + 3y)$

$= (x + 3y)(x + 9y)$

Now substitute this factorised form back into the expression for Breadth:

Breadth = $\frac{(x + 3y)(x + 9y)}{x + 9y}$

Assuming $x + 9y \neq 0$, we can cancel the common factor $(x + 9y)$ from the numerator and the denominator:

Breadth = $x + 3y$

Thus, the breadth of the rectangle is $(x + 3y)$.

Solution:

We need to divide $15 (y + 3) (y^2 – 16)$ by $5 (y^2 – y – 12)$.

We can write this as a fraction:

$\frac{15 (y + 3) (y^2 – 16)}{5 (y^2 – y – 12)}$

First, factorise the expressions in the numerator and the denominator.

Numerator: $15 (y + 3) (y^2 – 16)$

The term $y^2 – 16$ is a difference of squares, which can be factorised as $(y - 4)(y + 4)$.

So, the numerator is $15 (y + 3) (y - 4) (y + 4)$.

Denominator: $5 (y^2 – y – 12)$

The quadratic expression $y^2 – y – 12$ can be factorised. We look for two numbers that multiply to $-12$ and add up to $-1$. These numbers are $-4$ and $3$.

So, $y^2 – y – 12 = (y - 4)(y + 3)$.

The denominator is $5 (y - 4) (y + 3)$.

Now, substitute the factorised forms into the fraction:

$\frac{15 (y + 3) (y - 4) (y + 4)}{5 (y - 4) (y + 3)}$

Cancel out the common factors in the numerator and the denominator, assuming $y+3 \neq 0$ and $y-4 \neq 0$:

$\frac{\cancel{15}^{3} \cancel{(y + 3)} \cancel{(y - 4)} (y + 4)}{\cancel{5}_{1} \cancel{(y - 4)} \cancel{(y + 3)}}$

The remaining terms are $3$ in the numerator and $1$ in the denominator, multiplied by $(y+4)$.

So, the simplified expression is $3(y+4)$.

Therefore, the division results in $3(y+4)$, which can also be written as $3y + 12$.

Solution:

We are given the expression $x + \frac{1}{x} = 5$.

We want to evaluate $x^2 + \frac{1}{x^2}$.

Consider the given equation:

$x + \frac{1}{x} = 5$

Square both sides of the equation:

$\left(x + \frac{1}{x}\right)^2 = 5^2$

Expand the left side using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a=x$ and $b=\frac{1}{x}$.

$x^2 + 2 \left(x\right) \left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 25$

Simplify the middle term:

$x^2 + 2 \times 1 + \frac{1}{x^2} = 25$

$x^2 + 2 + \frac{1}{x^2} = 25$

Now, isolate the term $x^2 + \frac{1}{x^2}$ by subtracting 2 from both sides of the equation:

$x^2 + \frac{1}{x^2} = 25 - 2$

$x^2 + \frac{1}{x^2} = 23$

Thus, the value of $x^2 + \frac{1}{x^2}$ is 23.

Solution:

We need to find the value of $\frac{38^{2} \;-\; 22^{2}}{16}$.

The numerator, $38^2 - 22^2$, is in the form of a difference of squares, $a^2 - b^2$.

We use the algebraic identity: $a^2 - b^2 = (a - b)(a + b)$.

Here, $a = 38$ and $b = 22$.

Apply the identity to the numerator:

$38^2 - 22^2 = (38 - 22)(38 + 22)$

Calculate the terms inside the parentheses:

$38 - 22 = 16$

$38 + 22 = 60$

So, the numerator is $(16)(60)$.

Substitute this back into the given expression:

$\frac{38^{2} \;-\; 22^{2}}{16} = \frac{(16)(60)}{16}$

Assuming the denominator $16 \neq 0$, we can cancel out the common factor of 16 from the numerator and the denominator:

$\frac{\cancel{16} \times 60}{\cancel{16}}$

The expression simplifies to 60.

Therefore, the value of $\frac{38^{2} \;-\; 22^{2}}{16}$ is 60.

Solution:

We are given the equation:

$10000x = (9982)^2 – (18)^2$

We need to find the value of $x$.

Let's simplify the Right Hand Side (RHS) of the equation first.

RHS = $(9982)^2 – (18)^2$

This expression is in the form of a difference of squares, $a^2 - b^2$.

We use the algebraic identity: $a^2 - b^2 = (a - b)(a + b)$.

Here, $a = 9982$ and $b = 18$.

Apply the identity to the RHS:

RHS = $(9982 - 18)(9982 + 18)$

Calculate the terms inside the parentheses:

$9982 - 18 = 9964$

$9982 + 18 = 10000$

So, RHS = $(9964)(10000)$

Now, substitute this back into the original equation:

$10000x = 9964 \times 10000$

To find the value of $x$, divide both sides of the equation by 10000:

$x = \frac{9964 \times 10000}{10000}$

Cancel out the common factor of 10000 from the numerator and the denominator:

$x = \frac{9964 \times \cancel{10000}}{\cancel{10000}}$

$x = 9964$

Therefore, the value of $x$ is 9964.

Let a monomial be represented by a single term, say $M$.

Let a binomial be represented by the sum of two terms, say $B_1 + B_2$.

The product of the monomial and the binomial is $M \times (B_1 + B_2)$.

Using the distributive property of multiplication over addition, the product is $M \times B_1 + M \times B_2$.

Assuming $M$, $B_1$, and $B_2$ are non-zero terms such that $M \times B_1$ and $M \times B_2$ are unlike terms (which is the general case unless specified otherwise), the result $M B_1 + M B_2$ is an expression containing exactly two terms.

An algebraic expression with two terms is called a binomial.

Therefore, the product of a monomial and a binomial is a binomial.

The correct option is (b) binomial.

A polynomial is an algebraic expression that consists of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

For example, in the polynomial $3x^2 + 2x - 5$, the exponents of the variable $x$ are $2$, $1$ (since $x = x^1$), and $0$ (since $-5 = -5x^0$). These exponents are $2, 1, 0$, which are all non-negative integers.

Expressions like $x^{-1} = \frac{1}{x}$ or $x^{1/2} = \sqrt{x}$ are not considered polynomials because their exponents ($-1$ and $\frac{1}{2}$) are not non-negative integers.

Therefore, by the definition of a polynomial, the exponents of the variables must be non-negative integers ($0, 1, 2, 3, ...$).

The correct option is (c) non-negative integers.

Let's expand the expression $(a – b)^2$:

$(a – b)^2 = (a – b) \times (a – b)$

Using the distributive property (multiplying each term in the first parenthesis by each term in the second parenthesis):

$(a – b)^2 = a \times (a – b) - b \times (a – b)$

$(a – b)^2 = (a \times a) - (a \times b) - (b \times a) + (-b \times -b)$

$(a – b)^2 = a^2 - ab - ba + b^2$

Since $ab = ba$ (commutative property of multiplication), we can combine the middle terms:

Now let's examine the given options:

(a) $(a – b)^2 = a^2 + 2ab – b^2$ (Incorrect, based on equation (i))

(b) $(a – b)^2 = a^2 – 2ab + b^2$ (Correct, based on equation (i))

(c) $(a – b)^2 = a^2 – b^2$ (Incorrect, this is the expansion of $(a-b)(a+b)$)

(d) $(a + b)^2 = a^2 + 2ab – b^2$ (Incorrect, the correct expansion of $(a+b)^2$ is $a^2 + 2ab + b^2$)

The correct option is (b).

We are asked to find the sum of $-7pq$ and $2pq$.

The given terms, $-7pq$ and $2pq$, are like terms because they have the same variables ($p$ and $q$) raised to the same powers (power $1$ for both $p$ and $q$).

To add like terms, we add their numerical coefficients and keep the variable part the same.

The coefficients are $-7$ and $2$.

Sum of coefficients $= -7 + 2 = -5$.

The variable part is $pq$.

So, the sum of $-7pq$ and $2pq$ is $-5pq$.

The correct option is (d) – 5pq.

We need to subtract $-3x^2y^2$ from $x^2y^2$. This can be written as:

$x^2y^2 - (-3x^2y^2)$

Subtracting a negative value is the same as adding the corresponding positive value. So, $- (-3x^2y^2)$ becomes $+ 3x^2y^2$.

The expression becomes:

$x^2y^2 + 3x^2y^2$

The terms $x^2y^2$ and $3x^2y^2$ are like terms because they have the same variables ($x$ and $y$) raised to the same powers ($2$ for $x$ and $2$ for $y$).

To add like terms, we add their coefficients. The coefficient of $x^2y^2$ is $1$, and the coefficient of $3x^2y^2$ is $3$.

Sum of coefficients $= 1 + 3 = 4$.

So, the sum of the terms is $4$ times the common variable part $x^2y^2$.

$x^2y^2 + 3x^2y^2 = 4x^2y^2$

The correct option is (d) $4x^2y^2$.

Like terms are algebraic terms that have the same variables raised to the same powers. The numerical coefficients of like terms can be different.

The given term is $4m^3n^2$. The variables are $m$ and $n$. The power of $m$ is $3$ and the power of $n$ is $2$.

Let's examine the options:

(a) $4m^2n^2$: The power of $m$ is $2$, which is different from $3$. Not a like term.

(b) $-6m^3n^2$: The variables are $m$ and $n$. The power of $m$ is $3$ and the power of $n$ is $2$. These match the variables and powers of the given term. The coefficient is $-6$, which is a number. This is a like term.

(c) $6pm^3n^2$: This term contains the variable $p$, which is not present in the given term $4m^3n^2$. Not a like term.

(d) $4m^3n$: The variable $n$ has a power of $1$, which is different from $2$. Not a like term.

Only option (b) has the same variables with the same respective powers as the given term.

The correct option is (b) – 6m³n².

A binomial is an algebraic expression that contains exactly two terms.

Let's simplify each option to determine the number of terms:

(a) $7 \times a + a$

This simplifies to $7a + a$. Since $7a$ and $a$ are like terms, we can add them:

$7a + a = (7+1)a = 8a$

This expression has only one term ($8a$), so it is a monomial.

(b) $6a^2 + 7b + 2c$

This expression has three terms: $6a^2$, $7b$, and $2c$. These are unlike terms, so they cannot be combined further. This is a trinomial.

(c) $4a \times 3b \times 2c$

This expression involves multiplication of terms. We can multiply the coefficients and the variables:

$4a \times 3b \times 2c = (4 \times 3 \times 2) \times (a \times b \times c) = 24abc$

This expression has only one term ($24abc$), so it is a monomial.

(d) $6 (a^2 + b)$

This expression involves a number multiplied by a sum inside parentheses. We can use the distributive property:

$6 (a^2 + b) = 6 \times a^2 + 6 \times b = 6a^2 + 6b$

This expression has two terms: $6a^2$ and $6b$. These are unlike terms. Since it has exactly two terms, it is a binomial.

The correct option is (d) $6 (a^2 + b)$.

We need to find the sum of the three given expressions:

$(a – b + ab) + (b + c – bc) + (c – a – ac)$

Remove the parentheses and combine the terms:

$a – b + ab + b + c – bc + c – a – ac$

Group the like terms:

$(a - a) + (-b + b) + (c + c) + ab - bc - ac$

Combine the coefficients of the like terms:

$(1 - 1)a + (-1 + 1)b + (1 + 1)c + ab - bc - ac$

$0a + 0b + 2c + ab - bc - ac$

Simplify the expression:

$2c + ab - bc - ac$

Rearranging the terms to match the options:

$2c + ab - ac - bc$

Comparing this result with the given options, we find that it matches option (a).

The correct option is (a) $2c + ab – ac – bc$.

We need to find the product of the given monomials: $4p$, $-7q^3$, and $-7pq$.

Product $= (4p) \times (-7q^3) \times (-7pq)$

We multiply the numerical coefficients together and the variable parts together.

Product of coefficients $= 4 \times (-7) \times (-7)$

Product of coefficients $= -28 \times (-7)$

Product of coefficients $= 196$

Product of variable parts $= p \times q^3 \times pq$

Using the rule of exponents $a^m \times a^n = a^{m+n}$, we can combine the variables with the same base:

$p \times q^3 \times p^1 \times q^1 = (p^1 \times p^1) \times (q^3 \times q^1)$

$= p^{1+1} \times q^{3+1}$

$= p^2 \times q^4$

$= p^2q^4$

Now, combine the product of the coefficients and the product of the variable parts:

Product $= 196 \times p^2q^4 = 196p^2q^4$

Comparing this result with the given options, we find that it matches option (a).

The correct option is (a) $196 p^2q^4$.

The formula for the area of a rectangle is given by:

Area = Length $\times$ Breadth

Given:

Length = $4ab$

Breadth = $6b^2$

Substitute the given values into the formula:

Area $= (4ab) \times (6b^2)$

To multiply these monomials, we multiply the coefficients and the variable parts separately:

Multiply the coefficients: $4 \times 6 = 24$

Multiply the variable parts: $ab \times b^2 = (a^1 b^1) \times b^2$

Using the rule of exponents $x^m \times x^n = x^{m+n}$:

$a^1 \times (b^1 \times b^2) = a^1 \times b^{1+2} = a^1 b^3 = ab^3$

Combine the coefficient and variable part:

Area $= 24 \times ab^3 = 24ab^3$

Comparing this result with the given options, we find that it matches option (b).

The correct option is (b) $24ab^3$.

The formula for the volume of a rectangular box (cuboid) is given by:

Volume = Length $\times$ Breadth $\times$ Height

Given:

Length = $2ab$

Breadth = $3ac$

Height = $2ac$

Substitute the given values into the formula:

Volume $= (2ab) \times (3ac) \times (2ac)$

To multiply these monomials, we multiply the coefficients and the variable parts separately.

Multiply the coefficients: $2 \times 3 \times 2 = 6 \times 2 = 12$

Multiply the variable parts: $ab \times ac \times ac = (a^1 b^1) \times (a^1 c^1) \times (a^1 c^1)$

Using the rule of exponents $x^m \times x^n = x^{m+n}$, we combine the variables with the same base:

Variable part $= (a^1 \times a^1 \times a^1) \times b^1 \times (c^1 \times c^1)$

$= a^{1+1+1} \times b^1 \times c^{1+1}$

$= a^3 \times b^1 \times c^2$

$= a^3bc^2$

Combine the coefficient and variable part:

Volume $= 12 \times a^3bc^2 = 12a^3bc^2$

Comparing this result with the given options, we find that it matches option (a).

The correct option is (a) $12a^3bc^2$.

We need to find the product of the trinomial $6a^2 - 7b + 5ab$ and the monomial $2ab$.

We apply the distributive property, multiplying the monomial by each term of the trinomial:

$(6a^2 - 7b + 5ab) \times (2ab)$

$= (6a^2) \times (2ab) + (-7b) \times (2ab) + (5ab) \times (2ab)$

Multiply the first terms:

$(6a^2) \times (2ab) = (6 \times 2) \times (a^2 \times a^1 \times b^1) = 12 \times a^{2+1} \times b^1 = 12a^3b$

Multiply the second terms:

$(-7b) \times (2ab) = (-7 \times 2) \times (b^1 \times a^1 \times b^1) = -14 \times a^1 \times b^{1+1} = -14ab^2$

Multiply the third terms:

$(5ab) \times (2ab) = (5 \times 2) \times (a^1 \times b^1 \times a^1 \times b^1) = 10 \times a^{1+1} \times b^{1+1} = 10a^2b^2$

Now, add the results of the multiplications:

$12a^3b - 14ab^2 + 10a^2b^2$

Comparing this result with the given options, we find that it matches option (b).

The correct option is (b) $12a^3b – 14ab^2 + 10a^2b^2$.

We need to find the square of the binomial $(3x - 4y)$. Squaring means multiplying the expression by itself.

$(3x - 4y)^2 = (3x - 4y) \times (3x - 4y)$

We can use the identity $(a - b)^2 = a^2 - 2ab + b^2$, where $a = 3x$ and $b = 4y$.

Substitute the values of $a$ and $b$ into the identity:

$(3x - 4y)^2 = (3x)^2 - 2(3x)(4y) + (4y)^2$

Calculate each term:

$(3x)^2 = 3^2 \times x^2 = 9x^2$

$2(3x)(4y) = 2 \times 3 \times 4 \times x \times y = 24xy$

$(4y)^2 = 4^2 \times y^2 = 16y^2$

Substitute these back into the expanded identity:

$(3x - 4y)^2 = 9x^2 - 24xy + 16y^2$

Rearranging the terms to match the options:

$9x^2 + 16y^2 - 24xy$

Alternatively, we can perform the multiplication directly:

$(3x - 4y)(3x - 4y) = 3x(3x - 4y) - 4y(3x - 4y)$

$= (3x \times 3x) - (3x \times 4y) - (4y \times 3x) + (-4y \times -4y)$

$= 9x^2 - 12xy - 12xy + 16y^2$

Combine the like terms ($-12xy$ and $-12xy$):

$= 9x^2 - 24xy + 16y^2$

Comparing this result with the given options, we find that it matches option (d).

The correct option is (d) $9x^2 + 16y^2 – 24xy$.

Like terms are algebraic terms that have the same variables raised to the same powers. The numerical coefficients can be different.

We are given the term $5xyz^2$. The variables are $x, y, z$. The power of $x$ is $1$, the power of $y$ is $1$, and the power of $z$ is $2$.

Let's examine each pair of terms:

(a) $5xyz^2$ and $-3xy^2z$

In $5xyz^2$: $x^1, y^1, z^2$.

In $-3xy^2z$: $x^1, y^2, z^1$.

The powers of $y$ ($1$ vs $2$) and $z$ ($2$ vs $1$) are different. These are not like terms.

(b) $-5xyz^2$ and $7xyz^2$

In $-5xyz^2$: $x^1, y^1, z^2$.

In $7xyz^2$: $x^1, y^1, z^2$.

The variables and their respective powers are the same. These are like terms.

(c) $5xyz^2$ and $5x^2yz$

In $5xyz^2$: $x^1, y^1, z^2$.

In $5x^2yz$: $x^2, y^1, z^1$.

The powers of $x$ ($1$ vs $2$) and $z$ ($2$ vs $1$) are different. These are not like terms.

(d) $5xyz^2$ and $x^2y^2z^2$

In $5xyz^2$: $x^1, y^1, z^2$.

In $x^2y^2z^2$: $x^2, y^2, z^2$.

The powers of $x$, $y$, and $z$ are different. These are not like terms.

The only pair of like terms is in option (b).

The correct option is (b) $– 5xyz^2, 7xyz^2$.

The given term is $\frac{-y}{3}$.

The coefficient of a variable in a term is the numerical factor that multiplies the variable part.

We can rewrite the given term to show the numerical factor clearly:

$\frac{-y}{3} = -\frac{y}{3}$

This can also be written as:

$-\frac{1}{3} \times y$

In this form, the numerical factor multiplying the variable $y$ is $-\frac{1}{3}$.

Therefore, the coefficient of $y$ in the term $\frac{-y}{3}$ is $-\frac{1}{3}$.

The correct option is (c) $\frac{-1}{3}$.

We are asked to find the expression that is equal to $a^2 - b^2$. This is a common algebraic identity called the difference of squares.

Let's expand each of the given options to see which one matches $a^2 - b^2$.

(a) $(a – b)^2$

Using the identity $(x-y)^2 = x^2 - 2xy + y^2$:

$(a – b)^2 = a^2 - 2ab + b^2$

This is not equal to $a^2 - b^2$ (unless $2ab = 0$, which is not generally true).

(b) $(a – b)(a – b)$

This is the same as $(a – b)^2$, which we expanded in option (a):

$(a – b)(a – b) = a^2 - 2ab + b^2$

This is not equal to $a^2 - b^2$.

(c) $(a + b)(a – b)$

Using the distributive property (FOIL method):

$(a + b)(a – b) = a \times a + a \times (-b) + b \times a + b \times (-b)$

$= a^2 - ab + ba - b^2$

Since $ab = ba$, the middle terms cancel out:

$= a^2 - ab + ab - b^2$

$= a^2 + 0 - b^2$

This expression is equal to $a^2 - b^2$.

(d) $(a + b)(a + b)$

This is the same as $(a + b)^2$. Using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$(a + b)(a + b) = a^2 + 2ab + b^2$

This is not equal to $a^2 - b^2$ (unless $2ab = 0$).

Based on the expansions, the expression equal to $a^2 - b^2$ is $(a+b)(a-b)$.

The correct option is (c) $(a + b) (a – b)$.

To find the common factor of the given terms, we find the greatest common divisor (GCD) of the numerical coefficients and the lowest power of each common variable present in all the terms.

The given terms are: $17abc$, $34ab^2$, and $51a^2b$.

Numerical Coefficients: 17, 34, 51

We find the GCD of 17, 34, and 51.

Prime factorization of 17 is $17^1$.

Prime factorization of 34 is $2 \times 17$.

Prime factorization of 51 is $3 \times 17$.

The common prime factor is 17. The lowest power of 17 is $17^1 = 17$.

So, the GCD of the coefficients is 17.

Variable Parts: $abc$, $ab^2$, $a^2b$

Variable $a$ is present in all terms. The powers of $a$ are $a^1, a^1, a^2$. The lowest power is $a^1 = a$.

Variable $b$ is present in all terms. The powers of $b$ are $b^1, b^2, b^1$. The lowest power is $b^1 = b$.

Variable $c$ is present in $17abc$ but not in $34ab^2$ or $51a^2b$. Therefore, $c$ is not a common factor.

The common variables with their lowest powers are $a$ and $b$.

The common factor is the product of the GCD of the coefficients and the common variables raised to their lowest powers.

Common factor $= 17 \times a \times b = 17ab$.

Comparing this result with the given options, we find that it matches option (b).

The correct option is (b) 17ab.

We need to find the square of the expression $(9x - 7xy)$.

$(9x - 7xy)^2$

We can use the identity $(a - b)^2 = a^2 - 2ab + b^2$, where $a = 9x$ and $b = 7xy$.

Substitute the values of $a$ and $b$ into the identity:

$(9x - 7xy)^2 = (9x)^2 - 2(9x)(7xy) + (7xy)^2$

Calculate each term:

$(9x)^2 = 9^2 \times x^2 = 81x^2$

$2(9x)(7xy) = 2 \times 9 \times 7 \times x \times x \times y = 18 \times 7 \times x^{1+1} \times y^1 = 126x^2y$

$(7xy)^2 = 7^2 \times x^2 \times y^2 = 49x^2y^2$

Substitute these back into the expanded identity:

$(9x - 7xy)^2 = 81x^2 - 126x^2y + 49x^2y^2$

Rearranging the terms to match the options:

$81x^2 + 49x^2y^2 - 126x^2y$

Comparing this result with the given options, we find that it matches option (c).

The correct option is (c) $81x^2 + 49x^2y^2 –126x^2y$.

We need to factorise the expression $23xy – 46x + 54y – 108$. This expression has four terms, which suggests factorization by grouping.

Group the terms in pairs:

$(23xy – 46x) + (54y – 108)$

Factor out the common factor from the first pair:

The common factor of $23xy$ and $46x$ is $23x$.

$23xy - 46x = 23x(y) - 23x(2) = 23x(y - 2)$

Factor out the common factor from the second pair:

The common factor of $54y$ and $108$ is $54$ (since $108 = 2 \times 54$).

$54y - 108 = 54(y) - 54(2) = 54(y - 2)$

Now, substitute these back into the grouped expression:

$23x(y - 2) + 54(y - 2)$

We can see that $(y - 2)$ is a common binomial factor in both terms.

Factor out the common binomial factor $(y - 2)$:

$(y - 2)(23x + 54)$

This can also be written as $(23x + 54)(y - 2)$ due to the commutative property of multiplication.

Comparing this result with the given options, we find that it matches option (a).

The correct option is (a) $(23x + 54) (y – 2)$.

We need to factorise the quadratic expression $r^2 – 10r + 21$.

This is a quadratic trinomial of the form $ar^2 + br + c$, where $a=1$, $b=-10$, and $c=21$.

To factorise this expression, we look for two numbers that multiply to $c$ (the constant term) and add up to $b$ (the coefficient of the $r$ term).

We need two numbers, let's call them $p$ and $q$, such that:

Let's list pairs of integers that multiply to 21:

$(1, 21) \implies 1 + 21 = 22$ (Does not sum to -10)

$(-1, -21) \implies -1 + (-21) = -22$ (Does not sum to -10)

$(3, 7) \implies 3 + 7 = 10$ (Does not sum to -10)

$(-3, -7) \implies -3 + (-7) = -10$ (Sums to -10)

The two numbers we are looking for are $-3$ and $-7$.

So, we can rewrite the middle term $-10r$ as $-3r - 7r$ (or $-7r - 3r$).

$r^2 – 10r + 21 = r^2 - 3r - 7r + 21$

Now, we can factor by grouping the first two terms and the last two terms:

$= (r^2 - 3r) + (-7r + 21)$

Factor out the common factor from the first group:

$= r(r - 3) + (-7r + 21)$

Factor out the common factor from the second group. The common factor of $-7r$ and $21$ is $-7$ (since $21 = -7 \times -3$).

$= r(r - 3) - 7(r - 3)$

Now, we see that $(r - 3)$ is a common binomial factor in both terms.

Factor out the common binomial $(r - 3)$:

$= (r - 3)(r - 7)$

This is the factorised form.

Let's check option (b) by expanding it:

$(r - 7)(r - 3) = r(r - 3) - 7(r - 3)$

$= r \times r - r \times 3 - 7 \times r - 7 \times (-3)$

$= r^2 - 3r - 7r + 21$

$= r^2 - 10r + 21$

This matches the original expression.

The correct option is (b) $(r – 7) (r – 3)$.

We need to factorise the quadratic expression $p^2 – 17p – 38$.

This is a quadratic trinomial of the form $ap^2 + bp + c$, where $a=1$, $b=-17$, and $c=-38$.

To factorise this expression, we look for two numbers that multiply to $c$ (the constant term) and add up to $b$ (the coefficient of the $p$ term).

We need two numbers, let's call them $p$ and $q$, such that:

Since the product is negative, one number must be positive and the other negative. Since the sum is negative, the number with the larger absolute value must be negative.

Let's list pairs of integers that multiply to 38 and see if their difference (with the larger number being negative) is -17:

Pairs of factors of 38: $(1, 38), (2, 19)$.

Consider the pairs with one positive and one negative, with the negative number having the larger absolute value:

$(1, -38) \implies 1 + (-38) = -37$ (Does not sum to -17)

$(2, -19) \implies 2 + (-19) = -17$ (Sums to -17)

The two numbers we are looking for are $2$ and $-19$.

So, we can rewrite the middle term $-17p$ as $2p - 19p$.

$p^2 – 17p – 38 = p^2 + 2p - 19p - 38$

Now, we can factor by grouping the first two terms and the last two terms:

$= (p^2 + 2p) + (-19p - 38)$

Factor out the common factor from the first group:

$= p(p + 2) + (-19p - 38)$

Factor out the common factor from the second group. The common factor of $-19p$ and $-38$ is $-19$ (since $-38 = -19 \times 2$).

$= p(p + 2) - 19(p + 2)$

Now, we see that $(p + 2)$ is a common binomial factor in both terms.

Factor out the common binomial $(p + 2)$:

$= (p + 2)(p - 19)$

This is the factorised form. This can also be written as $(p - 19)(p + 2)$.

Let's check option (a) by expanding it:

$(p - 19)(p + 2) = p(p + 2) - 19(p + 2)$

$= p \times p + p \times 2 - 19 \times p - 19 \times 2$

$= p^2 + 2p - 19p - 38$

$= p^2 - 17p - 38$

This matches the original expression.

The correct option is (a) $(p – 19) (p + 2)$.

We need to divide the monomial $57p^2qr$ by the monomial $114pq$.

We can write this division as a fraction:

$\frac{57p^2qr}{114pq}$

We can separate the division of the numerical coefficients and the division of the variable parts:

$(\frac{57}{114}) \times (\frac{p^2qr}{pq})$

First, simplify the numerical fraction $\frac{57}{114}$. We can find the greatest common divisor of 57 and 114. Notice that $114 = 2 \times 57$.

$\frac{57}{114} = \frac{\cancel{57}^{1}}{\cancel{114}_{2}} = \frac{1}{2}$

Next, simplify the variable part $\frac{p^2qr}{pq}$. We use the rule of exponents $\frac{x^m}{x^n} = x^{m-n}$ for each variable:

$\frac{p^2}{p^1} = p^{2-1} = p^1 = p$

$\frac{q^1}{q^1} = q^{1-1} = q^0 = 1$ (assuming $q \neq 0$)

$\frac{r^1}{1} = r^1 = r$ (since $r$ is only in the numerator)

So, the variable part simplifies to:

$\frac{p^2qr}{pq} = p \times 1 \times r = pr$

Now, combine the simplified numerical and variable parts:

$\frac{1}{2} \times pr = \frac{1}{2}pr$

Comparing this result with the given options, we find that it matches option (c).

The correct option is (c) $\frac{1}{2}$ pr.

We need to divide $p (4p^2 – 16)$ by $4p (p – 2)$.

We can write this division as a fraction:

$\frac{p (4p^2 – 16)}{4p (p – 2)}$

First, simplify the expression in the numerator. Notice that $4p^2 - 16$ has a common factor of 4:

$4p^2 - 16 = 4(p^2 - 4)$

The expression $p^2 - 4$ is a difference of squares, $p^2 - 2^2$. It can be factorised as $(p - 2)(p + 2)$.

So, $4p^2 - 16 = 4(p - 2)(p + 2)$.

Substitute this back into the numerator of the fraction:

Numerator $= p \times 4(p - 2)(p + 2) = 4p(p - 2)(p + 2)$

Now, the fraction becomes:

$\frac{4p(p - 2)(p + 2)}{4p(p – 2)}$

We can cancel out the common factors in the numerator and the denominator, assuming $p \neq 0$ and $p - 2 \neq 0$ (i.e., $p \neq 2$).

The common factors are $4$, $p$, and $(p - 2)$.

$\frac{\cancel{4}\cancel{p}\cancel{(p - 2)}(p + 2)}{\cancel{4}\cancel{p}\cancel{(p – 2)}} = p + 2$

Thus, the result of the division is $p + 2$.

Comparing this result with the given options, we find that it matches option (c).

The correct option is (c) p + 2.

We need to find the common factor of the terms $3ab$ and $2cd$.

Term 1: $3ab$

Factors are $1, 3, a, b, ab, 3a, 3b, 3ab$.

Term 2: $2cd$

Factors are $1, 2, c, d, cd, 2c, 2d, 2cd$.

Let's look for factors that are common to both lists.

Numerical factors: The factors of 3 are 1 and 3. The factors of 2 are 1 and 2. The only common numerical factor is 1.

Variable factors: The variables in the first term are $a$ and $b$. The variables in the second term are $c$ and $d$. There are no variables common to both terms.

The only common factor is the numerical factor, which is 1.

Alternatively, we can find the GCD of the numerical coefficients and the lowest power of each common variable.

Numerical coefficients: 3 and 2. GCD(3, 2) = 1.

Variables: The variables $a$ and $b$ are in the first term. The variables $c$ and $d$ are in the second term. There are no common variables.

The common factor is the product of the GCD of coefficients and the common variables (none in this case). So, the common factor is 1.

Comparing this result with the given options, we find that it matches option (a).

The correct option is (a) 1.

An irreducible factor (over the integers or rational numbers) is a factor that cannot be factored further into simpler factors (other than 1, -1, or itself). For algebraic expressions, this usually refers to prime numbers and variables raised to the power of 1.

The given term is $24x^2y^2$.

Let's find the prime factorization of the coefficient 24:

$\begin{array}{c|cc} 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$.

The expression can be written as:

$24x^2y^2 = (2 \times 2 \times 2 \times 3) \times (x \times x) \times (y \times y)$

The irreducible factors of $24x^2y^2$ are the prime numbers and the variables raised to the power of 1.

Irreducible factors: $2, 2, 2, 3, x, x, y, y$.

The unique irreducible factors are $2, 3, x, y$.

Now let's examine the options:

(a) $x^2$: This is a factor, but it is not irreducible because $x^2 = x \times x$.

(b) $y^2$: This is a factor, but it is not irreducible because $y^2 = y \times y$.

(c) $x$: This is a factor, and it is irreducible as a variable term.

(d) $24x$: This is a factor, but it is not irreducible because $24x = 2 \times 2 \times 2 \times 3 \times x$.

Among the given options, only $x$ is an irreducible factor.

The correct option is (c) x.

The expression is $(a + b)^2$. This means $(a + b)$ multiplied by itself.

$(a + b)^2 = (a + b) \times (a + b)$

The factors of an expression include 1, the expression itself, and any other expressions that divide it evenly.

In this case, the factors are:

1 (since 1 divides any expression)

$(a + b)$ (which divides $(a+b)^2$ to give $(a+b)$)

$(a + b)^2$ (which divides itself to give 1)

Are there any other distinct factors? The expression $(a+b)^2$ can be written as the product of $(a+b)$ and $(a+b)$. The unique factors are 1, $(a+b)$, and $(a+b)^2$.

Thus, there are 3 distinct factors: $1$, $(a+b)$, and $(a+b)^2$.

Consider a numerical example: $(x+3)^2$. The factors are $1$, $(x+3)$, and $(x+3)^2$. There are 3 factors.

The correct option is (b) 3.

We need to factorise the expression $3x – 24$.

To factorise, we look for the greatest common factor (GCF) of the terms $3x$ and $-24$.

The terms are $3x$ and $-24$.

The numerical coefficients are 3 and -24.

The factors of 3 are 1 and 3.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.

The greatest common numerical factor of 3 and 24 is 3.

The variable in the first term is $x$. There is no variable in the second term. So, there is no common variable factor.

The greatest common factor of $3x$ and $-24$ is 3.

Now, factor out the GCF (3) from each term:

$3x = 3 \times x$

$-24 = 3 \times (-8)$

So, $3x – 24 = 3 \times x + 3 \times (-8)$

Using the distributive property in reverse (factoring out 3):

$3x – 24 = 3(x + (-8)) = 3(x - 8)$

Comparing this result with the given options, we find that it matches option (b).

The correct option is (b) 3 (x – 8).

We need to find the factors of the expression $x^2 – 4$.

This expression is a difference of squares. We can write it as $x^2 - 2^2$.

The difference of squares identity is $a^2 - b^2 = (a - b)(a + b)$.

Comparing $x^2 - 4$ with $a^2 - b^2$, we have $a = x$ and $b = 2$.

Using the identity, we can factorise $x^2 - 4$ as:

$x^2 - 4 = (x - 2)(x + 2)$

The factors of $x^2 - 4$ are $(x - 2)$ and $(x + 2)$. The order of the factors does not matter in multiplication, so they can also be listed as $(x + 2)$ and $(x - 2)$.

Let's check the options by multiplying the given factors:

(a) $(x – 2)(x – 2) = (x – 2)^2 = x^2 - 4x + 4$ (Incorrect)

(b) $(x + 2)(x – 2) = x^2 - 2^2 = x^2 - 4$ (Correct, using the difference of squares identity)

Alternatively, using FOIL: $(x + 2)(x - 2) = x \times x + x \times (-2) + 2 \times x + 2 \times (-2) = x^2 - 2x + 2x - 4 = x^2 - 4$.

(c) $(x + 2)(x + 2) = (x + 2)^2 = x^2 + 4x + 4$ (Incorrect)

(d) $(x – 4)(x – 4) = (x – 4)^2 = x^2 - 8x + 16$ (Incorrect)

The correct option is (b) $(x + 2), (x – 2)$.

We need to find the value of the expression $(– 27x^2y) \div (– 9xy)$.

We can write this division as a fraction:

$\frac{-27x^2y}{-9xy}$

We can separate the division of the numerical coefficients and the division of the variable parts:

$(\frac{-27}{-9}) \times (\frac{x^2y}{xy})$

First, simplify the numerical part $\frac{-27}{-9}$. Dividing a negative number by a negative number results in a positive number. $27 \div 9 = 3$.

$\frac{-27}{-9} = 3$

Next, simplify the variable part $\frac{x^2y}{xy}$. We use the rule of exponents $\frac{a^m}{a^n} = a^{m-n}$ for variables with the same base. We assume $x \neq 0$ and $y \neq 0$.

For the variable $x$: $\frac{x^2}{x^1} = x^{2-1} = x^1 = x$.

For the variable $y$: $\frac{y^1}{y^1} = y^{1-1} = y^0$. Any non-zero number raised to the power of 0 is 1. So, $y^0 = 1$.

The variable part simplifies to:

$\frac{x^2y}{xy} = \frac{x^2}{x} \times \frac{y}{y} = x \times 1 = x$

Now, combine the simplified numerical and variable parts:

$3 \times x = 3x$

Thus, $(– 27x^2y) \div (– 9xy) = 3x$.

Comparing this result with the given options, we find that it matches option (d).

The correct option is (d) 3x.

We need to divide the expression $(2x^2 + 4)$ by $2$.

We can write this as:

$\frac{2x^2 + 4}{2}$

To divide a sum by a number, we divide each term in the sum by that number.

$\frac{2x^2 + 4}{2} = \frac{2x^2}{2} + \frac{4}{2}$

Now, perform the division for each term:

$\frac{2x^2}{2} = \frac{\cancel{2}}{\cancel{2}} x^2 = 1 \times x^2 = x^2$

$\frac{4}{2} = 2$

Combine the results of the division:

$x^2 + 2$

Alternatively, we can factor out a common factor from the numerator first.

The common factor of $2x^2$ and $4$ is 2.

$2x^2 + 4 = 2(x^2 + 2)$

Now, substitute this back into the fraction:

$\frac{2(x^2 + 2)}{2}$

Cancel out the common factor of 2 in the numerator and the denominator:

$\frac{\cancel{2}(x^2 + 2)}{\cancel{2}} = x^2 + 2$

Comparing this result with the given options, we find that it matches option (b).

The correct option is (b) $x^2 + 2$.

We need to divide the expression $(3x^3 + 9x^2 + 27x)$ by $3x$.

We can write this as:

$\frac{3x^3 + 9x^2 + 27x}{3x}$

To divide a sum (or difference) of terms by a monomial, we divide each term in the sum (or difference) by the monomial.

$\frac{3x^3 + 9x^2 + 27x}{3x} = \frac{3x^3}{3x} + \frac{9x^2}{3x} + \frac{27x}{3x}$

Now, perform the division for each term, assuming $x \neq 0$:

For the first term: $\frac{3x^3}{3x} = (\frac{3}{3}) \times (\frac{x^3}{x^1}) = 1 \times x^{3-1} = x^2$

For the second term: $\frac{9x^2}{3x} = (\frac{9}{3}) \times (\frac{x^2}{x^1}) = 3 \times x^{2-1} = 3x^1 = 3x$

For the third term: $\frac{27x}{3x} = (\frac{27}{3}) \times (\frac{x^1}{x^1}) = 9 \times x^{1-1} = 9 \times x^0 = 9 \times 1 = 9$

Combine the results of the division:

$x^2 + 3x + 9$

Alternatively, we can factor out the common factor from the numerator first.

The common factor of $3x^3$, $9x^2$, and $27x$ is $3x$.

$3x^3 + 9x^2 + 27x = 3x(x^2) + 3x(3x) + 3x(9) = 3x(x^2 + 3x + 9)$

Now, substitute this back into the fraction:

$\frac{3x(x^2 + 3x + 9)}{3x}$

Cancel out the common factor of $3x$ in the numerator and the denominator (assuming $3x \neq 0$):

$\frac{\cancel{3x}(x^2 + 3x + 9)}{\cancel{3x}} = x^2 + 3x + 9$

Comparing this result with the given options, we find that it matches option (d).

The correct option is (d) $x^2 + 3x + 9$.

We need to find the value of $(a + b)^2 + (a – b)^2$.

We can expand each squared term using the standard algebraic identities:

The identity for the square of a sum is $(a + b)^2 = a^2 + 2ab + b^2$.

The identity for the square of a difference is $(a – b)^2 = a^2 – 2ab + b^2$.

Now, substitute these expanded forms into the given expression:

$(a + b)^2 + (a – b)^2 = (a^2 + 2ab + b^2) + (a^2 – 2ab + b^2)$

Remove the parentheses and combine the terms:

$= a^2 + 2ab + b^2 + a^2 – 2ab + b^2$

Group the like terms together:

$= (a^2 + a^2) + (2ab – 2ab) + (b^2 + b^2)$

Combine the like terms:

$= 2a^2 + 0 + 2b^2$

Simplify the expression:

$= 2a^2 + 2b^2$

Comparing this result with the given options, we find that it matches option (c).

The correct option is (c) $2a^2 + 2b^2$.

We need to find the value of $(a + b)^2 – (a – b)^2$.

We can expand each squared term using the standard algebraic identities:

The identity for the square of a sum is:

The identity for the square of a difference is:

Now, substitute these expanded forms into the given expression:

$(a + b)^2 – (a – b)^2 = (a^2 + 2ab + b^2) – (a^2 – 2ab + b^2)$

Remove the parentheses. Remember to change the sign of each term inside the second parentheses because of the subtraction:

$= a^2 + 2ab + b^2 - a^2 - (-2ab) - b^2$

$= a^2 + 2ab + b^2 - a^2 + 2ab - b^2$

Group the like terms together:

$= (a^2 - a^2) + (2ab + 2ab) + (b^2 - b^2)$

Combine the like terms:

$= 0 + 4ab + 0$

Simplify the expression:

$= 4ab$

Comparing this result with the given options, we find that it matches option (a).

The correct option is (a) 4ab.

Alternate Solution:

We can treat the expression as a difference of squares $X^2 - Y^2$, where $X = (a + b)$ and $Y = (a - b)$.

Using the identity $X^2 - Y^2 = (X - Y)(X + Y)$:

$(a + b)^2 – (a – b)^2 = [(a + b) – (a – b)][(a + b) + (a – b)]$

Simplify the first bracket:

$(a + b) – (a – b) = a + b – a + b = (a - a) + (b + b) = 0 + 2b = 2b$

Simplify the second bracket:

$(a + b) + (a – b) = a + b + a – b = (a + a) + (b - b) = 2a + 0 = 2a$

Now, multiply the results of the two brackets:

$(2b)(2a) = 4ab$

The result is $4ab$, which is the same as the previous method.

Consider two terms with like signs. This means either both terms are positive or both terms are negative.

Case 1: Both terms are positive.

$(+\text{term}) \times (+\text{term}) = (+\text{product})$

For example, $(+5) \times (+3) = +15$.

Case 2: Both terms are negative.

$(-\text{term}) \times (-\text{term}) = (+\text{product})$

For example, $(-5) \times (-3) = +15$.

In both cases where the signs are the same (like signs), the product is positive.

The product of two terms with like signs is a positive term.

Consider two terms with unlike signs. This means one term is positive and the other term is negative.

Case 1: The first term is positive and the second term is negative.

$(+\text{term}) \times (-\text{term}) = (-\text{product})$

For example, $(+5) \times (-3) = -15$.

Case 2: The first term is negative and the second term is positive.

$(-\text{term}) \times (+\text{term}) = (-\text{product})$

For example, $(-5) \times (+3) = -15$.

In both cases where the signs are different (unlike signs), the product is negative.

The product of two terms with unlike signs is a negative term.

The expression $a(b + c)$ represents the multiplication of the term $a$ by the binomial $(b + c)$.

According to the distributive property of multiplication over addition, the term outside the parentheses is multiplied by each term inside the parentheses.

So, $a(b + c) = a \times b + a \times c$.

This expands to $ab + ac$.

The given structure for filling the blanks is $a(b + c) = \text{ax \_\_\_\_} \times \text{ax \_\_\_\_}$.

Based on the distributive property, the terms inside the parentheses that are multiplied by $a$ are $b$ and $c$. The distributive process creates the terms $a \times b$ and $a \times c$, which are then added.

Assuming the blanks are intended to be filled with the terms that $a$ is multiplying from inside the parentheses, we place $b$ in the first blank and $c$ in the second blank.

The completed statement, following the structure provided in the question, is:

$a(b + c) = \text{ax}$ b $\times \text{ax}$ c.

Note: The standard result of the distributive property is $a(b+c) = ab+ac$. The structure provided in the question with a multiplication sign ($\times$) between the two parts ($\text{ax b} \times \text{ax c} = a^2x^2bc$) is generally not equivalent to $ab+ac$, suggesting a potential formatting inconsistency in the question. However, the blanks are filled based on the terms $b$ and $c$ being distributed by $a$.

The right side of the equation is $a^2 – 2ab + b^2$.

This is the expanded form of the square of a binomial.

The algebraic identity for the square of a difference is $(a - b)^2 = a^2 - 2ab + b^2$.

The expression $(a - b)^2$ means $(a - b)$ multiplied by itself, i.e., $(a - b) \times (a - b)$.

So, we have $(a – b) \times (a – b) = a^2 – 2ab + b^2$.

Comparing this with the given statement $(a – b) \_\_\_\_\_\_\_\_\_$ = $a^2 – 2ab + b^2$, the blank must be filled with $(a - b)$.

The completed statement is:

$(a – b)$ (a – b) $= a^2 – 2ab + b^2$

The left side of the equation is $a^2 – b^2$.

This expression is known as the difference of squares.

The algebraic identity for the difference of squares states that $a^2 – b^2$ can be factorised into the product of the sum and the difference of the terms being squared.

$a^2 – b^2 = (a + b)(a – b)$

The given statement is $a^2 – b^2 = (a + b ) \_\_\_\_\_\_\_\_\_\_$.

Comparing this with the identity, the blank must be filled with $(a - b)$.

The completed statement is:

$a^2 – b^2 = (a + b )$ (a – b).

We are given the equation $(a – b)^2 + \_\_\_\_\_\_\_\_\_\_\_\_ = a^2 – b^2$.

First, expand the term $(a – b)^2$ using the identity $(a-b)^2 = a^2 - 2ab + b^2$.

So the equation becomes:

$(a^2 - 2ab + b^2) + \_\_\_\_\_\_\_\_\_\_\_\_ = a^2 – b^2$

Let the expression in the blank be represented by $X$. The equation is:

$a^2 - 2ab + b^2 + X = a^2 - b^2$

To find $X$, we can subtract $(a^2 - 2ab + b^2)$ from both sides of the equation:

$X = (a^2 - b^2) - (a^2 - 2ab + b^2)$

Remove the parentheses, remembering to change the sign of each term inside the second parentheses:

$X = a^2 - b^2 - a^2 + 2ab - b^2$

Group the like terms:

$X = (a^2 - a^2) + 2ab + (-b^2 - b^2)$

Combine the like terms:

$X = 0 + 2ab - 2b^2$

$X = 2ab - 2b^2$

So, the expression that fills the blank is $2ab - 2b^2$.

The completed statement is:

$(a – b)^2 +$ $2ab - 2b^2$ $= a^2 – b^2$

We are given the equation $(a + b)^2 – 2ab = \_\_\_\_\_\_\_\_\_\_\_ + \_\_\_\_\_\_\_\_\_\_\_\_$.

First, expand the term $(a + b)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$.

Substitute this expansion into the left side of the equation:

$(a^2 + 2ab + b^2) – 2ab$

Now, combine the like terms in this expression. The terms $2ab$ and $-2ab$ are like terms.

$a^2 + (2ab – 2ab) + b^2$

$a^2 + 0 + b^2$

$a^2 + b^2$

So, the left side of the equation simplifies to $a^2 + b^2$.

The equation becomes $a^2 + b^2 = \_\_\_\_\_\_\_\_\_\_\_ + \_\_\_\_\_\_\_\_\_\_\_\_$.

The right side has two blanks added together. The simplified left side is the sum of $a^2$ and $b^2$.

Therefore, the blanks should be filled with $a^2$ and $b^2$ in any order.

The completed statement is:

$(a + b)^2 – 2ab =$ $a^2$ $+$ $b^2$.

Alternatively, it could be $(a + b)^2 – 2ab =$ $b^2$ $+$ $a^2$.

We are given the equation $(x + a) (x + b) = x^2 + (a + b) x + \_\_\_\_\_\_\_\_\_$.

We need to expand the left side of the equation, $(x + a)(x + b)$. We can use the distributive property (FOIL method):

$(x + a)(x + b) = x \times x + x \times b + a \times x + a \times b$

$= x^2 + bx + ax + ab$

Group the terms with $x$:

$= x^2 + (b + a)x + ab$

Since addition is commutative, $(b + a) = (a + b)$.

$= x^2 + (a + b)x + ab$

Comparing this expanded form with the right side of the given equation $x^2 + (a + b) x + \_\_\_\_\_\_\_\_\_$, we can see that the blank corresponds to the term $ab$.

This is the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.

The completed statement is:

$(x + a) (x + b) = x^2 + (a + b) x +$ ab.

Let's consider examples of multiplying two polynomials:

Example 1: Product of a monomial (degree 1) and a binomial (degree 1)

$x \times (x + 1) = x^2 + x$

The result is a binomial (which is a type of polynomial).

Example 2: Product of two binomials (degree 1 each)

$(x + 1)(x + 2) = x^2 + 2x + x + 2 = x^2 + 3x + 2$

The result is a trinomial (which is a type of polynomial).

Example 3: Product of a binomial (degree 1) and a trinomial (degree 2)

$(x + 1)(x^2 + x + 1) = x(x^2 + x + 1) + 1(x^2 + x + 1)$

$= x^3 + x^2 + x + x^2 + x + 1$

$= x^3 + 2x^2 + 2x + 1$

The result is a polynomial with four terms.

In general, when you multiply two polynomials, the result is always another polynomial.

The degree of the product polynomial is the sum of the degrees of the original polynomials.

The product of two polynomials is a polynomial.

We need to find the common factor of the terms $ax^2$ and $bx$.

Term 1: $ax^2 = a \times x \times x$

Term 2: $bx = b \times x$

We look for factors that are common to both terms.

The variable $x$ is present in both terms.

In the first term, the power of $x$ is $2$ ($x^2$).

In the second term, the power of $x$ is $1$ ($x^1$).

The lowest power of $x$ that is common to both terms is $x^1 = x$.

The coefficients are $a$ and $b$. Unless $a$ and $b$ share a common numerical factor or a common variable factor (which is not indicated here), their only common factor is 1 (or -1).

Assuming $a$ and $b$ are just coefficients and have no common factors other than 1, the common factor of $ax^2$ and $bx$ is the common variable factor, which is $x$.

So, we can factor out $x$ from both terms:

$ax^2 + bx = x(ax) + x(b) = x(ax + b)$

The common factor is $x$.

The completed statement is:

Common factor of $ax^2 + bx$ is x.

We need to find the factorised form of the expression $18mn + 10mnp$.

To factorise this binomial, we look for the greatest common factor (GCF) of the two terms, $18mn$ and $10mnp$.

Let's find the GCF of the numerical coefficients (18 and 10) and the variable parts ($mn$ and $mnp$).

Numerical coefficients: 18 and 10.

Factors of 18: 1, 2, 3, 6, 9, 18.

Factors of 10: 1, 2, 5, 10.

The greatest common numerical factor is 2.

Variable parts: $mn$ and $mnp$.

Both terms contain the variables $m$ and $n$.

The lowest power of $m$ is $m^1$.

The lowest power of $n$ is $n^1$.

The variable $p$ is only in the second term, so it is not a common variable factor.

The common variable factor is $m^1 n^1 = mn$.

The greatest common factor (GCF) of $18mn$ and $10mnp$ is the product of the GCF of coefficients and the common variable factors:

GCF $= 2 \times mn = 2mn$.

Now, we factor out the GCF ($2mn$) from each term:

$18mn = 2mn \times 9$

$10mnp = 2mn \times 5p$

So, $18mn + 10mnp = 2mn(9) + 2mn(5p)$

Using the distributive property in reverse:

$18mn + 10mnp = 2mn(9 + 5p)$

The completed statement is:

Factorised form of $18mn + 10mnp$ is $2mn(9 + 5p)$.

We need to find the factorised form of the trinomial $4y^2 – 12y + 9$.

This trinomial has the form $Ay^2 + By + C$, where $A=4$, $B=-12$, and $C=9$.

Notice that the first term $4y^2 = (2y)^2$ is a perfect square, and the last term $9 = 3^2$ is a perfect square. The middle term is $-12y$.

Let's check if this trinomial is a perfect square trinomial of the form $(ay - b)^2 = (ay)^2 - 2(ay)(b) + b^2$ or $(b - ay)^2 = b^2 - 2(b)(ay) + (ay)^2$.

Comparing $4y^2 – 12y + 9$ with $(ay - b)^2$, we can assume $ay = 2y$ and $b = 3$.

So, $a = 2$ and $b = 3$.

Let's check the middle term: $-2(ay)(b) = -2(2y)(3) = -2 \times 2 \times 3 \times y = -12y$.

This matches the middle term of the given trinomial.

Thus, the trinomial is a perfect square trinomial, and its factorised form is $(ay - b)^2 = (2y - 3)^2$.

Alternatively, we can factor by splitting the middle term. We need two numbers that multiply to $A \times C = 4 \times 9 = 36$ and add up to $B = -12$.

Let the two numbers be $p$ and $q$.

Since the product is positive and the sum is negative, both numbers must be negative.

Pairs of negative factors of 36: $(-1, -36), (-2, -18), (-3, -12), (-4, -9), (-6, -6)$.

Sums of these pairs:

$-1 + (-36) = -37$

$-2 + (-18) = -20$

$-3 + (-12) = -15$

$-4 + (-9) = -13$

$-6 + (-6) = -12$

The two numbers are -6 and -6.

Split the middle term $-12y$ into $-6y - 6y$:

$4y^2 – 12y + 9 = 4y^2 - 6y - 6y + 9$

Factor by grouping:

$= (4y^2 - 6y) + (-6y + 9)$

Factor out common factors from each group:

$= 2y(2y - 3) + (-3)(2y - 3)$

Factor out the common binomial factor $(2y - 3)$:

$= (2y - 3)(2y - 3)$

$= (2y - 3)^2$

The completed statement is:

Factorised form of $4y^2 – 12y + 9$ is $(2y – 3)^2$.

We need to find the value of $(38x^3y^2z) \div (19xy^2)$.

We can write this division as a fraction:

$\frac{38x^3y^2z}{19xy^2}$

Separate the division of numerical coefficients and the division of variable parts:

$(\frac{38}{19}) \times (\frac{x^3y^2z}{xy^2})$

Simplify the numerical part $\frac{38}{19}$. Since $38 = 2 \times 19$, $\frac{38}{19} = 2$.

Simplify the variable part $\frac{x^3y^2z}{xy^2}$. Use the rule $\frac{a^m}{a^n} = a^{m-n}$ for each variable, assuming $x \neq 0$ and $y \neq 0$.

For $x$: $\frac{x^3}{x^1} = x^{3-1} = x^2$.

For $y$: $\frac{y^2}{y^2} = y^{2-2} = y^0 = 1$.

For $z$: $\frac{z^1}{1} = z^1 = z$.

The variable part simplifies to $x^2 \times 1 \times z = x^2z$.

Combine the numerical and variable parts:

$2 \times x^2z = 2x^2z$

The completed statement is:

$38x^3y^2z \div 19xy^2$ is equal to $2x^2z$.

The formula for the volume of a rectangular box (cuboid) is:

Volume = Length $\times$ Breadth $\times$ Height

Given:

Length = $2x$

Breadth = $3y$

Height = $4z$

Substitute these values into the volume formula:

Volume $= (2x) \times (3y) \times (4z)$

To multiply these monomials, we multiply the numerical coefficients and the variable parts separately:

Product of coefficients $= 2 \times 3 \times 4 = 6 \times 4 = 24$

Product of variable parts $= x \times y \times z = xyz$

Combine the numerical and variable parts:

Volume $= 24 \times xyz = 24xyz$

The completed statement is:

Volume of a rectangular box with length 2x, breadth 3y and height 4z is $24xyz$.

We are asked to evaluate the expression $67^2 – 37^2$ using the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$.

In this case, $a = 67$ and $b = 37$.

According to the identity:

$67^2 – 37^2 = (67 - 37)(67 + 37)$

The given equation is $67^2 – 37^2 = (67 – 37) \times \_\_\_\_\_\_\_\_\_ = \_\_\_\_\_\_\_\_\_\_\_$.

Comparing $(67 - 37)(67 + 37)$ with $(67 – 37) \times \_\_\_\_\_\_\_\_\_$, the first blank must be filled with $(67 + 37)$.

Now, evaluate the terms inside the parentheses:

$67 - 37 = 30$

$67 + 37 = 104$

Substitute these values back into the expression:

$67^2 – 37^2 = 30 \times 104$

Now, calculate the product $30 \times 104$:

$30 \times 104 = 30 \times (100 + 4) = 30 \times 100 + 30 \times 4 = 3000 + 120 = 3120$

The value of $67^2 – 37^2$ is 3120.

Comparing $30 \times 104 = \_\_\_\_\_\_\_\_\_\_\_$, the second blank must be filled with 3120.

The completed statement is:

$67^2 – 37^2 = (67 – 37) \times$ (67 + 37) $=$ 3120.

We are asked to evaluate the expression $103^2 – 102^2$ using the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$.

In this case, $a = 103$ and $b = 102$.

According to the identity:

$103^2 – 102^2 = (103 - 102)(103 + 102)$

The given equation is $103^2 – 102^2 = \_\_\_\_\_\_\_\_\_ \times (103 – 102) = \_\_\_\_\_\_\_\_\_\_\_$.

Comparing $(103 - 102)(103 + 102)$ with $\_\_\_\_\_\_\_\_\_ \times (103 – 102)$, the first blank must be filled with $(103 + 102)$.

Now, evaluate the terms inside the parentheses:

$103 - 102 = 1$

$103 + 102 = 205$

Substitute these values back into the expression:

$103^2 – 102^2 = 205 \times 1$

Now, calculate the product $205 \times 1$:

$205 \times 1 = 205$

The value of $103^2 – 102^2$ is 205.

Comparing $205 \times 1 = \_\_\_\_\_\_\_\_\_\_\_$, the second blank must be filled with 205.

The completed statement is:

$103^2 – 102^2 =$ (103 + 102) $\times (103 – 102) =$ 205.

The area of a rectangle is calculated by multiplying its length and breadth.

Area = Length $\times$ Breadth

Given the sides of the rectangular plot are $4x^2$ and $3y^2$. We can consider one side as the length and the other as the breadth.

Let Length $= 4x^2$ and Breadth $= 3y^2$.

Area $= (4x^2) \times (3y^2)$

To multiply these monomials, we multiply the numerical coefficients and the variable parts separately.

Product of coefficients $= 4 \times 3 = 12$

Product of variable parts $= x^2 \times y^2 = x^2y^2$

Combine the numerical and variable parts:

Area $= 12 \times x^2y^2 = 12x^2y^2$

The completed statement is:

Area of a rectangular plot with sides $4x^2$ and $3y^2$ is $12x^2y^2$.

A rectangular box with equal length, breadth, and height is a cube.

The formula for the volume of a cube is given by: Volume $= \text{side}^3$.

Alternatively, for a rectangular box (cuboid), the volume is Length $\times$ Breadth $\times$ Height.

Given that $l = b = h = 2x$.

Volume $= l \times b \times h = (2x) \times (2x) \times (2x)$

To multiply these monomials, we multiply the numerical coefficients and the variable parts separately.

Product of coefficients $= 2 \times 2 \times 2 = 8$

Product of variable parts $= x \times x \times x = x^{1+1+1} = x^3$

Combine the numerical and variable parts:

Volume $= 8 \times x^3 = 8x^3$

The completed statement is:

Volume of a rectangular box with $l = b = h = 2x$ is $8x^3$.

In an algebraic term, the coefficient is the numerical factor that multiplies the variable part.

The given term is $-37abc$.

This term can be written as $(-37) \times (abc)$.

The numerical factor is $-37$.

The variable part is $abc$.

Therefore, the coefficient of the term $-37abc$ is the numerical factor, which is $-37$.

The completed statement is:

The coefficient in – 37abc is – 37.

An algebraic expression is made up of terms that are separated by addition (+) or subtraction (-) signs.

The given expression is $a^2 + bc \times d$.

Let's evaluate the multiplication part first: $bc \times d = bcd$.

So the expression is $a^2 + bcd$.

In this expression, the terms are $a^2$ and $bcd$. These two terms are separated by an addition sign.

Since there is an addition sign separating two distinct parts that cannot be combined further (assuming $a, b, c, d$ are different variables), the expression has two terms.

The completed statement is:

Number of terms in the expression $a^2 + bc \times d$ is two.

The area of a square with side length $s$ is given by the formula Area $= s^2$.

We have two squares.

Square 1 has side length $s_1 = 4a$.

The area of Square 1 is Area$_1 = (4a)^2 = 4^2 \times a^2 = 16a^2$.

Square 2 has side length $s_2 = 4b$.

The area of Square 2 is Area$_2 = (4b)^2 = 4^2 \times b^2 = 16b^2$.

We are asked to find the sum of the areas of the two squares.

Sum of areas = Area$_1$ + Area$_2$

Sum of areas $= 16a^2 + 16b^2$

We can also factor out the common factor 16 from the sum:

Sum of areas $= 16(a^2 + b^2)$

The completed statement is:

The sum of areas of two squares with sides 4a and 4b is $16a^2 + 16b^2$ or $16(a^2 + b^2)$.

The common factor method of factorisation involves identifying a factor that is common to all terms of a polynomial and then writing the polynomial as the product of the common factor and the remaining expression.

For example, consider the expression $ax + ay$. The common factor is $a$. Factoring out $a$, we get $a(x + y)$.

This process is the reverse of the distributive property of multiplication over addition (or subtraction).

The distributive property states that $a(x + y) = ax + ay$.

Factoring out the common factor $a$ from $ax + ay$ to get $a(x + y)$ is essentially applying the distributive property in reverse.

Therefore, the common factor method of factorisation is based on the distributive property.

The completed statement is:

The common factor method of factorisation for a polynomial is based on distributive property.

The area of a square with side length $s$ is given by the formula Area $= s^2$.

We are given that the area of the square is $9y^2$.

So, $s^2 = 9y^2$.

To find the side length $s$, we need to take the square root of the area.

$s = \sqrt{9y^2}$

Using the property of square roots $\sqrt{mn} = \sqrt{m} \times \sqrt{n}$ (for non-negative values):

$s = \sqrt{9} \times \sqrt{y^2}$

The square root of 9 is 3 (we consider the positive root for length).

The square root of $y^2$ is $|y|$. Since side length is typically considered a positive value, we assume $y$ is such that the side is positive. If $y$ represents a length or a variable that results in a positive side, we take $\sqrt{y^2} = y$ (assuming $y \ge 0$). If the variable can be negative, the side length is $|3y|$. However, in the context of problems like this, we usually assume the variable is such that the dimension is positive.

Assuming the side length must be a positive expression, we take the positive square root:

$s = 3 \times y = 3y$

Let's check: If the side is $3y$, the area is $(3y)^2 = 3^2 \times y^2 = 9y^2$, which matches the given area.

The completed statement is:

The side of the square of area $9y^2$ is $3y$.

We need to simplify the expression $\frac{3x \;+\; 3}{3}$.

To simplify a fraction where the numerator is a sum (or difference) of terms and the denominator is a single term, we can divide each term in the numerator by the denominator.

$\frac{3x \;+\; 3}{3} = \frac{3x}{3} + \frac{3}{3}$

Now, perform the division for each term:

$\frac{3x}{3} = \frac{\cancel{3}x}{\cancel{3}} = 1 \times x = x$

$\frac{3}{3} = 1$

Combine the results of the division:

$x + 1$

Alternatively, we can factor out the common factor from the numerator first.

The common factor of $3x$ and $3$ is 3.

$3x + 3 = 3(x) + 3(1) = 3(x + 1)$

Now, substitute this back into the fraction:

$\frac{3(x + 1)}{3}$

Cancel out the common factor of 3 in the numerator and the denominator (assuming $3 \neq 0$):

$\frac{\cancel{3}(x + 1)}{\cancel{3}} = x + 1$

The completed statement is:

On simplification $\frac{3x \;+\; 3}{3}$ = $x + 1$.

We need to find the factorisation of the expression $2x + 4y$.

To factorise this binomial, we look for the greatest common factor (GCF) of the two terms, $2x$ and $4y$.

Let's find the GCF of the numerical coefficients (2 and 4) and the variable parts ($x$ and $y$).

Numerical coefficients: 2 and 4.

Factors of 2: 1, 2.

Factors of 4: 1, 2, 4.

The greatest common numerical factor is 2.

Variable parts: $x$ and $y$.

The variable $x$ is in the first term, and the variable $y$ is in the second term. There are no common variables.

The greatest common factor (GCF) of $2x$ and $4y$ is the common numerical factor, which is 2.

Now, we factor out the GCF (2) from each term:

$2x = 2 \times x$

$4y = 2 \times 2y$

So, $2x + 4y = 2 \times x + 2 \times 2y$

Using the distributive property in reverse:

$2x + 4y = 2(x + 2y)$

The completed statement is:

The factorisation of $2x + 4y$ is $2(x + 2y)$.

The given statement is $(a+b)^2 = a^2 + b^2$.

We know the algebraic identity for the square of a sum of two terms:

$(a+b)^2 = a^2 + 2ab + b^2$

Comparing this identity with the given statement $(a+b)^2 = a^2 + b^2$, we see that the given statement implies $a^2 + 2ab + b^2 = a^2 + b^2$.

Subtracting $a^2 + b^2$ from both sides, we get:

$(a^2 + 2ab + b^2) - (a^2 + b^2) = 0$

$2ab = 0$

The equation $2ab = 0$ is only true if $a=0$ or $b=0$ (or both). The original statement $(a+b)^2 = a^2 + b^2$ is only true under these specific conditions.

However, the statement is claimed to be a general identity, which means it should hold true for all values of $a$ and $b$. Let's test with a simple example where $a \neq 0$ and $b \neq 0$.

Let $a=1$ and $b=1$.

Left Hand Side (LHS) = $(a+b)^2 = (1+1)^2 = 2^2 = 4$

Right Hand Side (RHS) = $a^2 + b^2 = 1^2 + 1^2 = 1 + 1 = 2$

Since LHS $\neq$ RHS ($4 \neq 2$) for this example, the statement $(a+b)^2 = a^2 + b^2$ is not universally true.

Therefore, the statement $(a+b)^2 = a^2 + b^2$ is False.

The given statement is $(a-b)^2 = a^2 - b^2$.

We know the algebraic identity for the square of a difference of two terms:

$(a-b)^2 = a^2 - 2ab + b^2$

Comparing this identity with the given statement $(a-b)^2 = a^2 - b^2$, we see that the given statement implies $a^2 - 2ab + b^2 = a^2 - b^2$.

Subtracting $a^2 - b^2$ from both sides, we get:

$(a^2 - 2ab + b^2) - (a^2 - b^2) = 0$

$-2ab + 2b^2 = 0$

$2b(b-a) = 0$

The equation $2b(b-a) = 0$ is only true if $b=0$ or $b-a=0$ (i.e., $b=a$). The original statement $(a-b)^2 = a^2 - b^2$ is only true under these specific conditions.

However, the statement is claimed to be a general identity, which means it should hold true for all values of $a$ and $b$. Let's test with a simple example where $a \neq b$ and $b \neq 0$.

Let $a=2$ and $b=1$.

Left Hand Side (LHS) = $(a-b)^2 = (2-1)^2 = 1^2 = 1$

Right Hand Side (RHS) = $a^2 - b^2 = 2^2 - 1^2 = 4 - 1 = 3$

Since LHS $\neq$ RHS ($1 \neq 3$) for this example, the statement $(a-b)^2 = a^2 - b^2$ is not universally true.

Therefore, the statement $(a-b)^2 = a^2 - b^2$ is False.

The given statement is $(a + b) (a – b) = a^2 – b^2$.

Let's expand the Left Hand Side (LHS) of the equation using the distributive property:

LHS = $(a + b) (a – b)$

LHS = $a(a - b) + b(a - b)$

LHS = $a \times a - a \times b + b \times a - b \times b$

LHS = $a^2 - ab + ba - b^2$

Since multiplication is commutative, $ab = ba$. Substituting this into the expression for LHS:

LHS = $a^2 - ab + ab - b^2$

LHS = $a^2 + (-ab + ab) - b^2$

LHS = $a^2 + 0 - b^2$

LHS = $a^2 - b^2$

Now, let's compare the expanded LHS with the Right Hand Side (RHS):

LHS = $a^2 - b^2$

RHS = $a^2 - b^2$

Since LHS = RHS, the given statement $(a + b)(a – b) = a^2 – b^2$ is an algebraic identity which holds true for all values of $a$ and $b$.

Therefore, the statement $(a + b)(a – b) = a^2 – b^2$ is True.

The given statement is "The product of two negative terms is a negative term".

Let's consider two negative terms, say $-a$ and $-b$, where $a > 0$ and $b > 0$.

The product of these two terms is $(-a) \times (-b)$.

According to the rules of multiplication of signed numbers, the product of two negative numbers is a positive number.

$(-) \times (-) = (+)$

Therefore, $(-a) \times (-b) = +(ab) = ab$. Since $a > 0$ and $b > 0$, their product $ab$ is positive.

For example, let the two negative terms be $-2$ and $-3$.

The product is $(-2) \times (-3) = 6$.

The result $6$ is a positive term, not a negative term.

Thus, the product of two negative terms is a positive term.

Therefore, the statement "The product of two negative terms is a negative term" is False.

The given statement is "The product of one negative and one positive term is a negative term".

Let's consider a negative term, say $-a$, where $a > 0$, and a positive term, say $b$, where $b > 0$.

The product of these two terms is $(-a) \times b$.

According to the rules of multiplication of signed numbers, the product of a negative number and a positive number is a negative number.

$(-) \times (+) = (-)$

$(+) \times (-) = (-)$

Therefore, $(-a) \times b = -(ab) = -ab$. Since $a > 0$ and $b > 0$, their product $ab$ is positive, so $-ab$ is negative.

For example, let the negative term be $-4$ and the positive term be $5$.

The product is $(-4) \times (5) = -20$.

The result $-20$ is a negative term.

This confirms that the product of one negative and one positive term is always a negative term.

Therefore, the statement "The product of one negative and one positive term is a negative term" is True.

The given term is $-6x^2y^2$.

In an algebraic term, the coefficient is the numerical factor that multiplies the variable part of the term.

In the term $-6x^2y^2$, the variable part is $x^2y^2$. The number multiplying this variable part is $-6$.

Therefore, the coefficient of the term $-6x^2y^2$ is $-6$.

The given statement says that the coefficient of the term $-6x^2y^2$ is $-6$, which matches our finding.

Therefore, the statement "The coefficient of the term – 6x²y² is – 6" is True.

The given expression is $p^2q + q^2r + r^2q$.

An algebraic expression is classified by the number of terms it contains. A term is a single number, variable, or the product of numbers and variables.

• A monomial has one term.
• A binomial has two terms.
• A trinomial has three terms.
• A polynomial has one or more terms.

Let's identify the terms in the given expression $p^2q + q^2r + r^2q$:

• The first term is $p^2q$.
• The second term is $q^2r$.
• The third term is $r^2q$.

These three terms ($p^2q$, $q^2r$, and $r^2q$) are not like terms because they have different combinations of variables and powers. Therefore, they cannot be combined into a single term.

The expression has 3 distinct terms.

Since the expression $p^2q + q^2r + r^2q$ contains exactly three terms, it is a trinomial.

The statement claims that the expression is a binomial, which has only two terms.

Therefore, the statement "p²q + q²r + r²q is a binomial" is False.

The given statement is that the factors of $a^2 – 2ab + b^2$ are $(a + b)$ and $(a + b)$.

We know the algebraic identity for the square of a difference of two terms:

$a^2 - 2ab + b^2 = (a-b)^2$

This means the expression $a^2 - 2ab + b^2$ can be factored as $(a-b) \times (a-b)$.

Thus, the factors of $a^2 - 2ab + b^2$ are $(a-b)$ and $(a-b)$.

The statement claims that the factors are $(a+b)$ and $(a+b)$. Let's check the product of $(a+b)$ and $(a+b)$:

$(a+b)(a+b) = (a+b)^2 = a^2 + 2ab + b^2$

This product $a^2 + 2ab + b^2$ is different from $a^2 - 2ab + b^2$ unless $ab=0$ (i.e., $a=0$ or $b=0$).

Since the factors of $a^2 - 2ab + b^2$ are $(a-b)$ and $(a-b)$, and not $(a+b)$ and $(a+b)$, the given statement is incorrect.

Therefore, the statement "The factors of a² – 2ab + b² are (a + b) and (a + b)" is False.

The given statement is that $h$ is a factor of $2\pi(h+r)$.

For a term or variable to be a factor of an expression, the entire expression must be perfectly divisible by that term or variable. This means that when you divide the expression by the factor, the result is another expression without any remainder, and typically without the factor in the denominator.

Let's expand the given expression:

$2\pi(h+r) = 2\pi h + 2\pi r$

Now, let's attempt to divide the expression by $h$ to see if it is perfectly divisible:

$\frac{2\pi(h+r)}{h} = \frac{2\pi h + 2\pi r}{h}$

We can split this into two fractions:

$\frac{2\pi h}{h} + \frac{2\pi r}{h}$

Simplifying the first term:

$\frac{2\pi \cancel{h}}{\cancel{h}} = 2\pi$

The second term cannot be simplified further in general:

$\frac{2\pi r}{h}$

So, $\frac{2\pi(h+r)}{h} = 2\pi + \frac{2\pi r}{h}$.

For $h$ to be a factor of $2\pi(h+r)$, the result of this division must be an expression that does not have $h$ in the denominator. Since the term $\frac{2\pi r}{h}$ contains $h$ in the denominator (unless $r=0$), $h$ is not a general factor of $2\pi(h+r)$.

The expression $2\pi h + 2\pi r$ is a sum of two terms. While $h$ is a factor of the first term ($2\pi h = h \times 2\pi$), it is not a factor of the second term ($2\pi r$), unless $r=0$. For $h$ to be a factor of the entire expression, it must be a factor of every term (or be able to be factored out from the entire expression). Since $2\pi r$ does not have $h$ as a factor (for $r \neq 0$), $h$ is not a factor of the entire expression $2\pi h + 2\pi r$.

Therefore, the statement "h is a factor of 2π (h + r)" is False.

The given expression is $\frac{n^2}{2} + \frac{n}{2}$.

We need to find the factors of this expression. We can factor out the common terms from both parts of the sum.

The expression can be written as $\frac{1}{2}n^2 + \frac{1}{2}n$.

We observe that both terms $\frac{1}{2}n^2$ and $\frac{1}{2}n$ have a common numerical factor of $\frac{1}{2}$ and a common variable factor of $n$.

We can factor out $\frac{1}{2}n$ from the expression:

$\frac{n^2}{2} + \frac{n}{2} = \frac{1}{2}n(n) + \frac{1}{2}n(1)$

= $\frac{1}{2}n(n + 1)$

The factored form of the expression is $\frac{1}{2} \times n \times (n+1)$.

The factors of the expression are the terms that are multiplied together to form the expression. From the factored form, the factors are $\frac{1}{2}$, $n$, and $(n+1)$.

The statement claims that some of the factors of the expression are $\frac{1}{2}$, $n$, and $(n+1)$. Since these three terms are indeed the factors of the expression, the statement is correct.

Therefore, the statement "Some of the factors of $\frac{n^2}{2} + \frac{n}{2}$ are $\frac{1}{2}$ , n and (n + 1)" is True.

The given statement is "An equation is true for all values of its variables."

An equation is a mathematical statement that shows two expressions are equal.

For example, $x + 5 = 10$ is an equation.

To solve an equation means to find the specific value(s) of the variable(s) that make the equation true.

For the equation $x + 5 = 10$, the only value of $x$ that makes the statement true is $x=5$. If we substitute any other value for $x$ (e.g., $x=1$), the statement $1+5=10$ (which is $6=10$) is false.

A statement of equality that is true for *all* permissible values of its variables is called an identity.

For example, $(a+b)^2 = a^2 + 2ab + b^2$ is an identity because it is true for any values of $a$ and $b$.

The statement "An equation is true for all values of its variables" would mean that every equation is an identity. As shown by the example $x+5=10$, this is not the case. Many equations are only true for specific values of the variables, which are called the solutions of the equation.

Therefore, the statement "An equation is true for all values of its variables" is False.

The given statement is the equality $x^2 + (a + b)x + ab = (a + b) (x + ab)$.

Let's analyze the Left Hand Side (LHS) of the equation:

LHS = $x^2 + (a + b)x + ab$

This expression is a quadratic in $x$. It is a standard form that can be factored:

$x^2 + (a + b)x + ab = (x+a)(x+b)$

Now, let's analyze the Right Hand Side (RHS) of the equation and expand it:

RHS = $(a + b) (x + ab)$

Using the distributive property:

RHS = $a(x + ab) + b(x + ab)$

RHS = $a \times x + a \times ab + b \times x + b \times ab$

RHS = $ax + a^2b + bx + ab^2$

RHS = $ax + bx + a^2b + ab^2$

The given statement claims that LHS = RHS for all values of $x$, $a$, and $b$. This means it claims the following is an identity:

$(x+a)(x+b) = ax + bx + a^2b + ab^2$

Expanding the LHS: $(x+a)(x+b) = x^2 + bx + ax + ab$

So, the statement claims:

$x^2 + ax + bx + ab = ax + bx + a^2b + ab^2$

Subtracting $ax + bx$ from both sides:

$x^2 + ab = a^2b + ab^2$

This simplified equation $x^2 + ab = a^2b + ab^2$ must be true for all values of $x$, $a$, and $b$ for the original statement to be true. However, this is clearly not an identity.

For example, if we choose $a=1$ and $b=1$, the simplified equation becomes:

$x^2 + (1)(1) = (1)^2(1) + (1)(1)^2$

$x^2 + 1 = 1 + 1$

$x^2 + 1 = 2$

$x^2 = 1$

This equation $x^2=1$ is only true for $x=1$ or $x=-1$. It is not true for all values of $x$. Therefore, the original statement is not true for all values of $x$, $a$, and $b$.

Therefore, the statement "x² + (a + b)x + ab = (a + b) (x + ab)" is False.

The given statement is that the common factor of $11pq^2$, $121p^2q^3$, and $1331p^2q$ is $11p^2q^2$. To verify this, we need to find the Greatest Common Factor (GCF) of the three terms.

The three terms are:

1. $11pq^2$

2. $121p^2q^3$

3. $1331p^2q$

Let's find the GCF by considering the numerical coefficients and the variable parts separately.

Numerical Coefficients:

The numerical coefficients are 11, 121, and 1331.

We find the prime factorization of each coefficient:

$11 = 11^1$

$121 = 11 \times 11 = 11^2$

$1331 = 11 \times 11 \times 11 = 11^3$

The GCF of the numerical coefficients is the lowest power of the common prime factor, which is $11^1 = 11$.

Variable Part p:

The powers of $p$ in the terms are $p^1$, $p^2$, and $p^2$.

The lowest power of $p$ among the terms is $p^1 = p$.

The GCF of the variable $p$ is $p$.

Variable Part q:

The powers of $q$ in the terms are $q^2$, $q^3$, and $q^1$.

The lowest power of $q$ among the terms is $q^1 = q$.

The GCF of the variable $q$ is $q$.

To find the GCF of the entire terms, we multiply the GCFs of the numerical and variable parts:

GCF = (GCF of coefficients) $\times$ (GCF of $p$) $\times$ (GCF of $q$)

GCF = $11 \times p \times q = 11pq$

The greatest common factor of the terms $11pq^2$, $121p^2q^3$, and $1331p^2q$ is $11pq$.

The given statement claims the common factor is $11p^2q^2$. Since $11pq \neq 11p^2q^2$, the statement is incorrect.

Therefore, the statement "Common factor of 11pq², 121p²q³, 1331p²q is 11p²q²" is False.

The given expression is $12a^2b^2 + 4ab^2 – 32$. The terms in this expression are $12a^2b^2$, $4ab^2$, and $-32$.

To find the common factor of the expression, we need to find the Greatest Common Factor (GCF) of all the terms.

Let's find the factors for each term:

1. Factors of $12a^2b^2$: Include numerical factors of 12 (1, 2, 3, 4, 6, 12) and variable factors involving $a$ and $b$ ($a, a^2, b, b^2, ab, a^2b, ab^2, a^2b^2$).

2. Factors of $4ab^2$: Include numerical factors of 4 (1, 2, 4) and variable factors involving $a$ and $b$ ($a, b, b^2, ab, ab^2$).

3. Factors of $-32$: Include numerical factors of 32 (1, 2, 4, 8, 16, 32) and their negative counterparts. There are no variable factors $a$ or $b$ in this term.

Now, let's find the common factors among all three terms:

• Look at the numerical coefficients: 12, 4, and 32. The common numerical factors are the numbers that divide all three coefficients. These are 1, 2, and 4. The greatest among these is 4.
• Look at the variable parts: The variables present are $a$ and $b$. The first term has $a^2b^2$, the second term has $ab^2$, but the third term has no variables ($a$ or $b$). For a variable to be a common factor of the entire expression, it must be a factor of every term. Since $a$ is not a factor of $-32$ and $b$ is not a factor of $-32$, there are no common variable factors other than 1.

The common factor of the expression is the product of the common numerical factor and the common variable factor. In this case, the common factor is the greatest common numerical factor, which is 4.

GCF(12, 4, 32) = 4

GCF($a^2b^2$, $ab^2$, 1) = 1

Common Factor = $4 \times 1 = 4$

The statement says that the common factor of the expression is 4. Our calculation confirms this.

Therefore, the statement "Common factor of 12a²b² + 4ab² – 32 is 4" is True.

The given expression is $-3a^2 + 3ab + 3ac$.

To factor the expression, we look for common factors among the terms $-3a^2$, $3ab$, and $3ac$.

The numerical coefficients are $-3$, $3$, and $3$. The greatest common numerical factor is $3$.

The variable parts are $a^2$, $ab$, and $ac$. The common variable factor is $a$ (the lowest power of $a$ present in all terms).

So, the common factor of the expression is $3a$.

Now we factor out $3a$ from each term:

$-3a^2 \div (3a) = -a$

$3ab \div (3a) = b$

$3ac \div (3a) = c$

Therefore, the factorization of $-3a^2 + 3ab + 3ac$ is $3a(-a + b + c)$.

The statement claims that the factorization is $3a(-a - b - c)$. Let's expand this expression to check:

$3a(-a - b - c) = (3a) \times (-a) + (3a) \times (-b) + (3a) \times (-c)$

= $-3a^2 - 3ab - 3ac$

This expanded form $-3a^2 - 3ab - 3ac$ is not equal to the original expression $-3a^2 + 3ab + 3ac$.

The correct factorization is $3a(-a + b + c)$, which can also be written as $-3a(a - b - c)$ by factoring out $-1$ from the parenthesis.

Since the claimed factorization $3a(-a - b - c)$ does not result in the original expression, the statement is incorrect.

Therefore, the statement "Factorisation of – 3a² + 3ab + 3ac is 3a (–a – b – c)" is False.

The given statement claims that the factorised form of $p^2 + 30p + 216$ is $(p + 18) (p – 12)$.

To check if this factorization is correct, we can multiply the proposed factors $(p + 18)$ and $(p – 12)$ and see if the product is equal to the original expression $p^2 + 30p + 216$.

Using the FOIL method or distributive property:

$(p + 18)(p – 12) = p(p – 12) + 18(p – 12)$

= $p \times p + p \times (-12) + 18 \times p + 18 \times (-12)$

= $p^2 - 12p + 18p - 216$

Combine the like terms (the terms with $p$):

$-12p + 18p = (-12 + 18)p = 6p$

So, the product is:

$p^2 + 6p - 216$

Comparing the product $p^2 + 6p - 216$ with the original expression $p^2 + 30p + 216$, we see that they are not the same. The coefficient of $p$ is different ($6$ versus $30$) and the constant term is different ($-216$ versus $216$).

Alternatively, we can attempt to factor the expression $p^2 + 30p + 216$ directly. We need to find two numbers that multiply to 216 and add up to 30.

Let the two numbers be $m$ and $n$. We need $m \times n = 216$ and $m + n = 30$.

Since the sum and product are positive, both numbers must be positive.

Let's list pairs of factors of 216 and check their sum:

• 1 and 216 (Sum = 217)
• 2 and 108 (Sum = 110)
• 3 and 72 (Sum = 75)
• 4 and 54 (Sum = 58)
• 6 and 36 (Sum = 42)
• 8 and 27 (Sum = 35)
• 9 and 24 (Sum = 33)
• 12 and 18 (Sum = 30)

The numbers are 12 and 18. So, the factorization is $(p + 12)(p + 18)$.

The correct factorized form of $p^2 + 30p + 216$ is $(p + 12)(p + 18)$. The statement claims it is $(p + 18)(p – 12)$, which is incorrect.

Therefore, the statement "Factorised form of p² + 30p + 216 is (p + 18) (p – 12)" is False.

The given statement is "The difference of the squares of two consecutive numbers is their sum."

Let the two consecutive numbers be represented by $n$ and $n+1$, where $n$ is an integer.

The squares of these numbers are $n^2$ and $(n+1)^2$.

The difference of the squares of these two consecutive numbers is $(n+1)^2 - n^2$.

Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$, where $a = (n+1)$ and $b = n$, we have:

$(n+1)^2 - n^2 = ((n+1) - n)((n+1) + n)$

$(n+1)^2 - n^2 = (n + 1 - n)(n + 1 + n)$

$(n+1)^2 - n^2 = (1)(2n + 1)$

$(n+1)^2 - n^2 = 2n + 1$

Now, let's find the sum of the two consecutive numbers:

Sum = $n + (n+1)$

Sum = $n + n + 1$

Sum = $2n + 1$

Comparing the difference of the squares ($2n+1$) and the sum of the numbers ($2n+1$), we see that they are equal for any integer value of $n$.

Therefore, the statement "The difference of the squares of two consecutive numbers is their sum" is True.

The given statement is that the expression $abc + bca + cab$ is a monomial.

An algebraic expression is classified based on the number of terms it contains after combining like terms.

A monomial is an expression with exactly one term.

The given expression is $abc + bca + cab$. The individual parts separated by '+' signs are $abc$, $bca$, and $cab$. These are the terms before simplification.

Let's examine the terms:

• The first term is $abc$. It is the product of the variables $a$, $b$, and $c$.
• The second term is $bca$. By the commutative property of multiplication, $bca = abc$.
• The third term is $cab$. By the commutative property of multiplication, $cab = abc$.

Since $bca$ and $cab$ are equal to $abc$, all three terms in the expression are like terms.

We can combine these like terms by adding their coefficients (which are all 1):

$abc + bca + cab = 1 \cdot abc + 1 \cdot abc + 1 \cdot abc$

= $(1 + 1 + 1)abc$

= $3abc$

The simplified form of the expression is $3abc$. This simplified expression consists of a single term ($3abc$).

Since the expression simplifies to a single term, it is a monomial.

Therefore, the statement "abc + bca + cab is a monomial" is True.

The given statement is about the quotient of dividing $\frac{p}{3}$ by $\frac{3}{p}$.

We need to perform the division:

$\frac{p}{3} \div \frac{3}{p}$

Dividing by a fraction is the same as multiplying by its reciprocal. The reciprocal of $\frac{3}{p}$ is $\frac{p}{3}$.

So, the division becomes:

$\frac{p}{3} \times \frac{p}{3}$

Now, we multiply the numerators and the denominators:

Quotient = $\frac{p \times p}{3 \times 3} = \frac{p^2}{9}$

The calculated quotient is $\frac{p^2}{9}$.

The statement claims that the quotient is 9. This would mean $\frac{p^2}{9} = 9$.

Multiplying both sides by 9 gives $p^2 = 81$. This equation is only true for $p = 9$ or $p = -9$. It is not true for all possible values of $p$ (assuming $p \neq 0$, as $\frac{3}{p}$ is undefined if $p=0$).

Since the quotient $\frac{p^2}{9}$ is not equal to 9 for all values of $p$ (where $p \neq 0$), the statement is incorrect.

Therefore, the statement "On dividing $\frac{p}{3}$ by $\frac{3}{p}$ , the quotient is 9" is False.

The given equation is $51^2 – 49^2 = 100p$. We need to find the value of $p$ that satisfies this equation and check if it is 2.

Let's evaluate the left side of the equation: $51^2 – 49^2$.

We can use the difference of squares identity: $a^2 - b^2 = (a - b)(a + b)$.

Here, $a = 51$ and $b = 49$.

$51^2 – 49^2 = (51 - 49)(51 + 49)$

Calculate the values within the parentheses:

$51 - 49 = 2$

$51 + 49 = 100$

Now substitute these values back into the identity:

$51^2 – 49^2 = (2)(100)$

$51^2 – 49^2 = 200$

Now substitute this value back into the original equation:

$200 = 100p$

To find the value of $p$, divide both sides of the equation by 100:

$\frac{200}{100} = \frac{100p}{100}$

$2 = p$

So, the value of $p$ that satisfies the equation $51^2 – 49^2 = 100p$ is 2.

The given statement says that the value of $p$ is 2, which matches our calculated value.

Therefore, the statement "The value of p for 51² – 49² = 100p is 2" is True.

The given statement claims that $(9x – 51) \div 9$ is equal to $x – 51$.

To evaluate this, we need to perform the division of the expression $(9x - 51)$ by 9.

$(9x – 51) \div 9 = \frac{9x - 51}{9}$

We can divide each term in the numerator by the denominator:

$\frac{9x - 51}{9} = \frac{9x}{9} - \frac{51}{9}$

Now, we simplify each fraction:

$\frac{9x}{9} = x$

$\frac{51}{9} = \frac{3 \times 17}{3 \times 3} = \frac{17}{3}$

So, the result of the division is:

$(9x – 51) \div 9 = x - \frac{17}{3}$

The statement claims that the result is $x - 51$. Our calculation shows the result is $x - \frac{17}{3}$.

Since $x - \frac{17}{3} \neq x - 51$ (because $\frac{17}{3} \neq 51$), the statement is incorrect.

Therefore, the statement "(9x – 51) ÷ 9 is x – 51" is False.

The given statement claims that the value of $(a + 1) (a – 1) (a^2 + 1)$ is $a^4 – 1$.

Let's evaluate the expression $(a + 1) (a – 1) (a^2 + 1)$.

We can start by multiplying the first two factors, $(a + 1)$ and $(a – 1)$. This is a product of a sum and a difference, which follows the difference of squares identity: $(x+y)(x-y) = x^2 - y^2$.

Here, $x = a$ and $y = 1$.

So, $(a + 1)(a – 1) = a^2 - 1^2 = a^2 - 1$.

Now, substitute this result back into the original expression:

$(a + 1) (a – 1) (a^2 + 1) = (a^2 - 1)(a^2 + 1)$

Again, we have a product of a difference and a sum, $(a^2 - 1)(a^2 + 1)$, which is also in the form of the difference of squares identity: $(x-y)(x+y) = x^2 - y^2$.

Here, $x = a^2$ and $y = 1$.

So, $(a^2 - 1)(a^2 + 1) = (a^2)^2 - 1^2$.

Simplify the powers:

$(a^2)^2 = a^{2 \times 2} = a^4$

$1^2 = 1$

Therefore, the expression simplifies to:

$(a + 1) (a – 1) (a^2 + 1) = a^4 - 1$

The calculated value of the expression is $a^4 - 1$. The given statement claims that the value is $a^4 - 1$, which matches our result.

Therefore, the statement "The value of (a + 1) (a – 1) (a² + 1) is a⁴ – 1" is True.

(i) We need to add the expressions $7a^2bc$, $-3abc^2$, $3a^2bc$, and $2abc^2$.

Let's group the like terms:

Terms with $a^2bc$: $7a^2bc$ and $3a^2bc$

Terms with $abc^2$: $-3abc^2$ and $2abc^2$

Now, add the coefficients of the like terms:

$(7 + 3)a^2bc + (-3 + 2)abc^2$

$10a^2bc + (-1)abc^2$

$10a^2bc - abc^2$

The sum is $\mathbf{10a^2bc - abc^2}$.

(ii) We need to add the expressions $9ax + 3by – cz$ and $- 5by + ax + 3cz$.

Write the expressions one below the other, aligning like terms:

$+9ax +3by -1cz$

$+1ax -5by +3cz$

Add the coefficients of each column of like terms:

For $ax$: $9 + 1 = 10$

For $by$: $3 + (-5) = 3 - 5 = -2$

For $cz$: $-1 + 3 = 2$

The sum is $\mathbf{10ax - 2by + 2cz}$.

(iii) We need to add the expressions $xy^2z^2 + 3x^2y^2z – 4x^2yz^2$ and $- 9x^2y^2z + 3xy^2z^2 + x^2yz^2$.

Group the like terms from both expressions:

Terms with $xy^2z^2$: $xy^2z^2$ and $3xy^2z^2$

Terms with $x^2y^2z$: $3x^2y^2z$ and $-9x^2y^2z$

Terms with $x^2yz^2$: $-4x^2yz^2$ and $x^2yz^2$

Add the coefficients of the like terms:

For $xy^2z^2$: $(1 + 3)xy^2z^2 = 4xy^2z^2$

For $x^2y^2z$: $(3 + (-9))x^2y^2z = (3 - 9)x^2y^2z = -6x^2y^2z$

For $x^2yz^2$: $(-4 + 1)x^2yz^2 = -3x^2yz^2$

The sum is $\mathbf{4xy^2z^2 - 6x^2y^2z - 3x^2yz^2}$.

(iv) We need to add the expressions $5x^2 – 3xy + 4y^2 – 9$ and $7y^2 + 5xy – 2x^2 + 13$.

Write the expressions one below the other, aligning like terms:

$+5x^2 -3xy +4y^2 -9$

$-2x^2 +5xy +7y^2 +13$

Add the coefficients of each column of like terms:

For $x^2$: $5 + (-2) = 5 - 2 = 3$

For $xy$: $-3 + 5 = 2$

For $y^2$: $4 + 7 = 11$

For constants: $-9 + 13 = 4$

The sum is $\mathbf{3x^2 + 2xy + 11y^2 + 4}$.

(v) We need to add the expressions $2p^4 – 3p^3 + p^2 – 5p + 7$ and $-3p^4 – 7p^3 – 3p^2 – p – 12$.

Write the expressions one below the other, aligning like terms by powers of $p$:

$+2p^4 -3p^3 +1p^2 -5p +7$

$-3p^4 -7p^3 -3p^2 -1p -12$

Add the coefficients of each column of like terms:

For $p^4$: $2 + (-3) = 2 - 3 = -1$

For $p^3$: $-3 + (-7) = -3 - 7 = -10$

For $p^2$: $1 + (-3) = 1 - 3 = -2$

For $p$: $-5 + (-1) = -5 - 1 = -6$

For constants: $7 + (-12) = 7 - 12 = -5$

The sum is $\mathbf{-p^4 - 10p^3 - 2p^2 - 6p - 5}$.

(vi) We need to add the expressions $3a (a – b + c)$ and $2b (a – b + c)$.

First, expand each expression using the distributive property:

$3a (a – b + c) = 3a \times a - 3a \times b + 3a \times c = 3a^2 - 3ab + 3ac$

$2b (a – b + c) = 2b \times a - 2b \times b + 2b \times c = 2ab - 2b^2 + 2bc$

Now, add the expanded expressions: $(3a^2 - 3ab + 3ac) + (2ab - 2b^2 + 2bc)$.

Group the like terms:

$3a^2 + (-3ab + 2ab) + 3ac - 2b^2 + 2bc$

Combine the like terms:

$3a^2 + (-1ab) + 3ac - 2b^2 + 2bc$

$3a^2 - ab + 3ac - 2b^2 + 2bc$

Arrange the terms (optional, usually in alphabetical order or by degree):

$\mathbf{3a^2 - 2b^2 - ab + 3ac + 2bc}$

(vii) We need to add the expressions $3a (2b + 5c)$ and $3c (2a + 2b)$.

First, expand each expression using the distributive property:

$3a (2b + 5c) = 3a \times 2b + 3a \times 5c = 6ab + 15ac$

$3c (2a + 2b) = 3c \times 2a + 3c \times 2b = 6ac + 6bc$

Now, add the expanded expressions: $(6ab + 15ac) + (6ac + 6bc)$.

Group the like terms:

$6ab + (15ac + 6ac) + 6bc$

Combine the like terms:

$6ab + 21ac + 6bc$

Arrange the terms (optional, usually in alphabetical order or by degree):

$\mathbf{6ab + 6bc + 21ac}$

(i) We need to subtract $5a^2b^2c^2$ from $-7a^2b^2c^2$.

This is equivalent to calculating $-7a^2b^2c^2 - (5a^2b^2c^2)$.

Combine the coefficients of the like terms ($a^2b^2c^2$):

$(-7 - 5)a^2b^2c^2$

$-12a^2b^2c^2$

The difference is $\mathbf{-12a^2b^2c^2}$.

(ii) We need to subtract $6x^2 – 4xy + 5y^2$ from $8y^2 + 6xy – 3x^2$.

This is equivalent to calculating $(8y^2 + 6xy – 3x^2) - (6x^2 – 4xy + 5y^2)$.

Rearrange the second expression and change the sign of each term, then add:

$(8y^2 + 6xy – 3x^2) + (-6x^2 + 4xy - 5y^2)$

Group like terms:

$(-3x^2 - 6x^2) + (6xy + 4xy) + (8y^2 - 5y^2)$

Combine the coefficients of the like terms:

$(-3 - 6)x^2 + (6 + 4)xy + (8 - 5)y^2$

$-9x^2 + 10xy + 3y^2$

The difference is $\mathbf{-9x^2 + 10xy + 3y^2}$.

(iii) We need to subtract $2ab^2c^2 + 4a^2b^2c – 5a^2bc^2$ from $-10a^2b^2c + 4ab^2c^2 + 2a^2bc^2$.

This is equivalent to calculating $(-10a^2b^2c + 4ab^2c^2 + 2a^2bc^2) - (2ab^2c^2 + 4a^2b^2c – 5a^2bc^2)$.

Change the sign of each term in the second expression, then add:

$(-10a^2b^2c + 4ab^2c^2 + 2a^2bc^2) + (-2ab^2c^2 - 4a^2b^2c + 5a^2bc^2)$

Group like terms:

$(-10a^2b^2c - 4a^2b^2c) + (4ab^2c^2 - 2ab^2c^2) + (2a^2bc^2 + 5a^2bc^2)$

Combine the coefficients of the like terms:

$(-10 - 4)a^2b^2c + (4 - 2)ab^2c^2 + (2 + 5)a^2bc^2$

$-14a^2b^2c + 2ab^2c^2 + 7a^2bc^2$

Arrange the terms (optional):

$\mathbf{2ab^2c^2 - 14a^2b^2c + 7a^2bc^2}$

(iv) We need to subtract $3t^4 – 4t^3 + 2t^2 – 6t + 6$ from $- 4t^4 + 8t^3 – 4t^2 – 2t + 11$.

This is equivalent to calculating $(- 4t^4 + 8t^3 – 4t^2 – 2t + 11) - (3t^4 – 4t^3 + 2t^2 – 6t + 6)$.

Change the sign of each term in the second expression, then add:

$(- 4t^4 + 8t^3 – 4t^2 – 2t + 11) + (-3t^4 + 4t^3 - 2t^2 + 6t - 6)$

Group like terms:

$(-4t^4 - 3t^4) + (8t^3 + 4t^3) + (-4t^2 - 2t^2) + (-2t + 6t) + (11 - 6)$

Combine the coefficients of the like terms:

$(-4 - 3)t^4 + (8 + 4)t^3 + (-4 - 2)t^2 + (-2 + 6)t + (11 - 6)$

$-7t^4 + 12t^3 - 6t^2 + 4t + 5$

The difference is $\mathbf{-7t^4 + 12t^3 - 6t^2 + 4t + 5}$.

(v) We need to subtract $2ab + 5bc – 7ac$ from $5ab – 2bc – 2ac + 10abc$.

This is equivalent to calculating $(5ab – 2bc – 2ac + 10abc) - (2ab + 5bc – 7ac)$.

Change the sign of each term in the second expression, then add:

$(5ab – 2bc – 2ac + 10abc) + (-2ab - 5bc + 7ac)$

Group like terms:

$(5ab - 2ab) + (-2bc - 5bc) + (-2ac + 7ac) + 10abc$

Combine the coefficients of the like terms:

$(5 - 2)ab + (-2 - 5)bc + (-2 + 7)ac + 10abc$

$3ab - 7bc + 5ac + 10abc$

Arrange the terms (optional):

$\mathbf{10abc + 3ab - 7bc + 5ac}$

(vi) We need to subtract $7p (3q + 7p)$ from $8p (2p – 7q)$.

First, expand both expressions:

$7p (3q + 7p) = 7p \times 3q + 7p \times 7p = 21pq + 49p^2$

$8p (2p – 7q) = 8p \times 2p - 8p \times 7q = 16p^2 - 56pq$

Now, subtract the first expanded expression from the second expanded expression:

$(16p^2 - 56pq) - (49p^2 + 21pq)$

Change the sign of each term in the second expression, then add:

$(16p^2 - 56pq) + (-49p^2 - 21pq)$

Group like terms:

$(16p^2 - 49p^2) + (-56pq - 21pq)$

Combine the coefficients of the like terms:

$(16 - 49)p^2 + (-56 - 21)pq$

$-33p^2 - 77pq$

The difference is $\mathbf{-33p^2 - 77pq}$.

(vii) We need to subtract $-3p^2 + 3pq + 3px$ from $3p(– p – a – r)$.

First, expand the second expression:

$3p(– p – a – r) = 3p \times (-p) + 3p \times (-a) + 3p \times (-r)$

= $-3p^2 - 3ap - 3pr$

Now, subtract the first expression from the second expanded expression:

$(-3p^2 - 3ap - 3pr) - (-3p^2 + 3pq + 3px)$

Change the sign of each term in the second expression, then add:

$(-3p^2 - 3ap - 3pr) + (3p^2 - 3pq - 3px)$

Group like terms:

$(-3p^2 + 3p^2) - 3ap - 3pr - 3pq - 3px$

Combine the coefficients of the like terms:

$(-3 + 3)p^2 - 3ap - 3pr - 3pq - 3px$

$0p^2 - 3ap - 3pr - 3pq - 3px$

$-3ap - 3pr - 3pq - 3px$

The difference is $\mathbf{-3ap - 3pr - 3pq - 3px}$.

(i) Multiply $-7pq^2r^3$ by $-13p^3q^2r$.

$(-7pq^2r^3) \times (-13p^3q^2r) = (-7) \times (-13) \times p^{1+3} \times q^{2+2} \times r^{3+1}$

$= 91p^4q^4r^4$

(ii) Multiply $3x^2y^2z^2$ by $17xyz$.

$(3x^2y^2z^2) \times (17xyz) = 3 \times 17 \times x^{2+1} \times y^{2+1} \times z^{2+1}$

$= 51x^3y^3z^3$

(iii) Multiply $15xy^2$ by $17yz^2$.

$(15xy^2) \times (17yz^2) = 15 \times 17 \times x^1 \times y^{2+1} \times z^2$

$= 255xy^3z^2$

(iv) Multiply $-5a^2bc$, $11ab$, and $13abc^2$.

$(-5a^2bc) \times (11ab) \times (13abc^2) = (-5) \times 11 \times 13 \times a^{2+1+1} \times b^{1+1+1} \times c^{1+1+2}$

$= -715a^4b^3c^4$

(v) Multiply $-3x^2y$ by $(5y – xy)$.

$(-3x^2y)(5y - xy) = (-3x^2y)(5y) + (-3x^2y)(-xy)$

$= -15x^2y^2 + 3x^3y^2$

(vi) Multiply $abc$ by $(bc + ca)$.

$abc(bc + ca) = abc \times bc + abc \times ca$

$= ab^2c^2 + a^2bc^2$

(vii) Multiply $7pqr$ by $(p – q + r)$.

$7pqr(p - q + r) = 7pqr \times p - 7pqr \times q + 7pqr \times r$

$= 7p^2qr - 7pq^2r + 7pqr^2$

(viii) Multiply $x^2y^2z^2$ by $(xy – yz + zx)$.

$x^2y^2z^2(xy - yz + zx) = x^2y^2z^2 \times xy - x^2y^2z^2 \times yz + x^2y^2z^2 \times zx$

$= x^3y^3z^2 - x^2y^3z^3 + x^3y^2z^3$

(ix) Multiply $(p + 6)$ by $(q – 7)$.

$(p + 6)(q - 7) = p(q - 7) + 6(q - 7)$

$= pq - 7p + 6q - 42$

(x) Multiply $6mn$ by $0mn$.

$(6mn) \times (0mn) = 0$ (Since anything multiplied by zero is zero)

(xi) Multiply $a$, $a^5$, and $a^6$.

$a \times a^5 \times a^6 = a^{1+5+6} = a^{12}$

(xii) Multiply $-7st$, $-1$, and $-13st^2$.

$(-7st) \times (-1) \times (-13st^2) = (-7) \times (-1) \times (-13) \times s^{1+1} \times t^{1+2}$

$= 7 \times (-13) \times s^2 \times t^3$

$= -91s^2t^3$

(xiii) Multiply $b^3$, $3b^2$, and $7ab^5$.

$b^3 \times 3b^2 \times 7ab^5 = 1 \times 3 \times 7 \times a^1 \times b^{3+2+5}$

$= 21ab^{10}$

(xiv) Multiply $\frac{100}{9}$ rs by $\frac{3}{4}$ r³s².

$\left( \frac{100}{9} rs \right) \times \left( \frac{3}{4} r^3s^2 \right) = \left( \frac{100}{9} \times \frac{3}{4} \right) \times (r^1 \times r^3) \times (s^1 \times s^2)$

$= \left( \frac{\cancel{100}^{25}}{\cancel{9}_{3}} \times \frac{\cancel{3}^1}{\cancel{4}_1} \right) r^{1+3} s^{1+2}$

$= \left( \frac{25 \times 1}{3 \times 1} \right) r^4 s^3$

$= \frac{25}{3}r^4s^3$

(xv) Multiply $(a^2 – b^2)$ by $(a^2 + b^2)$.

Using the identity $(x-y)(x+y) = x^2 - y^2$ with $x=a^2$ and $y=b^2$:

$(a^2 - b^2)(a^2 + b^2) = (a^2)^2 - (b^2)^2$

$= a^4 - b^4$

(xvi) Multiply $(ab + c)$ by $(ab + c)$.

$(ab + c)(ab + c) = (ab + c)^2$

Using the identity $(x+y)^2 = x^2 + 2xy + y^2$ with $x=ab$ and $y=c$:

$(ab + c)^2 = (ab)^2 + 2(ab)(c) + c^2$

$= a^2b^2 + 2abc + c^2$

(xvii) Multiply $(pq – 2r)$ by $(pq – 2r)$.

$(pq – 2r)(pq – 2r) = (pq – 2r)^2$

Using the identity $(x-y)^2 = x^2 - 2xy + y^2$ with $x=pq$ and $y=2r$:

$(pq - 2r)^2 = (pq)^2 - 2(pq)(2r) + (2r)^2$

$= p^2q^2 - 4pqr + 4r^2$

(xviii) Multiply $\left( \frac{3}{4}x - \frac{4}{3}y \right)$ by $\left( \frac{2}{3}x + \frac{3}{2}y \right)$.

$\left( \frac{3}{4}x - \frac{4}{3}y \right) \left( \frac{2}{3}x + \frac{3}{2}y \right)$

$= \frac{3}{4}x \left( \frac{2}{3}x + \frac{3}{2}y \right) - \frac{4}{3}y \left( \frac{2}{3}x + \frac{3}{2}y \right)$

$= \left( \frac{3}{4}x \times \frac{2}{3}x \right) + \left( \frac{3}{4}x \times \frac{3}{2}y \right) - \left( \frac{4}{3}y \times \frac{2}{3}x \right) - \left( \frac{4}{3}y \times \frac{3}{2}y \right)$

$= \left( \frac{6}{12} \right)x^2 + \left( \frac{9}{8} \right)xy - \left( \frac{8}{9} \right)xy - \left( \frac{12}{6} \right)y^2$

$= \frac{1}{2}x^2 + \frac{9}{8}xy - \frac{8}{9}xy - 2y^2$

Combine the $xy$ terms: $\frac{9}{8} - \frac{8}{9} = \frac{9 \times 9 - 8 \times 8}{72} = \frac{81 - 64}{72} = \frac{17}{72}$

$= \frac{1}{2}x^2 + \frac{17}{72}xy - 2y^2$

(xix) Multiply $\frac{3}{2}$ p² + $\frac{2}{3}$ q² by (2p² – 3q²).

$\left( \frac{3}{2} p^2 + \frac{2}{3} q^2 \right) (2p^2 – 3q^2)$

$= \frac{3}{2}p^2(2p^2 - 3q^2) + \frac{2}{3}q^2(2p^2 - 3q^2)$

$= \left( \frac{3}{2}p^2 \times 2p^2 \right) + \left( \frac{3}{2}p^2 \times (-3q^2) \right) + \left( \frac{2}{3}q^2 \times 2p^2 \right) + \left( \frac{2}{3}q^2 \times (-3q^2) \right)$

$= 3p^4 - \frac{9}{2}p^2q^2 + \frac{4}{3}p^2q^2 - 2q^4$

Combine the $p^2q^2$ terms: $-\frac{9}{2} + \frac{4}{3} = \frac{-27 + 8}{6} = -\frac{19}{6}$

$= 3p^4 - \frac{19}{6}p^2q^2 - 2q^4$

(xx) Multiply $(x^2 – 5x + 6)$ by $(2x + 7)$.

$(x^2 – 5x + 6)(2x + 7)$

$= x^2(2x + 7) - 5x(2x + 7) + 6(2x + 7)$

$= (2x^3 + 7x^2) - (10x^2 + 35x) + (12x + 42)$

$= 2x^3 + 7x^2 - 10x^2 - 35x + 12x + 42$

Group and combine like terms:

$= 2x^3 + (7 - 10)x^2 + (-35 + 12)x + 42$

$= 2x^3 - 3x^2 - 23x + 42$

(xxi) Multiply $(3x^2 + 4x – 8)$ by $(2x^2 – 4x + 3)$.

$(3x^2 + 4x – 8)(2x^2 – 4x + 3)$

$= 3x^2(2x^2 – 4x + 3) + 4x(2x^2 – 4x + 3) - 8(2x^2 – 4x + 3)$

$= (6x^4 - 12x^3 + 9x^2) + (8x^3 - 16x^2 + 12x) - (16x^2 - 32x + 24)$

$= 6x^4 - 12x^3 + 9x^2 + 8x^3 - 16x^2 + 12x - 16x^2 + 32x - 24$

Group and combine like terms:

$= 6x^4 + (-12 + 8)x^3 + (9 - 16 - 16)x^2 + (12 + 32)x - 24$

$= 6x^4 - 4x^3 + (9 - 32)x^2 + 44x - 24$

$= 6x^4 - 4x^3 - 23x^2 + 44x - 24$

(xxii) Multiply $(2x – 2y – 3)$ by $(x + y + 5)$.

$(2x – 2y – 3)(x + y + 5)$

$= 2x(x + y + 5) - 2y(x + y + 5) - 3(x + y + 5)$

$= (2x^2 + 2xy + 10x) - (2xy + 2y^2 + 10y) - (3x + 3y + 15)$

$= 2x^2 + 2xy + 10x - 2xy - 2y^2 - 10y - 3x - 3y - 15$

Group and combine like terms:

$= 2x^2 - 2y^2 + (2xy - 2xy) + (10x - 3x) + (-10y - 3y) - 15$

$= 2x^2 - 2y^2 + 0xy + 7x - 13y - 15$

$= 2x^2 - 2y^2 + 7x - 13y - 15$

(i) We need to simplify $(3x + 2y)^2 + (3x – 2y)^2$.

We can use the identity $(A+B)^2 + (A-B)^2 = 2(A^2 + B^2)$.

Here, $A = 3x$ and $B = 2y$.

So, $(3x + 2y)^2 + (3x – 2y)^2 = 2((3x)^2 + (2y)^2)$

$= 2(9x^2 + 4y^2)$

$= 18x^2 + 8y^2$

The simplified expression is $\mathbf{18x^2 + 8y^2}$.

(ii) We need to simplify $(3x + 2y)^2 – (3x – 2y)^2$.

We can use the identity $(A+B)^2 - (A-B)^2 = 4AB$.

Here, $A = 3x$ and $B = 2y$.

So, $(3x + 2y)^2 – (3x – 2y)^2 = 4(3x)(2y)$

$= 4 \times 3 \times 2 \times xy$

$= 24xy$

The simplified expression is $\mathbf{24xy}$.

(iii) We need to simplify $\left( \frac{7}{9}a + \frac{9}{7}b \right)^2$ - ab.

First, expand the square using the identity $(A+B)^2 = A^2 + 2AB + B^2$.

Here, $A = \frac{7}{9}a$ and $B = \frac{9}{7}b$.

$\left( \frac{7}{9}a + \frac{9}{7}b \right)^2 = \left( \frac{7}{9}a \right)^2 + 2 \left( \frac{7}{9}a \right) \left( \frac{9}{7}b \right) + \left( \frac{9}{7}b \right)^2$

$= \frac{49}{81}a^2 + 2 \left( \frac{\cancel{7}}{\cancel{9}}a \right) \left( \frac{\cancel{9}}{\cancel{7}}b \right) + \frac{81}{49}b^2$

$= \frac{49}{81}a^2 + 2ab + \frac{81}{49}b^2$

Now, subtract $ab$ from this result:

$\left( \frac{49}{81}a^2 + 2ab + \frac{81}{49}b^2 \right) - ab$

$= \frac{49}{81}a^2 + (2ab - ab) + \frac{81}{49}b^2$

$= \frac{49}{81}a^2 + ab + \frac{81}{49}b^2$

The simplified expression is $\mathbf{\frac{49}{81}a^2 + ab + \frac{81}{49}b^2}$.

(iv) We need to simplify $\left( \frac{3}{4}x - \frac{4}{3}y \right)^2$ + 2xy.

First, expand the square using the identity $(A-B)^2 = A^2 - 2AB + B^2$.

Here, $A = \frac{3}{4}x$ and $B = \frac{4}{3}y$.

$\left( \frac{3}{4}x - \frac{4}{3}y \right)^2 = \left( \frac{3}{4}x \right)^2 - 2 \left( \frac{3}{4}x \right) \left( \frac{4}{3}y \right) + \left( \frac{4}{3}y \right)^2$

$= \frac{9}{16}x^2 - 2 \left( \frac{\cancel{3}}{\cancel{4}}x \right) \left( \frac{\cancel{4}}{\cancel{3}}y \right) + \frac{16}{9}y^2$

$= \frac{9}{16}x^2 - 2xy + \frac{16}{9}y^2$

Now, add $2xy$ to this result:

$\left( \frac{9}{16}x^2 - 2xy + \frac{16}{9}y^2 \right) + 2xy$

$= \frac{9}{16}x^2 + (-2xy + 2xy) + \frac{16}{9}y^2$

$= \frac{9}{16}x^2 + 0xy + \frac{16}{9}y^2$

$= \frac{9}{16}x^2 + \frac{16}{9}y^2$

The simplified expression is $\mathbf{\frac{9}{16}x^2 + \frac{16}{9}y^2}$.

(v) We need to simplify $(1.5p + 1.2q)^2 – (1.5p – 1.2q)^2$.

We can use the identity $(A+B)^2 - (A-B)^2 = 4AB$.

Here, $A = 1.5p$ and $B = 1.2q$.

So, $(1.5p + 1.2q)^2 – (1.5p – 1.2q)^2 = 4(1.5p)(1.2q)$

$= 4 \times 1.5 \times 1.2 \times pq$

$= 6 \times 1.2 \times pq$

$= 7.2pq$

The simplified expression is $\mathbf{7.2pq}$.

(vi) We need to simplify $(2.5m + 1.5q)^2 + (2.5m – 1.5q)^2$.

We can use the identity $(A+B)^2 + (A-B)^2 = 2(A^2 + B^2)$.

Here, $A = 2.5m$ and $B = 1.5q$.

So, $(2.5m + 1.5q)^2 + (2.5m – 1.5q)^2 = 2((2.5m)^2 + (1.5q)^2)$

$= 2(6.25m^2 + 2.25q^2)$

$= 12.5m^2 + 4.5q^2$

The simplified expression is $\mathbf{12.5m^2 + 4.5q^2}$.

(vii) We need to simplify $(x^2 – 4) + (x^2 + 4) + 16$.

Remove the parentheses and combine like terms:

$(x^2 – 4) + (x^2 + 4) + 16 = x^2 - 4 + x^2 + 4 + 16$

$= (x^2 + x^2) + (-4 + 4) + 16$

$= 2x^2 + 0 + 16$

$= 2x^2 + 16$

The simplified expression is $\mathbf{2x^2 + 16}$.

(viii) We need to simplify $(ab – c)^2 + 2abc$.

First, expand the square using the identity $(A-B)^2 = A^2 - 2AB + B^2$.

Here, $A = ab$ and $B = c$.

$(ab – c)^2 = (ab)^2 - 2(ab)(c) + c^2$

$= a^2b^2 - 2abc + c^2$

Now, add $2abc$ to this result:

$(a^2b^2 - 2abc + c^2) + 2abc$

$= a^2b^2 + (-2abc + 2abc) + c^2$

$= a^2b^2 + 0abc + c^2$

$= a^2b^2 + c^2$

The simplified expression is $\mathbf{a^2b^2 + c^2}$.

(ix) We need to simplify $(a – b) (a^2 + b^2 + ab) – (a + b) (a^2 + b^2 – ab)$.

Recognize the patterns related to the sum and difference of cubes identities:

$(x-y)(x^2+xy+y^2) = x^3 - y^3$

$(x+y)(x^2-xy+y^2) = x^3 + y^3$

The first part of the expression is $(a – b) (a^2 + ab + b^2)$, which equals $a^3 - b^3$.

The second part of the expression is $(a + b) (a^2 - ab + b^2)$, which equals $a^3 + b^3$.

So, the expression becomes $(a^3 - b^3) - (a^3 + b^3)$.

$(a^3 - b^3) - (a^3 + b^3) = a^3 - b^3 - a^3 - b^3$

$= (a^3 - a^3) + (-b^3 - b^3)$

$= 0 - 2b^3$

$= -2b^3$

The simplified expression is $\mathbf{-2b^3}$.

(x) We need to simplify $(b^2 – 49) (b + 7) + 343$.

First, expand the product $(b^2 – 49) (b + 7)$:

$(b^2 – 49) (b + 7) = b^2(b + 7) - 49(b + 7)$

$= (b^2 \times b + b^2 \times 7) - (49 \times b + 49 \times 7)$

$= b^3 + 7b^2 - 49b - 343$

Now, add 343 to this result:

$(b^3 + 7b^2 - 49b - 343) + 343$

$= b^3 + 7b^2 - 49b + (-343 + 343)$

$= b^3 + 7b^2 - 49b + 0$

$= b^3 + 7b^2 - 49b$

The simplified expression is $\mathbf{b^3 + 7b^2 - 49b}$.

(xi) We need to simplify $(4.5a + 1.5b)^2 + (4.5b + 1.5a)^2$.

Expand each square using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$(4.5a + 1.5b)^2 = (4.5a)^2 + 2(4.5a)(1.5b) + (1.5b)^2$

$= 20.25a^2 + 13.5ab + 2.25b^2$

$(4.5b + 1.5a)^2 = (4.5b)^2 + 2(4.5b)(1.5a) + (1.5a)^2$

$= 20.25b^2 + 13.5ab + 2.25a^2$

Now, add the two expanded expressions:

$(20.25a^2 + 13.5ab + 2.25b^2) + (20.25b^2 + 13.5ab + 2.25a^2)$

Group like terms:

$= (20.25a^2 + 2.25a^2) + (13.5ab + 13.5ab) + (2.25b^2 + 20.25b^2)$

Combine like terms:

$= 22.5a^2 + 27ab + 22.5b^2$

The simplified expression is $\mathbf{22.5a^2 + 27ab + 22.5b^2}$.

(xii) We need to simplify $(pq – qr)^2 + 4pq^2r$.

Expand the square using the identity $(A-B)^2 = A^2 - 2AB + B^2$.

Here, $A = pq$ and $B = qr$.

$(pq – qr)^2 = (pq)^2 - 2(pq)(qr) + (qr)^2$

$= p^2q^2 - 2p q^2 r + q^2r^2$

Now, add $4pq^2r$ to this result:

$(p^2q^2 - 2pq^2r + q^2r^2) + 4pq^2r$

Group like terms:

$= p^2q^2 + (-2pq^2r + 4pq^2r) + q^2r^2$

Combine like terms:

$= p^2q^2 + 2pq^2r + q^2r^2$

This expression is in the form $A^2 + 2AB + B^2 = (A+B)^2$, where $A=pq$ and $B=qr$.

$= (pq + qr)^2$

Alternatively, factor out $q^2$ from the simplified trinomial:

$= q^2(p^2 + 2pr + r^2)$

Recognize the perfect square in the parenthesis:

$= q^2(p+r)^2$

The simplified expression is $\mathbf{(pq + qr)^2}$ or $\mathbf{q^2(p+r)^2}$.

(xiii) We need to simplify $(s^2t + tq^2)^2 – (2stq)^2$.

Recognize this as a difference of squares: $A^2 - B^2 = (A-B)(A+B)$.

Here, $A = s^2t + tq^2$ and $B = 2stq$.

$(s^2t + tq^2)^2 – (2stq)^2 = ((s^2t + tq^2) - (2stq))((s^2t + tq^2) + (2stq))$

$= (s^2t + tq^2 - 2stq)(s^2t + tq^2 + 2stq)$

Factor out $t$ from the terms inside each parenthesis:

$= t(s^2 + q^2 - 2sq) \times t(s^2 + q^2 + 2sq)$

$= t^2 (s^2 - 2sq + q^2)(s^2 + 2sq + q^2)$

Recognize the perfect square trinomials:

$(s^2 - 2sq + q^2) = (s-q)^2$

$(s^2 + 2sq + q^2) = (s+q)^2$

Substitute the factored forms:

$= t^2 (s-q)^2 (s+q)^2$

This can also be written as:

$= t^2 ((s-q)(s+q))^2$

Using the difference of squares identity $(s-q)(s+q) = s^2 - q^2$:

$= t^2 (s^2 - q^2)^2$

Alternatively, expand the squares directly:

$(s^2t + tq^2)^2 = (s^2t)^2 + 2(s^2t)(tq^2) + (tq^2)^2 = s^4t^2 + 2s^2t^2q^2 + t^2q^4$

$(2stq)^2 = 4s^2t^2q^2$

Subtract the second from the first:

$(s^4t^2 + 2s^2t^2q^2 + t^2q^4) - 4s^2t^2q^2$

$= s^4t^2 + (2s^2t^2q^2 - 4s^2t^2q^2) + t^2q^4$

$= s^4t^2 - 2s^2t^2q^2 + t^2q^4$

Factor out $t^2$:

$= t^2(s^4 - 2s^2q^2 + q^4)$

Recognize the perfect square: $(s^2)^2 - 2s^2q^2 + (q^2)^2 = (s^2 - q^2)^2$

$= t^2(s^2 - q^2)^2$

The simplified expression is $\mathbf{t^2(s^2 - q^2)^2}$.

(i) Expand $(xy + yz)^2$.

Using the identity $(A+B)^2 = A^2 + 2AB + B^2$, with $A=xy$ and $B=yz$:

$(xy + yz)^2 = (xy)^2 + 2(xy)(yz) + (yz)^2$

$= x^2y^2 + 2xy^2z + y^2z^2$

(ii) Expand $(x^2y – xy^2)^2$.

Using the identity $(A-B)^2 = A^2 - 2AB + B^2$, with $A=x^2y$ and $B=xy^2$:

$(x^2y – xy^2)^2 = (x^2y)^2 - 2(x^2y)(xy^2) + (xy^2)^2$

$= x^4y^2 - 2x^{2+1}y^{1+2} + x^2y^4$

$= x^4y^2 - 2x^3y^3 + x^2y^4$

(iii) Expand $\left( \frac{4}{5}a + \frac{5}{4}b \right)^2$.

Using the identity $(A+B)^2 = A^2 + 2AB + B^2$, with $A=\frac{4}{5}a$ and $B=\frac{5}{4}b$:

$\left( \frac{4}{5}a + \frac{5}{4}b \right)^2 = \left( \frac{4}{5}a \right)^2 + 2 \left( \frac{4}{5}a \right) \left( \frac{5}{4}b \right) + \left( \frac{5}{4}b \right)^2$

$= \frac{16}{25}a^2 + 2 \left( \frac{\cancel{4}}{\cancel{5}}a \right) \left( \frac{\cancel{5}}{\cancel{4}}b \right) + \frac{25}{16}b^2$

$= \frac{16}{25}a^2 + 2ab + \frac{25}{16}b^2$

(iv) Expand $\left( \frac{2}{3}x - \frac{3}{2}y \right)^2$.

Using the identity $(A-B)^2 = A^2 - 2AB + B^2$, with $A=\frac{2}{3}x$ and $B=\frac{3}{2}y$:

$\left( \frac{2}{3}x - \frac{3}{2}y \right)^2 = \left( \frac{2}{3}x \right)^2 - 2 \left( \frac{2}{3}x \right) \left( \frac{3}{2}y \right) + \left( \frac{3}{2}y \right)^2$

$= \frac{4}{9}x^2 - 2 \left( \frac{\cancel{2}}{\cancel{3}}x \right) \left( \frac{\cancel{3}}{\cancel{2}}y \right) + \frac{9}{4}y^2$

$= \frac{4}{9}x^2 - 2xy + \frac{9}{4}y^2$

(v) Expand $\left( \frac{4}{5}p + \frac{5}{3}q \right)^2$.

Using the identity $(A+B)^2 = A^2 + 2AB + B^2$, with $A=\frac{4}{5}p$ and $B=\frac{5}{3}q$:

$\left( \frac{4}{5}p + \frac{5}{3}q \right)^2 = \left( \frac{4}{5}p \right)^2 + 2 \left( \frac{4}{5}p \right) \left( \frac{5}{3}q \right) + \left( \frac{5}{3}q \right)^2$

$= \frac{16}{25}p^2 + 2 \left( \frac{4}{\cancel{5}}p \right) \left( \frac{\cancel{5}}{3}q \right) + \frac{25}{9}q^2$

$= \frac{16}{25}p^2 + \frac{8}{3}pq + \frac{25}{9}q^2$

(vi) Expand $(x + 3) (x + 7)$.

Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$, with $a=3$ and $b=7$:

$(x + 3) (x + 7) = x^2 + (3+7)x + (3)(7)$

$= x^2 + 10x + 21$

(vii) Expand $(2x + 9) (2x – 7)$.

Using a variation of the identity $(X+a)(X+b) = X^2 + (a+b)X + ab$, with $X=2x$, $a=9$ and $b=-7$:

$(2x + 9) (2x – 7) = (2x)^2 + (9 + (-7))(2x) + (9)(-7)$

$= 4x^2 + (2)(2x) - 63$

$= 4x^2 + 4x - 63$

(viii) Expand $\left( \frac{4x}{5} + \frac{y}{4} \right) \left( \frac{4x}{5} + \frac{3y}{4} \right)$.

Using the identity $(X+A)(X+B) = X^2 + (A+B)X + AB$, with $X=\frac{4x}{5}$, $A=\frac{y}{4}$, $B=\frac{3y}{4}$:

$\left( \frac{4x}{5} + \frac{y}{4} \right) \left( \frac{4x}{5} + \frac{3y}{4} \right) = \left( \frac{4x}{5} \right)^2 + \left( \frac{y}{4} + \frac{3y}{4} \right) \left( \frac{4x}{5} \right) + \left( \frac{y}{4} \right) \left( \frac{3y}{4} \right)$

$= \frac{16x^2}{25} + \left( \frac{y+3y}{4} \right) \left( \frac{4x}{5} \right) + \frac{3y^2}{16}$

$= \frac{16x^2}{25} + \left( \frac{4y}{4} \right) \left( \frac{4x}{5} \right) + \frac{3y^2}{16}$

$= \frac{16x^2}{25} + (y) \left( \frac{4x}{5} \right) + \frac{3y^2}{16}$

$= \frac{16x^2}{25} + \frac{4xy}{5} + \frac{3y^2}{16}$

(ix) Expand $\left( \frac{2x}{3} - \frac{2}{3} \right) \left( \frac{2x}{3} + \frac{2a}{3} \right)$.

Using direct multiplication (distributive property):

$\left( \frac{2x}{3} - \frac{2}{3} \right) \left( \frac{2x}{3} + \frac{2a}{3} \right) = \frac{2x}{3} \left( \frac{2x}{3} + \frac{2a}{3} \right) - \frac{2}{3} \left( \frac{2x}{3} + \frac{2a}{3} \right)$

$= \left( \frac{2x}{3} \times \frac{2x}{3} \right) + \left( \frac{2x}{3} \times \frac{2a}{3} \right) - \left( \frac{2}{3} \times \frac{2x}{3} \right) - \left( \frac{2}{3} \times \frac{2a}{3} \right)$

$= \frac{4x^2}{9} + \frac{4ax}{9} - \frac{4x}{9} - \frac{4a}{9}$

(x) Expand $(2x – 5y) (2x – 5y)$.

This is equivalent to $(2x - 5y)^2$.

Using the identity $(A-B)^2 = A^2 - 2AB + B^2$, with $A=2x$ and $B=5y$:

$(2x – 5y)^2 = (2x)^2 - 2(2x)(5y) + (5y)^2$

$= 4x^2 - 20xy + 25y^2$

(xi) Expand $\left( \frac{2a}{3} + \frac{b}{3} \right) \left( \frac{2a}{3} - \frac{b}{3} \right)$.

Using the identity $(A+B)(A-B) = A^2 - B^2$, with $A=\frac{2a}{3}$ and $B=\frac{b}{3}$:

$\left( \frac{2a}{3} + \frac{b}{3} \right) \left( \frac{2a}{3} - \frac{b}{3} \right) = \left( \frac{2a}{3} \right)^2 - \left( \frac{b}{3} \right)^2$

$= \frac{4a^2}{9} - \frac{b^2}{9}$

(xii) Expand $(x^2 + y^2) (x^2 – y^2)$.

Using the identity $(A+B)(A-B) = A^2 - B^2$, with $A=x^2$ and $B=y^2$:

$(x^2 + y^2) (x^2 – y^2) = (x^2)^2 - (y^2)^2$

$= x^4 - y^4$

(xiii) Expand $(a^2 + b^2)^2$.

Using the identity $(A+B)^2 = A^2 + 2AB + B^2$, with $A=a^2$ and $B=b^2$:

$(a^2 + b^2)^2 = (a^2)^2 + 2(a^2)(b^2) + (b^2)^2$

$= a^4 + 2a^2b^2 + b^4$

(xiv) Expand (7x + 5)².

Using the identity $(A+B)^2 = A^2 + 2AB + B^2$, with $A=7x$ and $B=5$:

$(7x + 5)^2 = (7x)^2 + 2(7x)(5) + (5)^2$

$= 49x^2 + 70x + 25$

(xv) Expand (0.9p – 0.5q)².

Using the identity $(A-B)^2 = A^2 - 2AB + B^2$, with $A=0.9p$ and $B=0.5q$:

$(0.9p – 0.5q)^2 = (0.9p)^2 - 2(0.9p)(0.5q) + (0.5q)^2$

$= 0.81p^2 - 2(0.45pq) + 0.25q^2$

$= 0.81p^2 - 0.9pq + 0.25q^2$

(xvi) Expand $(xy)^2$. (Assuming the question implies expanding the term $(xy)^2$)

Using the property $(ab)^m = a^m b^m$:

$(xy)^2 = x^2 y^2$

(xvii) Multiply $(pq – 2r)$ by $(pq – 2r)$.

This is equivalent to $(pq - 2r)^2$.

Using the identity $(A-B)^2 = A^2 - 2AB + B^2$, with $A=pq$ and $B=2r$:

$(pq – 2r)^2 = (pq)^2 - 2(pq)(2r) + (2r)^2$

$= p^2q^2 - 4pqr + 4r^2$

(xviii) Multiply $\left( \frac{3}{4}x - \frac{4}{3}y \right)$ by $\left( \frac{2}{3}x + \frac{3}{2}y \right)$.

Using direct multiplication (distributive property):

$\left( \frac{3}{4}x - \frac{4}{3}y \right) \left( \frac{2}{3}x + \frac{3}{2}y \right) = \frac{3}{4}x \left( \frac{2}{3}x + \frac{3}{2}y \right) - \frac{4}{3}y \left( \frac{2}{3}x + \frac{3}{2}y \right)$

$= \left( \frac{3}{4}x \times \frac{2}{3}x \right) + \left( \frac{3}{4}x \times \frac{3}{2}y \right) - \left( \frac{4}{3}y \times \frac{2}{3}x \right) - \left( \frac{4}{3}y \times \frac{3}{2}y \right)$

$= \frac{6}{12}x^2 + \frac{9}{8}xy - \frac{8}{9}xy - \frac{12}{6}y^2$

$= \frac{1}{2}x^2 + \left( \frac{9}{8} - \frac{8}{9} \right)xy - 2y^2$

$\frac{9}{8} - \frac{8}{9} = \frac{9 \times 9 - 8 \times 8}{72} = \frac{81 - 64}{72} = \frac{17}{72}$

$= \frac{1}{2}x^2 + \frac{17}{72}xy - 2y^2$

(xix) Multiply $\left( \frac{3}{2} p^2 + \frac{2}{3} q^2 \right)$ by $(2p^2 – 3q^2)$.

Using direct multiplication (distributive property):

$\left( \frac{3}{2} p^2 + \frac{2}{3} q^2 \right) (2p^2 – 3q^2) = \frac{3}{2}p^2 (2p^2 – 3q^2) + \frac{2}{3}q^2 (2p^2 – 3q^2)$

$= \left( \frac{3}{2}p^2 \times 2p^2 \right) + \left( \frac{3}{2}p^2 \times (-3q^2) \right) + \left( \frac{2}{3}q^2 \times 2p^2 \right) + \left( \frac{2}{3}q^2 \times (-3q^2) \right)$

$= 3p^4 - \frac{9}{2}p^2q^2 + \frac{4}{3}p^2q^2 - 2q^4$

$= 3p^4 + \left( -\frac{9}{2} + \frac{4}{3} \right)p^2q^2 - 2q^4$

$-\frac{9}{2} + \frac{4}{3} = \frac{-27 + 8}{6} = -\frac{19}{6}$

$= 3p^4 - \frac{19}{6}p^2q^2 - 2q^4$

(xx) Multiply $(x^2 – 5x + 6)$ by $(2x + 7)$.

Using direct multiplication (distributive property):

$(x^2 – 5x + 6)(2x + 7) = x^2(2x + 7) - 5x(2x + 7) + 6(2x + 7)$

$= (x^2 \times 2x + x^2 \times 7) - (5x \times 2x + 5x \times 7) + (6 \times 2x + 6 \times 7)$

$= (2x^3 + 7x^2) - (10x^2 + 35x) + (12x + 42)$

$= 2x^3 + 7x^2 - 10x^2 - 35x + 12x + 42$

Group and combine like terms:

$= 2x^3 + (7 - 10)x^2 + (-35 + 12)x + 42$

$= 2x^3 - 3x^2 - 23x + 42$

(xxi) Multiply $(3x^2 + 4x – 8)$ by $(2x^2 – 4x + 3)$.

Using direct multiplication (distributive property):

$(3x^2 + 4x – 8)(2x^2 – 4x + 3) = 3x^2(2x^2 – 4x + 3) + 4x(2x^2 – 4x + 3) - 8(2x^2 – 4x + 3)$

$= (3x^2 \times 2x^2 - 3x^2 \times 4x + 3x^2 \times 3) + (4x \times 2x^2 - 4x \times 4x + 4x \times 3) - (8 \times 2x^2 - 8 \times 4x + 8 \times 3)$

$= (6x^4 - 12x^3 + 9x^2) + (8x^3 - 16x^2 + 12x) - (16x^2 - 32x + 24)$

$= 6x^4 - 12x^3 + 9x^2 + 8x^3 - 16x^2 + 12x - 16x^2 + 32x - 24$

Group and combine like terms:

$= 6x^4 + (-12 + 8)x^3 + (9 - 16 - 16)x^2 + (12 + 32)x - 24$

$= 6x^4 - 4x^3 + (9 - 32)x^2 + 44x - 24$

$= 6x^4 - 4x^3 - 23x^2 + 44x - 24$

(xxii) Multiply $(2x – 2y – 3)$ by $(x + y + 5)$.

Using direct multiplication (distributive property):

$(2x – 2y – 3)(x + y + 5) = 2x(x + y + 5) - 2y(x + y + 5) - 3(x + y + 5)$

$= (2x \times x + 2x \times y + 2x \times 5) - (2y \times x + 2y \times y + 2y \times 5) - (3 \times x + 3 \times y + 3 \times 5)$

$= (2x^2 + 2xy + 10x) - (2xy + 2y^2 + 10y) - (3x + 3y + 15)$

$= 2x^2 + 2xy + 10x - 2xy - 2y^2 - 10y - 3x - 3y - 15$

Group and combine like terms:

$= 2x^2 - 2y^2 + (2xy - 2xy) + (10x - 3x) + (-10y - 3y) - 15$

$= 2x^2 - 2y^2 + 0xy + 7x - 13y - 15$

$= 2x^2 - 2y^2 + 7x - 13y - 15$

(i) Evaluate $(52)^2$.

We can write $52$ as $50 + 2$. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$ with $a=50$ and $b=2$:

$(52)^2 = (50+2)^2 = 50^2 + 2(50)(2) + 2^2$

$= 2500 + 200 + 4$

$= 2704$

(ii) Evaluate $(49)^2$.

We can write $49$ as $50 - 1$. Using the identity $(a-b)^2 = a^2 - 2ab + b^2$ with $a=50$ and $b=1$:

$(49)^2 = (50-1)^2 = 50^2 - 2(50)(1) + 1^2$

$= 2500 - 100 + 1$

$= 2400 + 1$

$= 2401$

(iii) Evaluate $(103)^2$.

We can write $103$ as $100 + 3$. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$ with $a=100$ and $b=3$:

$(103)^2 = (100+3)^2 = 100^2 + 2(100)(3) + 3^2$

$= 10000 + 600 + 9$

$= 10609$

(iv) Evaluate $(98)^2$.

We can write $98$ as $100 - 2$. Using the identity $(a-b)^2 = a^2 - 2ab + b^2$ with $a=100$ and $b=2$:

$(98)^2 = (100-2)^2 = 100^2 - 2(100)(2) + 2^2$

$= 10000 - 400 + 4$

$= 9600 + 4$

$= 9604$

(v) Evaluate $(1005)^2$.

We can write $1005$ as $1000 + 5$. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$ with $a=1000$ and $b=5$:

$(1005)^2 = (1000+5)^2 = 1000^2 + 2(1000)(5) + 5^2$

$= 1000000 + 10000 + 25$

$= 1010025$

(vi) Evaluate $(995)^2$.

We can write $995$ as $1000 - 5$. Using the identity $(a-b)^2 = a^2 - 2ab + b^2$ with $a=1000$ and $b=5$:

$(995)^2 = (1000-5)^2 = 1000^2 - 2(1000)(5) + 5^2$

$= 1000000 - 10000 + 25$

$= 990000 + 25$

$= 990025$

(vii) Evaluate $47 \times 53$.

We can write $47$ as $50 - 3$ and $53$ as $50 + 3$. Using the identity $(a-b)(a+b) = a^2 - b^2$ with $a=50$ and $b=3$:

$47 \times 53 = (50 - 3)(50 + 3) = 50^2 - 3^2$

$= 2500 - 9$

$= 2491$

(viii) Evaluate $52 \times 53$.

We can write $52$ as $50 + 2$ and $53$ as $50 + 3$. Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $x=50$, $a=2$, and $b=3$:

$52 \times 53 = (50 + 2)(50 + 3) = 50^2 + (2+3)(50) + (2)(3)$

$= 2500 + (5)(50) + 6$

$= 2500 + 250 + 6$

$= 2750 + 6$

$= 2756$

(ix) Evaluate $105 \times 95$.

We can write $105$ as $100 + 5$ and $95$ as $100 - 5$. Using the identity $(a+b)(a-b) = a^2 - b^2$ with $a=100$ and $b=5$:

$105 \times 95 = (100 + 5)(100 - 5) = 100^2 - 5^2$

$= 10000 - 25$

$= 9975$

(x) Evaluate $104 \times 97$.

We can write $104$ as $100 + 4$ and $97$ as $100 - 3$. Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $x=100$, $a=4$, and $b=-3$:

$104 \times 97 = (100 + 4)(100 - 3) = 100^2 + (4 + (-3))(100) + (4)(-3)$

$= 10000 + (1)(100) - 12$

$= 10000 + 100 - 12$

$= 10100 - 12$

$= 10088$

(xi) Evaluate $101 \times 103$.

We can write $101$ as $100 + 1$ and $103$ as $100 + 3$. Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $x=100$, $a=1$, and $b=3$:

$101 \times 103 = (100 + 1)(100 + 3) = 100^2 + (1+3)(100) + (1)(3)$

$= 10000 + (4)(100) + 3$

$= 10000 + 400 + 3$

$= 10403$

(xii) Evaluate $98 \times 103$.

We can write $98$ as $100 - 2$ and $103$ as $100 + 3$. Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $x=100$, $a=-2$, and $b=3$:

$98 \times 103 = (100 - 2)(100 + 3) = 100^2 + (-2+3)(100) + (-2)(3)$

$= 10000 + (1)(100) - 6$

$= 10000 + 100 - 6$

$= 10100 - 6$

$= 10094$

(xiii) Evaluate $(9.9)^2$.

We can write $9.9$ as $10 - 0.1$. Using the identity $(a-b)^2 = a^2 - 2ab + b^2$ with $a=10$ and $b=0.1$:

$(9.9)^2 = (10 - 0.1)^2 = 10^2 - 2(10)(0.1) + (0.1)^2$

$= 100 - 2(1) + 0.01$

$= 100 - 2 + 0.01$

$= 98 + 0.01$

$= 98.01$

(xiv) Evaluate $9.8 \times 10.2$.

We can write $9.8$ as $10 - 0.2$ and $10.2$ as $10 + 0.2$. Using the identity $(a-b)(a+b) = a^2 - b^2$ with $a=10$ and $b=0.2$:

$9.8 \times 10.2 = (10 - 0.2)(10 + 0.2) = 10^2 - (0.2)^2$

$= 100 - 0.04$

$= 99.96$

(xv) Evaluate $10.1 \times 10.2$.

We can write $10.1$ as $10 + 0.1$ and $10.2$ as $10 + 0.2$. Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$ with $x=10$, $a=0.1$, and $b=0.2$:

$10.1 \times 10.2 = (10 + 0.1)(10 + 0.2) = 10^2 + (0.1 + 0.2)(10) + (0.1)(0.2)$

$= 100 + (0.3)(10) + 0.02$

$= 100 + 3 + 0.02$

$= 103.02$

(xvi) Evaluate $(35.4)^2 – (14.6)^2$.

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=35.4$ and $b=14.6$:

$(35.4)^2 – (14.6)^2 = (35.4 - 14.6)(35.4 + 14.6)$

$= (20.8)(50)$

$= 20.8 \times 50$

$= 1040$

(xvii) Evaluate (69.3)² – (30.7)².

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=69.3$ and $b=30.7$:

$(69.3)^2 – (30.7)^2 = (69.3 - 30.7)(69.3 + 30.7)$

$= (38.6)(100)$

$= 3860$

(xviii) Evaluate (9.7)² – (0.3)².

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=9.7$ and $b=0.3$:

$(9.7)^2 – (0.3)^2 = (9.7 - 0.3)(9.7 + 0.3)$

$= (9.4)(10.0)$

$= 94$

(xix) Evaluate (132)² – (68)².

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=132$ and $b=68$:

$(132)^2 – (68)^2 = (132 - 68)(132 + 68)$

$= (64)(200)$

$= 12800$

(xx) Evaluate (339)² – (161)².

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=339$ and $b=161$:

$(339)^2 – (161)^2 = (339 - 161)(339 + 161)$

$= (178)(500)$

$178 \times 500 = 178 \times 5 \times 100 = 890 \times 100 = 89000$

$= 89000$

(xxi) Evaluate (729)² – (271)².

Using the identity $a^2 - b^2 = (a - b)(a + b)$ with $a=729$ and $b=271$:

$(729)^2 – (271)^2 = (729 - 271)(729 + 271)$

$= (458)(1000)$

$= 458000$

(i) Find the GCF of $-18a^2$ and $108a$.

Numerical coefficients: $|-18| = 18$ and $108$.

Prime factors of 18: $2 \times 3^2$

Prime factors of 108: $2^2 \times 3^3$

GCF(18, 108) = $2^1 \times 3^2 = 2 \times 9 = 18$. (The sign of the GCF is usually taken as positive).

Variable parts: $a^2$ and $a^1$.

GCF($a^2$, $a^1$) = $a^1 = a$ (take the lowest power).

GCF of the terms = GCF(coefficients) $\times$ GCF(variable part)

= $18 \times a = 18a$.

(ii) Find the GCF of $3x^2y$, $18xy^2$, and $-6xy$.

Numerical coefficients: $|3| = 3$, $|18| = 18$, $|-6| = 6$.

Prime factors of 3: 3

Prime factors of 18: $2 \times 3^2$

Prime factors of 6: $2 \times 3$

GCF(3, 18, 6) = $3^1 = 3$.

Variable part $x$: $x^2$, $x^1$, $x^1$. Lowest power is $x^1 = x$. GCF($x^2$, $x$, $x$) = $x$.

Variable part $y$: $y^1$, $y^2$, $y^1$. Lowest power is $y^1 = y$. GCF($y$, $y^2$, $y$) = $y$.

GCF of the terms = GCF(coefficients) $\times$ GCF($x$) $\times$ GCF($y$)

= $3 \times x \times y = 3xy$.

(iii) Find the GCF of $2xy$, $-y^2$, and $2x^2y$.

Numerical coefficients: $|2| = 2$, $|-1| = 1$, $|2| = 2$.

GCF(2, 1, 2) = 1.

Variable part $x$: $x^1$, no $x$ ($x^0$), $x^2$. Lowest power is $x^0 = 1$. GCF($x$, $x^0$, $x^2$) = $1$.

Variable part $y$: $y^1$, $y^2$, $y^1$. Lowest power is $y^1 = y$. GCF($y$, $y^2$, $y$) = $y$.

GCF of the terms = GCF(coefficients) $\times$ GCF($x$) $\times$ GCF($y$)

= $1 \times 1 \times y = y$.

(iv) Find the GCF of $l^2m^2n$, $lm^2n^2$, and $l^2mn^2$.

Numerical coefficients: 1, 1, 1. GCF(1, 1, 1) = 1.

Variable part $l$: $l^2$, $l^1$, $l^2$. Lowest power is $l^1 = l$. GCF($l^2$, $l$, $l^2$) = $l$.

Variable part $m$: $m^2$, $m^2$, $m^1$. Lowest power is $m^1 = m$. GCF($m^2$, $m^2$, $m$) = $m$.

Variable part $n$: $n^1$, $n^2$, $n^2$. Lowest power is $n^1 = n$. GCF($n$, $n^2$, $n^2$) = $n$.

GCF of the terms = $1 \times l \times m \times n = lmn$.

(v) Find the GCF of $21pqr$, $-7p^2q^2r^2$, and $49p^2qr$.

Numerical coefficients: $|21| = 21$, $|-7| = 7$, $|49| = 49$.

Prime factors of 21: $3 \times 7$

Prime factors of 7: 7

Prime factors of 49: $7^2$

GCF(21, 7, 49) = $7^1 = 7$.

Variable part $p$: $p^1$, $p^2$, $p^2$. Lowest power is $p^1 = p$. GCF($p$, $p^2$, $p^2$) = $p$.

Variable part $q$: $q^1$, $q^2$, $q^1$. Lowest power is $q^1 = q$. GCF($q$, $q^2$, $q$) = $q$.

Variable part $r$: $r^1$, $r^2$, $r^1$. Lowest power is $r^1 = r$. GCF($r$, $r^2$, $r$) = $r$.

GCF of the terms = $7 \times p \times q \times r = 7pqr$.

(vi) Find the GCF of $qrxy$, $pryz$, and $rxyz$.

Numerical coefficients: 1, 1, 1. GCF(1, 1, 1) = 1.

Variable part $x$: $x^1$, no $x$ ($x^0$), $x^1$. Lowest power is $x^0 = 1$. GCF($x$, $x^0$, $x$) = $1$.

Variable part $y$: $y^1$, $y^1$, $y^1$. Lowest power is $y^1 = y$. GCF($y$, $y$, $y$) = $y$.

Variable part $z$: no $z$ ($z^0$), $z^1$, $z^1$. Lowest power is $z^0 = 1$. GCF($z^0$, $z$, $z$) = $1$.

Variable part $q$: $q^1$, no $q$ ($q^0$), no $q$ ($q^0$). Lowest power is $q^0 = 1$. GCF($q$, $q^0$, $q^0$) = $1$.

Variable part $r$: $r^1$, $r^1$, $r^1$. Lowest power is $r^1 = r$. GCF($r$, $r$, $r$) = $r$.

Variable part $p$: no $p$ ($p^0$), $p^1$, no $p$ ($p^0$). Lowest power is $p^0 = 1$. GCF($p^0$, $p$, $p^0$) = $1$.

Only the variable $r$ and $y$ are common to all terms. Let's recheck the terms: $qrxy$, $pryz$, $rxyz$.

• Variable $x$: appears in $qrxy$ and $rxyz$, but not in $pryz$. Not a common factor.
• Variable $y$: appears in $qrxy$ and $pryz$ and $rxyz$. Common factor is $y$.
• Variable $z$: appears in $pryz$ and $rxyz$, but not in $qrxy$. Not a common factor.
• Variable $q$: appears in $qrxy$, but not in $pryz$ and $rxyz$. Not a common factor.
• Variable $r$: appears in $qrxy$ and $pryz$ and $rxyz$. Common factor is $r$.
• Variable $p$: appears in $pryz$, but not in $qrxy$ and $rxyz$. Not a common factor.

The only common variables are $r$ and $y$. The lowest power of $r$ is $r^1$ and the lowest power of $y$ is $y^1$.

GCF of the terms = $1 \times r \times y = ry$. (Or $yr$).

(vii) Find the GCF of $3x^3y^2z$, $-6xy^3z^2$, and $12x^2yz^3$.

Numerical coefficients: $|3| = 3$, $|-6| = 6$, $|12| = 12$.

Prime factors of 3: 3

Prime factors of 6: $2 \times 3$

Prime factors of 12: $2^2 \times 3$

GCF(3, 6, 12) = $3^1 = 3$.

Variable part $x$: $x^3$, $x^1$, $x^2$. Lowest power is $x^1 = x$. GCF($x^3$, $x$, $x^2$) = $x$.

Variable part $y$: $y^2$, $y^3$, $y^1$. Lowest power is $y^1 = y$. GCF($y^2$, $y^3$, $y$) = $y$.

Variable part $z$: $z^1$, $z^2$, $z^3$. Lowest power is $z^1 = z$. GCF($z$, $z^2$, $z^3$) = $z$.

GCF of the terms = $3 \times x \times y \times z = 3xyz$.

(viii) Find the GCF of $63p^2a^2r^2s$, $-9pq^2r^2s^2$, $15p^2qr^2s^2$, and $-60p^2a^2rs^2$.

Numerical coefficients: $|63|=63$, $|-9|=9$, $|15|=15$, $|-60|=60$.

Prime factors of 63: $3^2 \times 7$

Prime factors of 9: $3^2$

Prime factors of 15: $3 \times 5$

Prime factors of 60: $2^2 \times 3 \times 5$

GCF(63, 9, 15, 60) = $3^1 = 3$.

Variable part $p$: $p^2$, $p^1$, $p^2$, $p^2$. Lowest power is $p^1 = p$. GCF($p^2$, $p$, $p^2$, $p^2$) = $p$.

Variable part $a$: $a^2$, no $a$ ($a^0$), no $a$ ($a^0$), $a^2$. Lowest power is $a^0 = 1$. GCF($a^2$, $a^0$, $a^0$, $a^2$) = $1$.

Variable part $q$: no $q$ ($q^0$), $q^2$, $q^1$, no $q$ ($q^0$). Lowest power is $q^0 = 1$. GCF($q^0$, $q^2$, $q$, $q^0$) = $1$.

Variable part $r$: $r^2$, $r^2$, $r^2$, $r^1$. Lowest power is $r^1 = r$. GCF($r^2$, $r^2$, $r^2$, $r$) = $r$.

Variable part $s$: $s^1$, $s^2$, $s^2$, $s^2$. Lowest power is $s^1 = s$. GCF($s$, $s^2$, $s^2$, $s^2$) = $s$.

GCF of the terms = $3 \times p \times 1 \times 1 \times r \times s = 3prs$.

(ix) Find the GCF of $13x^2y$ and $169xy$.

Numerical coefficients: 13 and 169.

Prime factors of 13: 13

Prime factors of 169: $13^2$

GCF(13, 169) = $13^1 = 13$.

Variable part $x$: $x^2$, $x^1$. Lowest power is $x^1 = x$. GCF($x^2$, $x$) = $x$.

Variable part $y$: $y^1$, $y^1$. Lowest power is $y^1 = y$. GCF($y$, $y$) = $y$.

GCF of the terms = $13 \times x \times y = 13xy$.

(x) Find the GCF of $11x^2$ and $12y^2$.

Numerical coefficients: 11 and 12.

Prime factors of 11: 11

Prime factors of 12: $2^2 \times 3$

GCF(11, 12) = 1 (since they have no common prime factors).

Variable part $x$: $x^2$, no $x$ ($x^0$). Lowest power is $x^0 = 1$. GCF($x^2$, $x^0$) = $1$.

Variable part $y$: no $y$ ($y^0$), $y^2$. Lowest power is $y^0 = 1$. GCF($y^0$, $y^2$) = $1$.

GCF of the terms = $1 \times 1 \times 1 = 1$.

(i) Factorise $6ab + 12bc$.

The terms are $6ab$ and $12bc$. The GCF of 6 and 12 is 6. Both terms share the variable $b$.

GCF($6ab$, $12bc$) = $6b$.

Factor out $6b$:

$6ab \div 6b = a$

$12bc \div 6b = 2c$

$6ab + 12bc = 6b(a + 2c)$.

(ii) Factorise –xy – ay.

The terms are $-xy$ and $-ay$. Both terms share the variable $y$. Both terms have a negative sign, so we can factor out $-y$ (or $y$). Let's factor out $-y$.

GCF($-xy$, $-ay$) = $y$ or $-y$. Let's take $-y$.

$-xy \div (-y) = x$

$-ay \div (-y) = a$

$-xy – ay = -y(x + a)$.

Alternatively, factoring out $y$: $-xy - ay = y(-x - a)$.

(iii) Factorise $ax^3 – bx^2 + cx$.

The terms are $ax^3$, $-bx^2$, and $cx$. All terms share the variable $x$. The lowest power of $x$ is $x^1 = x$. There are no common numerical factors other than 1.

GCF($ax^3$, $-bx^2$, $cx$) = $x$.

Factor out $x$:

$ax^3 \div x = ax^2$

$-bx^2 \div x = -bx$

$cx \div x = c$

$ax^3 – bx^2 + cx = x(ax^2 - bx + c)$.

(iv) Factorise $l^2m^2n – lm^2n^2 – l^2mn^2$.

The terms are $l^2m^2n$, $-lm^2n^2$, and $-l^2mn^2$.

Common variables: $l$, $m$, $n$.

Lowest power of $l$: $l^1$

Lowest power of $m$: $m^1$

Lowest power of $n$: $n^1$

GCF = $lmn$.

Factor out $lmn$:

$l^2m^2n \div lmn = l^{2-1}m^{2-1}n^{1-1} = l^1m^1n^0 = lm$

$-lm^2n^2 \div lmn = -l^{1-1}m^{2-1}n^{2-1} = -l^0m^1n^1 = -mn$

$-l^2mn^2 \div lmn = -l^{2-1}m^{1-1}n^{2-1} = -l^1m^0n^1 = -ln$

$l^2m^2n – lm^2n^2 – l^2mn^2 = lmn(lm - mn - ln)$.

(v) Factorise $3pqr – 6p^2q^2r^2 – 15r^2$.

The terms are $3pqr$, $-6p^2q^2r^2$, and $-15r^2$.

Numerical coefficients: 3, -6, -15. GCF(3, 6, 15) = 3.

Variable $p$: $p^1$, $p^2$, no $p$. Not common to all. (GCF is $p^0=1$).

Variable $q$: $q^1$, $q^2$, no $q$. Not common to all. (GCF is $q^0=1$).

Variable $r$: $r^1$, $r^2$, $r^2$. Lowest power is $r^1 = r$. GCF($r$, $r^2$, $r^2$) = $r$.

GCF = $3r$.

Factor out $3r$:

$3pqr \div 3r = pq$

$-6p^2q^2r^2 \div 3r = -2p^2q^2r$

$-15r^2 \div 3r = -5r$

$3pqr – 6p^2q^2r^2 – 15r^2 = 3r(pq - 2p^2q^2r - 5r)$.

(vi) Factorise $x^3y^2 + x^2y^3 – xy^4 + xy$.

The terms are $x^3y^2$, $x^2y^3$, $-xy^4$, and $xy$. All terms share $x$ and $y$.

Lowest power of $x$: $x^1$.

Lowest power of $y$: $y^1$.

GCF = $xy$.

Factor out $xy$:

$x^3y^2 \div xy = x^2y$

$x^2y^3 \div xy = x^1y^2 = xy^2$

$-xy^4 \div xy = -y^3$

$xy \div xy = 1$

$x^3y^2 + x^2y^3 – xy^4 + xy = xy(x^2y + xy^2 - y^3 + 1)$.

(vii) Factorise $4xy^2 – 10x^2y + 16x^2y^2 + 2xy$.

The terms are $4xy^2$, $-10x^2y$, $16x^2y^2$, and $2xy$.

Numerical coefficients: 4, -10, 16, 2. GCF(4, 10, 16, 2) = 2.

Variable $x$: $x^1$, $x^2$, $x^2$, $x^1$. Lowest power is $x^1 = x$.

Variable $y$: $y^2$, $y^1$, $y^2$, $y^1$. Lowest power is $y^1 = y$.

GCF = $2xy$.

Factor out $2xy$:

$4xy^2 \div 2xy = 2y$

$-10x^2y \div 2xy = -5x$

$16x^2y^2 \div 2xy = 8xy$

$2xy \div 2xy = 1$

$4xy^2 – 10x^2y + 16x^2y^2 + 2xy = 2xy(2y - 5x + 8xy + 1)$.

(viii) Factorise $2a^3 – 3a^2b + 5ab^2 – ab$.

The terms are $2a^3$, $-3a^2b$, $5ab^2$, and $-ab$. All terms share the variable $a$. The lowest power of $a$ is $a^1 = a$. There are no common numerical factors other than 1. Variable $b$ is not in all terms.

GCF = $a$.

Factor out $a$:

$2a^3 \div a = 2a^2$

$-3a^2b \div a = -3ab$

$5ab^2 \div a = 5b^2$

$-ab \div a = -b$

$2a^3 – 3a^2b + 5ab^2 – ab = a(2a^2 - 3ab + 5b^2 - b)$.

(ix) Factorise $63p^2q^2r^2s – 9pq^2r^2s^2 + 15p^2qr^2s^2 – 60p^2q^2rs^2$.

The terms are $63p^2q^2r^2s$, $-9pq^2r^2s^2$, $15p^2qr^2s^2$, and $-60p^2q^2rs^2$.

Numerical coefficients: 63, -9, 15, -60. GCF(63, 9, 15, 60) = 3.

Variable $p$: $p^2$, $p^1$, $p^2$, $p^2$. Lowest power is $p^1 = p$.

Variable $q$: $q^2$, $q^2$, $q^1$, $q^2$. Lowest power is $q^1 = q$.

Variable $r$: $r^2$, $r^2$, $r^2$, $r^1$. Lowest power is $r^1 = r$.

Variable $s$: $s^1$, $s^2$, $s^2$, $s^2$. Lowest power is $s^1 = s$.

GCF = $3pqr s$.

Factor out $3pqrs$:

$63p^2q^2r^2s \div 3pqrs = 21pr$

$-9pq^2r^2s^2 \div 3pqrs = -3qrs$

$15p^2qr^2s^2 \div 3pqrs = 5pr s$

$-60p^2q^2rs^2 \div 3pqrs = -20pq s$

$63p^2q^2r^2s – 9pq^2r^2s^2 + 15p^2qr^2s^2 – 60p^2q^2rs^2 = 3pqrs(21pr - 3qrs + 5prs - 20pqs)$.

(x) Factorise $24x^2yz^3 – 6xy^3z^2 + 15x^2y^2z – 5xyz$.

The terms are $24x^2yz^3$, $-6xy^3z^2$, $15x^2y^2z$, and $-5xyz$.

Numerical coefficients: 24, -6, 15, -5. GCF(24, 6, 15, 5) = 1.

Variable $x$: $x^2$, $x^1$, $x^2$, $x^1$. Lowest power is $x^1 = x$.

Variable $y$: $y^1$, $y^3$, $y^2$, $y^1$. Lowest power is $y^1 = y$.

Variable $z$: $z^3$, $z^2$, $z^1$, $z^1$. Lowest power is $z^1 = z$.

GCF = $xyz$.

Factor out $xyz$:

$24x^2yz^3 \div xyz = 24xz^2$

$-6xy^3z^2 \div xyz = -6y^2z$

$15x^2y^2z \div xyz = 15xy$

$-5xyz \div xyz = -5$

$24x^2yz^3 – 6xy^3z^2 + 15x^2y^2z – 5xyz = xyz(24xz^2 - 6y^2z + 15xy - 5)$.

(xi) Factorise $a^3 + a^2 + a + 1$ by grouping.

Group the terms: $(a^3 + a^2) + (a + 1)$.

Factor the first group: $a^2(a + 1)$.

Factor the second group: $1(a + 1)$.

The expression becomes $a^2(a + 1) + 1(a + 1)$.

Now, factor out the common binomial factor $(a+1)$:

$(a + 1)(a^2 + 1)$.

(xii) Factorise $lx + my + mx + ly$ by grouping.

Rearrange the terms to group those with common factors: $lx + mx + ly + my$.

Group the terms: $(lx + mx) + (ly + my)$.

Factor the first group: $x(l + m)$.

Factor the second group: $y(l + m)$.

The expression becomes $x(l + m) + y(l + m)$.

Now, factor out the common binomial factor $(l+m)$:

$(l + m)(x + y)$.

(xiii) Factorise $a^3x – x^4 + a^2x^2 – ax^3$ by grouping.

Rearrange the terms: $a^3x + a^2x^2 - ax^3 - x^4$.

Group the terms: $(a^3x + a^2x^2) + (-ax^3 - x^4)$.

Factor the first group: GCF is $a^2x$. $a^2x(a + x)$.

Factor the second group: GCF is $-x^3$. $-x^3(a + x)$.

The expression becomes $a^2x(a + x) - x^3(a + x)$.

Now, factor out the common binomial factor $(a+x)$:

$(a + x)(a^2x - x^3)$.

Further factor out $x$ from the second binomial:

$(a + x) x(a^2 - x^2)$.

Recognize the difference of squares in the second binomial:

$(a^2 - x^2) = (a - x)(a + x)$.

So, the factorization is $x(a + x)(a - x)(a + x) = x(a+x)^2(a-x)$.

Alternatively, factor out $x$ first from the entire expression:

$x(a^3 - x^3 + a^2x - ax^2)$

Group inside the parenthesis: $x((a^3 - x^3) + (a^2x - ax^2))$

$= x((a-x)(a^2+ax+x^2) + ax(a-x))$

$= x(a-x)((a^2+ax+x^2) + ax)$

$= x(a-x)(a^2 + 2ax + x^2)$

$= x(a-x)(a+x)^2$.

(xiv) Factorise $2x^2 – 2y + 4xy – x$ by grouping.

Rearrange the terms: $2x^2 - x + 4xy - 2y$.

Group the terms: $(2x^2 - x) + (4xy - 2y)$.

Factor the first group: $x(2x - 1)$.

Factor the second group: $2y(2x - 1)$.

The expression becomes $x(2x - 1) + 2y(2x - 1)$.

Now, factor out the common binomial factor $(2x-1)$:

$(2x - 1)(x + 2y)$.

(xv) Factorise $y^2 + 8zx – 2xy – 4yz$ by grouping.

Rearrange the terms: $y^2 - 2xy - 4yz + 8zx$.

Group the terms: $(y^2 - 2xy) + (-4yz + 8zx)$.

Factor the first group: $y(y - 2x)$.

Factor the second group: $-4z(y - 2x)$. Note that we factor out $-4z$ to get the common binomial factor $(y-2x)$.

The expression becomes $y(y - 2x) - 4z(y - 2x)$.

Now, factor out the common binomial factor $(y-2x)$:

$(y - 2x)(y - 4z)$.

(xvi) Factorise $ax^2y – bxyz – ax^2z + bxy^2$ by grouping.

Rearrange the terms: $ax^2y - ax^2z + bxy^2 - bxyz$.

Group the terms: $(ax^2y - ax^2z) + (bxy^2 - bxyz)$.

Factor the first group: $ax^2(y - z)$.

Factor the second group: $bxy(y - z)$.

The expression becomes $ax^2(y - z) + bxy(y - z)$.

Now, factor out the common binomial factor $(y-z)$:

$(y - z)(ax^2 + bxy)$.

Further factor out $x$ from the second binomial:

$(y - z)x(ax + by)$.

The factorization is $x(y-z)(ax+by)$.

(xvii) Factorise $a^2b + a^2c + ab + ac + b^2c + c^2b$ by grouping.

Group the terms in pairs that share factors:

$(a^2b + a^2c) + (ab + ac) + (b^2c + c^2b)$.

Factor each group:

$a^2(b + c) + a(b + c) + bc(b + c)$.

Now, factor out the common binomial factor $(b+c)$ from all three terms:

$(b + c)(a^2 + a + bc)$.

(xviii) Factorise $2ax^2 + 4axy + 3bx^2 + 2ay^2 + 6bxy + 3by^2$ by grouping.

Group terms sharing factors, e.g., those with $a$ and those with $b$, or those with $x^2, xy, y^2$ with $a$ and $b$ coefficients.

Let's group by the presence of $a$ and $b$ coefficients:

Terms with $a$: $2ax^2 + 4axy + 2ay^2 = 2a(x^2 + 2xy + y^2) = 2a(x+y)^2$.

Terms with $b$: $3bx^2 + 6bxy + 3by^2 = 3b(x^2 + 2xy + y^2) = 3b(x+y)^2$.

The original expression is the sum of these two groups:

$2a(x+y)^2 + 3b(x+y)^2$.

Now, factor out the common binomial factor $(x+y)^2$:

$(x+y)^2 (2a + 3b)$.

Alternatively, group by powers of $x$ and $y$ with respective coefficients:

$(2ax^2 + 3bx^2) + (4axy + 6bxy) + (2ay^2 + 3by^2)$

$x^2(2a + 3b) + 2xy(2a + 3b) + y^2(2a + 3b)$

Now, factor out the common binomial factor $(2a+3b)$:

$(2a + 3b)(x^2 + 2xy + y^2)$.

Recognize the perfect square trinomial: $x^2 + 2xy + y^2 = (x+y)^2$.

$(2a + 3b)(x+y)^2$.

We use the identity $A^2 + 2AB + B^2 = (A + B)^2$.

(i) Factorise $x^2 + 6x + 9$.

The first term $x^2$ is the square of $x$. The last term $9$ is the square of $3$.

Let $A = x$ and $B = 3$. Check the middle term: $2AB = 2(x)(3) = 6x$.

This matches the given expression $x^2 + 6x + 9$.

So, $x^2 + 6x + 9 = (x + 3)^2$.

(ii) Factorise $x^2 + 12x + 36$.

The first term $x^2$ is the square of $x$. The last term $36$ is the square of $6$.

Let $A = x$ and $B = 6$. Check the middle term: $2AB = 2(x)(6) = 12x$.

This matches the given expression $x^2 + 12x + 36$.

So, $x^2 + 12x + 36 = (x + 6)^2$.

(iii) Factorise $x^2 + 14x + 49$.

The first term $x^2$ is the square of $x$. The last term $49$ is the square of $7$.

Let $A = x$ and $B = 7$. Check the middle term: $2AB = 2(x)(7) = 14x$.

This matches the given expression $x^2 + 14x + 49$.

So, $x^2 + 14x + 49 = (x + 7)^2$.

(iv) Factorise $x^2 + 2x + 1$.

The first term $x^2$ is the square of $x$. The last term $1$ is the square of $1$.

Let $A = x$ and $B = 1$. Check the middle term: $2AB = 2(x)(1) = 2x$.

This matches the given expression $x^2 + 2x + 1$.

So, $x^2 + 2x + 1 = (x + 1)^2$.

(v) Factorise $4x^2 + 4x + 1$.

The first term $4x^2$ is the square of $2x$. The last term $1$ is the square of $1$.

Let $A = 2x$ and $B = 1$. Check the middle term: $2AB = 2(2x)(1) = 4x$.

This matches the given expression $4x^2 + 4x + 1$.

So, $4x^2 + 4x + 1 = (2x + 1)^2$.

(vi) Factorise $a^2x^2 + 2ax + 1$.

The first term $a^2x^2$ is the square of $ax$. The last term $1$ is the square of $1$.

Let $A = ax$ and $B = 1$. Check the middle term: $2AB = 2(ax)(1) = 2ax$.

This matches the given expression $a^2x^2 + 2ax + 1$.

So, $a^2x^2 + 2ax + 1 = (ax + 1)^2$.

(vii) Factorise $a^2x^2 + 2abx + b^2$.

The first term $a^2x^2$ is the square of $ax$. The last term $b^2$ is the square of $b$.

Let $A = ax$ and $B = b$. Check the middle term: $2AB = 2(ax)(b) = 2abx$.

This matches the given expression $a^2x^2 + 2abx + b^2$.

So, $a^2x^2 + 2abx + b^2 = (ax + b)^2$.

(viii) Factorise $a^2x^2 + 2abxy + b^2y^2$.

The first term $a^2x^2$ is the square of $ax$. The last term $b^2y^2$ is the square of $by$.

Let $A = ax$ and $B = by$. Check the middle term: $2AB = 2(ax)(by) = 2abxy$.

This matches the given expression $a^2x^2 + 2abxy + b^2y^2$.

So, $a^2x^2 + 2abxy + b^2y^2 = (ax + by)^2$.

(ix) Factorise $4x^2 + 12x + 9$.

The first term $4x^2$ is the square of $2x$. The last term $9$ is the square of $3$.

Let $A = 2x$ and $B = 3$. Check the middle term: $2AB = 2(2x)(3) = 12x$.

This matches the given expression $4x^2 + 12x + 9$.

So, $4x^2 + 12x + 9 = (2x + 3)^2$.

(x) Factorise $16x^2 + 40x +25$.

The first term $16x^2$ is the square of $4x$. The last term $25$ is the square of $5$.

Let $A = 4x$ and $B = 5$. Check the middle term: $2AB = 2(4x)(5) = 40x$.

This matches the given expression $16x^2 + 40x +25$.

So, $16x^2 + 40x +25 = (4x + 5)^2$.

(xi) Factorise $9x^2 + 24x + 16$.

The first term $9x^2$ is the square of $3x$. The last term $16$ is the square of $4$.

Let $A = 3x$ and $B = 4$. Check the middle term: $2AB = 2(3x)(4) = 24x$.

This matches the given expression $9x^2 + 24x + 16$.

So, $9x^2 + 24x + 16 = (3x + 4)^2$.

(xii) Factorise $9x^2 + 30x + 25$.

The first term $9x^2$ is the square of $3x$. The last term $25$ is the square of $5$.

Let $A = 3x$ and $B = 5$. Check the middle term: $2AB = 2(3x)(5) = 30x$.

This matches the given expression $9x^2 + 30x + 25$.

So, $9x^2 + 30x + 25 = (3x + 5)^2$.

(xiii) Factorise $2x^3 + 24x^2 + 72x$.

First, factor out the greatest common factor (GCF). The GCF of the numerical coefficients (2, 24, 72) is 2. The GCF of the variable parts ($x^3, x^2, x$) is $x$. So, the GCF of the expression is $2x$.

$2x^3 + 24x^2 + 72x = 2x(x^2 + 12x + 36)$.

Now, factorise the trinomial inside the parenthesis, $x^2 + 12x + 36$. From part (ii), we know this is a perfect square trinomial, $(x+6)^2$.

So, $2x^3 + 24x^2 + 72x = 2x(x + 6)^2$.

(xiv) Factorise $a^2x^3 + 2abx^2 + b^2x$.

First, factor out the greatest common factor (GCF). The GCF of the variable parts ($x^3, x^2, x$) is $x$. There are no common numerical factors other than 1.

$a^2x^3 + 2abx^2 + b^2x = x(a^2x^2 + 2abx + b^2)$.

Now, factorise the trinomial inside the parenthesis, $a^2x^2 + 2abx + b^2$. From part (vii), we know this is a perfect square trinomial, $(ax+b)^2$.

So, $a^2x^3 + 2abx^2 + b^2x = x(ax + b)^2$.

(xv) Factorise $4x^4 + 12x^3 + 9x^2$.

First, factor out the greatest common factor (GCF). The GCF of the numerical coefficients (4, 12, 9) is 1. The GCF of the variable parts ($x^4, x^3, x^2$) is $x^2$. So, the GCF of the expression is $x^2$.

$4x^4 + 12x^3 + 9x^2 = x^2(4x^2 + 12x + 9)$.

Now, factorise the trinomial inside the parenthesis, $4x^2 + 12x + 9$. From part (ix), we know this is a perfect square trinomial, $(2x+3)^2$.

So, $4x^4 + 12x^3 + 9x^2 = x^2(2x + 3)^2$.

(xvi) Factorise $\frac{x^2}{4}$ + 2x + 4.

The first term $\frac{x^2}{4} = \left(\frac{x}{2}\right)^2$ is the square of $\frac{x}{2}$. The last term $4$ is the square of $2$.

Let $A = \frac{x}{2}$ and $B = 2$. Check the middle term: $2AB = 2\left(\frac{x}{2}\right)(2) = 2x$.

This matches the given expression $\frac{x^2}{4}$ + 2x + 4.

So, $\frac{x^2}{4}$ + 2x + 4 = $\left(\frac{x}{2} + 2\right)^2$.

(xvii) Factorise $9x^2 + 2xy + \frac{y^2}{9}$.

The first term $9x^2$ is the square of $3x$. The last term $\frac{y^2}{9} = \left(\frac{y}{3}\right)^2$ is the square of $\frac{y}{3}$.

Let $A = 3x$ and $B = \frac{y}{3}$. Check the middle term: $2AB = 2(3x)\left(\frac{y}{3}\right) = 2\left(\cancel{3}x\right)\left(\frac{y}{\cancel{3}}\right) = 2xy$.

This matches the given expression $9x^2 + 2xy + \frac{y^2}{9}$.

So, $9x^2 + 2xy + \frac{y^2}{9} = \left(3x + \frac{y}{3}\right)^2$.

We use the identity $A^2 - 2AB + B^2 = (A - B)^2$ for factorisation.

(i) Factorise $x^2 – 8x + 16$.

$x^2 = (x)^2$ and $16 = (4)^2$. Let $A=x$ and $B=4$.

Check the middle term: $-2AB = -2(x)(4) = -8x$. This matches the expression.

So, $x^2 – 8x + 16 = (x - 4)^2$.

(ii) Factorise $x^2 – 10x + 25$.

$x^2 = (x)^2$ and $25 = (5)^2$. Let $A=x$ and $B=5$.

Check the middle term: $-2AB = -2(x)(5) = -10x$. This matches the expression.

So, $x^2 – 10x + 25 = (x - 5)^2$.

(iii) Factorise $y^2 – 14y + 49$.

$y^2 = (y)^2$ and $49 = (7)^2$. Let $A=y$ and $B=7$.

Check the middle term: $-2AB = -2(y)(7) = -14y$. This matches the expression.

So, $y^2 – 14y + 49 = (y - 7)^2$.

(iv) Factorise $p^2 – 2p + 1$.

$p^2 = (p)^2$ and $1 = (1)^2$. Let $A=p$ and $B=1$.

Check the middle term: $-2AB = -2(p)(1) = -2p$. This matches the expression.

So, $p^2 – 2p + 1 = (p - 1)^2$.

(v) Factorise $4a^2 – 4ab + b^2$.

$4a^2 = (2a)^2$ and $b^2 = (b)^2$. Let $A=2a$ and $B=b$.

Check the middle term: $-2AB = -2(2a)(b) = -4ab$. This matches the expression.

So, $4a^2 – 4ab + b^2 = (2a - b)^2$.

(vi) Factorise $p^2y^2 – 2py + 1$.

$p^2y^2 = (py)^2$ and $1 = (1)^2$. Let $A=py$ and $B=1$.

Check the middle term: $-2AB = -2(py)(1) = -2py$. This matches the expression.

So, $p^2y^2 – 2py + 1 = (py - 1)^2$.

(vii) Factorise $a^2y^2 – 2aby + b^2$.

$a^2y^2 = (ay)^2$ and $b^2 = (b)^2$. Let $A=ay$ and $B=b$.

Check the middle term: $-2AB = -2(ay)(b) = -2aby$. This matches the expression.

So, $a^2y^2 – 2aby + b^2 = (ay - b)^2$.

(viii) Factorise $9x^2 – 12x + 4$.

$9x^2 = (3x)^2$ and $4 = (2)^2$. Let $A=3x$ and $B=2$.

Check the middle term: $-2AB = -2(3x)(2) = -12x$. This matches the expression.

So, $9x^2 – 12x + 4 = (3x - 2)^2$.

(ix) Factorise $4y^2 – 12y + 9$.

$4y^2 = (2y)^2$ and $9 = (3)^2$. Let $A=2y$ and $B=3$.

Check the middle term: $-2AB = -2(2y)(3) = -12y$. This matches the expression.

So, $4y^2 – 12y + 9 = (2y - 3)^2$.

(x) Factorise $\frac{x^2}{4}$ - 2x + 4.

$\frac{x^2}{4} = \left(\frac{x}{2}\right)^2$ and $4 = (2)^2$. Let $A=\frac{x}{2}$ and $B=2$.

Check the middle term: $-2AB = -2\left(\frac{x}{2}\right)(2) = -2x$. This matches the expression.

So, $\frac{x^2}{4}$ - 2x + 4 = $\left(\frac{x}{2} - 2\right)^2$.

(xi) Factorise $a^2y^3 – 2aby^2 + b^2y$.

First, factor out the common factor $y$ from all terms:

$y(a^2y^2 – 2aby + b^2)$.

Now, factorise the trinomial $a^2y^2 – 2aby + b^2$.

$a^2y^2 = (ay)^2$ and $b^2 = (b)^2$. Let $A=ay$ and $B=b$.

Check the middle term: $-2AB = -2(ay)(b) = -2aby$. This matches the trinomial.

So, $a^2y^2 – 2aby + b^2 = (ay - b)^2$.

Therefore, the factorization of the original expression is $y(ay - b)^2$.

(xii) Factorise $9y^2 – 4xy + \frac{4x^2}{9}$.

$9y^2 = (3y)^2$ and $\frac{4x^2}{9} = \left(\frac{2x}{3}\right)^2$. Let $A=3y$ and $B=\frac{2x}{3}$.

Check the middle term: $-2AB = -2(3y)\left(\frac{2x}{3}\right) = -2\left(\cancel{3}y\right)\left(\frac{2x}{\cancel{3}}\right) = -4xy$. This matches the expression.

So, $9y^2 – 4xy + \frac{4x^2}{9} = \left(3y - \frac{2x}{3}\right)^2$.

To factorise quadratic trinomials of the form $z^2 + bz + c$, we find two numbers $m$ and $n$ such that $m \times n = c$ and $m + n = b$. The factorisation is $(z+m)(z+n)$.

(i) Factorise $x^2 + 15x + 26$.

We need two numbers that multiply to 26 and add up to 15.

Factors of 26 are 1 and 26, 2 and 13.

Sum of 1 and 26 is 27.

Sum of 2 and 13 is 15.

The numbers are 2 and 13.

$x^2 + 15x + 26 = (x + 2)(x + 13)$.

(ii) Factorise $x^2 + 9x + 20$.

We need two numbers that multiply to 20 and add up to 9.

Factors of 20 are 1 and 20, 2 and 10, 4 and 5.

Sum of 1 and 20 is 21.

Sum of 2 and 10 is 12.

Sum of 4 and 5 is 9.

The numbers are 4 and 5.

$x^2 + 9x + 20 = (x + 4)(x + 5)$.

(iii) Factorise $y^2 + 18x + 65$. Assuming the question has a typo and meant $y^2 + 18y + 65$.

Factorise $y^2 + 18y + 65$.

We need two numbers that multiply to 65 and add up to 18.

Factors of 65 are 1 and 65, 5 and 13.

Sum of 1 and 65 is 66.

Sum of 5 and 13 is 18.

The numbers are 5 and 13.

$y^2 + 18y + 65 = (y + 5)(y + 13)$.

(iv) Factorise $p^2 + 14p + 13$.

We need two numbers that multiply to 13 and add up to 14.

Factors of 13 are 1 and 13.

Sum of 1 and 13 is 14.

The numbers are 1 and 13.

$p^2 + 14p + 13 = (p + 1)(p + 13)$.

(v) Factorise $y^2 + 4y – 21$.

We need two numbers that multiply to -21 and add up to 4.

Since the product is negative, one number is positive and the other is negative. Since the sum is positive, the positive number has a larger absolute value.

Factors of 21 are (1, 21), (3, 7).

Possible pairs with product -21 are (-1, 21), (1, -21), (-3, 7), (3, -7).

Sums are 20, -20, 4, -4.

The numbers are -3 and 7.

$y^2 + 4y – 21 = (y - 3)(y + 7)$.

(vi) Factorise $y^2 – 2y – 15$.

We need two numbers that multiply to -15 and add up to -2.

Since the product is negative, one number is positive and the other is negative. Since the sum is negative, the negative number has a larger absolute value.

Factors of 15 are (1, 15), (3, 5).

Possible pairs with product -15 are (-1, 15), (1, -15), (-3, 5), (3, -5).

Sums are 14, -14, 2, -2.

The numbers are 3 and -5.

$y^2 – 2y – 15 = (y + 3)(y - 5)$.

(vii) Factorise $18 + 11x + x^2$.

Rearrange in standard quadratic form: $x^2 + 11x + 18$.

We need two numbers that multiply to 18 and add up to 11.

Factors of 18 are (1, 18), (2, 9), (3, 6).

Sums are 19, 11, 9.

The numbers are 2 and 9.

$x^2 + 11x + 18 = (x + 2)(x + 9)$.

(viii) Factorise $x^2 – 10x + 21$.

We need two numbers that multiply to 21 and add up to -10.

Since the product is positive and the sum is negative, both numbers are negative.

Negative factors of 21 are (-1, -21), (-3, -7).

Sums are -22, -10.

The numbers are -3 and -7.

$x^2 – 10x + 21 = (x - 3)(x - 7)$.

(ix) Factorise $x^2 - 17x + 60$.

We need two numbers that multiply to 60 and add up to -17.

Since the product is positive and the sum is negative, both numbers are negative.

Negative factors of 60: (-1, -60), (-2, -30), (-3, -20), (-4, -15), (-5, -12), (-6, -10).

Sums are -61, -32, -23, -19, -17, -16.

The numbers are -5 and -12.

$x^2 - 17x + 60 = (x - 5)(x - 12)$.

(x) Factorise $x^2 + 4x – 77$.

We need two numbers that multiply to -77 and add up to 4.

Since the product is negative, one positive and one negative. Since the sum is positive, the positive number has a larger absolute value.

Factors of 77 are (1, 77), (7, 11).

Possible pairs are (-1, 77), (1, -77), (-7, 11), (7, -11).

Sums are 76, -76, 4, -4.

The numbers are -7 and 11.

$x^2 + 4x – 77 = (x - 7)(x + 11)$.

(xi) Factorise $y^2 + 7y + 12$.

We need two numbers that multiply to 12 and add up to 7.

Factors of 12 are (1, 12), (2, 6), (3, 4).

Sums are 13, 8, 7.

The numbers are 3 and 4.

$y^2 + 7y + 12 = (y + 3)(y + 4)$.

(xii) Factorise $p^2 – 13p – 30$.

We need two numbers that multiply to -30 and add up to -13.

Since the product is negative, one positive and one negative. Since the sum is negative, the negative number has a larger absolute value.

Factors of 30 are (1, 30), (2, 15), (3, 10), (5, 6).

Possible pairs are (-1, 30), (1, -30), (-2, 15), (2, -15), (-3, 10), (3, -10), (-5, 6), (5, -6).

Sums are 29, -29, 13, -13, 7, -7, 1, -1.

The numbers are 2 and -15.

$p^2 – 13p – 30 = (p + 2)(p - 15)$.

(xiii) Factorise $a^2 – 16p – 80$. Assuming the question has a typo and meant $a^2 - 16a - 80$.

Factorise $a^2 - 16a - 80$.

We need two numbers that multiply to -80 and add up to -16.

Since the product is negative, one positive and one negative. Since the sum is negative, the negative number has a larger absolute value.

Factors of 80 are (1, 80), (2, 40), (4, 20), (5, 16), (8, 10).

Possible pairs are (-1, 80), (1, -80), (-2, 40), (2, -40), (-4, 20), (4, -20), (-5, 16), (5, -16), (-8, 10), (8, -10).

Sums are 79, -79, 38, -38, 16, -16, 11, -11, 2, -2.

The numbers are 4 and -20.

$a^2 - 16a - 80 = (a + 4)(a - 20)$.

We use the identity $A^2 - B^2 = (A + B)(A - B)$ for factorisation.

(i) Factorise $x^2 – 9$.

$x^2 - 9 = x^2 - 3^2$

Using the identity with $A=x$ and $B=3$:

$x^2 - 3^2 = (x + 3)(x - 3)$.

(ii) Factorise $4x^2 – 25y^2$.

$4x^2 - 25y^2 = (2x)^2 - (5y)^2$

Using the identity with $A=2x$ and $B=5y$:

$(2x)^2 - (5y)^2 = (2x + 5y)(2x - 5y)$.

(iii) Factorise $4x^2 – 49y^2$.

$4x^2 - 49y^2 = (2x)^2 - (7y)^2$

Using the identity with $A=2x$ and $B=7y$:

$(2x)^2 - (7y)^2 = (2x + 7y)(2x - 7y)$.

(iv) Factorise $3a^2b^3 – 27a^4b$.

First, factor out the GCF of the terms $3a^2b^3$ and $27a^4b$.

GCF(3, 27) = 3.

GCF($a^2$, $a^4$) = $a^2$.

GCF($b^3$, $b$) = $b$.

GCF = $3a^2b$.

$3a^2b^3 – 27a^4b = 3a^2b(b^2 - 9a^2)$

Now, factorise the expression in the parenthesis using the identity:

$b^2 - 9a^2 = b^2 - (3a)^2$

Using the identity with $A=b$ and $B=3a$:

$b^2 - (3a)^2 = (b + 3a)(b - 3a)$.

So, $3a^2b^3 – 27a^4b = 3a^2b(b + 3a)(b - 3a)$.

(v) Factorise $28ay^2 – 175ax^2$.

First, factor out the GCF of the terms $28ay^2$ and $175ax^2$.

GCF(28, 175). $28 = 4 \times 7$. $175 = 7 \times 25$. GCF(28, 175) = 7.

GCF($a$, $a$) = $a$.

GCF($y^2$, $x^2$) = 1 (no common variables other than powers of 1).

GCF = $7a$.

$28ay^2 – 175ax^2 = 7a(4y^2 - 25x^2)$

Now, factorise the expression in the parenthesis using the identity:

$4y^2 - 25x^2 = (2y)^2 - (5x)^2$

Using the identity with $A=2y$ and $B=5x$:

$(2y)^2 - (5x)^2 = (2y + 5x)(2y - 5x)$.

So, $28ay^2 – 175ax^2 = 7a(2y + 5x)(2y - 5x)$.

(vi) Factorise $9x^2 – 1$.

$9x^2 - 1 = (3x)^2 - 1^2$

Using the identity with $A=3x$ and $B=1$:

$(3x)^2 - 1^2 = (3x + 1)(3x - 1)$.

(vii) Factorise $25ax^2 – 25a$.

First, factor out the GCF of the terms $25ax^2$ and $25a$.

GCF(25, 25) = 25.

GCF($a$, $a$) = $a$.

GCF($x^2$, 1) = 1.

GCF = $25a$.

$25ax^2 – 25a = 25a(x^2 - 1)$

Now, factorise the expression in the parenthesis using the identity:

$x^2 - 1 = x^2 - 1^2$

Using the identity with $A=x$ and $B=1$:

$x^2 - 1^2 = (x + 1)(x - 1)$.

So, $25ax^2 – 25a = 25a(x + 1)(x - 1)$.

(viii) Factorise $\frac{x^2}{9}$ - $\frac{y^2}{25}$.

$\frac{x^2}{9} - \frac{y^2}{25} = \left(\frac{x}{3}\right)^2 - \left(\frac{y}{5}\right)^2$

Using the identity with $A=\frac{x}{3}$ and $B=\frac{y}{5}$:

$\left(\frac{x}{3}\right)^2 - \left(\frac{y}{5}\right)^2 = \left(\frac{x}{3} + \frac{y}{5}\right)\left(\frac{x}{3} - \frac{y}{5}\right)$.

(ix) Factorise $\frac{2p^2}{25}$ - 32q².

First, factor out the GCF of the numerical coefficients $\frac{2}{25}$ and $-32$. The GCF of the numerators (2, 32) is 2. The denominators are 25 and 1. GCF is 2.

$\frac{2p^2}{25} - 32q^2 = 2\left(\frac{p^2}{25} - 16q^2\right)$

Now, factorise the expression in the parenthesis using the identity:

$\frac{p^2}{25} - 16q^2 = \left(\frac{p}{5}\right)^2 - (4q)^2$

Using the identity with $A=\frac{p}{5}$ and $B=4q$:

$\left(\frac{p}{5}\right)^2 - (4q)^2 = \left(\frac{p}{5} + 4q\right)\left(\frac{p}{5} - 4q\right)$.

So, $\frac{2p^2}{25}$ - 32q² = $2\left(\frac{p}{5} + 4q\right)\left(\frac{p}{5} - 4q\right)$.

(x) Factorise $49x^2 – 36y^2$.

$49x^2 - 36y^2 = (7x)^2 - (6y)^2$

Using the identity with $A=7x$ and $B=6y$:

$(7x)^2 - (6y)^2 = (7x + 6y)(7x - 6y)$.

(xi) Factorise $y^3 - \frac{y}{9}$.

First, factor out the common factor $y$:

$y^3 - \frac{y}{9} = y\left(y^2 - \frac{1}{9}\right)$

Now, factorise the expression in the parenthesis using the identity:

$y^2 - \frac{1}{9} = y^2 - \left(\frac{1}{3}\right)^2$

Using the identity with $A=y$ and $B=\frac{1}{3}$:

$y^2 - \left(\frac{1}{3}\right)^2 = \left(y + \frac{1}{3}\right)\left(y - \frac{1}{3}\right)$.

So, $y^3 - \frac{y}{9} = y\left(y + \frac{1}{3}\right)\left(y - \frac{1}{3}\right)$.

(xii) Factorise $\frac{x^2}{25}$ - 625.

$\frac{x^2}{25} - 625 = \left(\frac{x}{5}\right)^2 - (25)^2$

Using the identity with $A=\frac{x}{5}$ and $B=25$:

$\left(\frac{x}{5}\right)^2 - (25)^2 = \left(\frac{x}{5} + 25\right)\left(\frac{x}{5} - 25\right)$.

(xiii) Factorise $\frac{x^2}{8}$ - $\frac{y^2}{18}$.

First, factor out a common numerical factor to make the terms perfect squares. Notice $8 = 2 \times 4$ and $18 = 2 \times 9$. We can factor out $\frac{1}{2}$.

$\frac{x^2}{8} - \frac{y^2}{18} = \frac{1}{2}\left(\frac{x^2}{4} - \frac{y^2}{9}\right)$

Now, factorise the expression in the parenthesis using the identity:

$\frac{x^2}{4} - \frac{y^2}{9} = \left(\frac{x}{2}\right)^2 - \left(\frac{y}{3}\right)^2$

Using the identity with $A=\frac{x}{2}$ and $B=\frac{y}{3}$:

$\left(\frac{x}{2}\right)^2 - \left(\frac{y}{3}\right)^2 = \left(\frac{x}{2} + \frac{y}{3}\right)\left(\frac{x}{2} - \frac{y}{3}\right)$.

So, $\frac{x^2}{8}$ - $\frac{y^2}{18}$ = $\frac{1}{2}\left(\frac{x}{2} + \frac{y}{3}\right)\left(\frac{x}{2} - \frac{y}{3}\right)$.

(xiv) Factorise $\frac{4x^2}{9}$ - $\frac{9y^2}{16}$.

$\frac{4x^2}{9} - \frac{9y^2}{16} = \left(\frac{2x}{3}\right)^2 - \left(\frac{3y}{4}\right)^2$

Using the identity with $A=\frac{2x}{3}$ and $B=\frac{3y}{4}$:

$\left(\frac{2x}{3}\right)^2 - \left(\frac{3y}{4}\right)^2 = \left(\frac{2x}{3} + \frac{3y}{4}\right)\left(\frac{2x}{3} - \frac{3y}{4}\right)$.

(xv) Factorise $\frac{x^3y}{9}$ - $\frac{xy^3}{16}$.

First, factor out the GCF of the terms $\frac{x^3y}{9}$ and $\frac{xy^3}{16}$. The common variable factor is $xy$.

$\frac{x^3y}{9}$ - $\frac{xy^3}{16} = xy\left(\frac{x^2}{9} - \frac{y^2}{16}\right)$

Now, factorise the expression in the parenthesis using the identity:

$\frac{x^2}{9} - \frac{y^2}{16} = \left(\frac{x}{3}\right)^2 - \left(\frac{y}{4}\right)^2$

Using the identity with $A=\frac{x}{3}$ and $B=\frac{y}{4}$:

$\left(\frac{x}{3}\right)^2 - \left(\frac{y}{4}\right)^2 = \left(\frac{x}{3} + \frac{y}{4}\right)\left(\frac{x}{3} - \frac{y}{4}\right)$.

So, $\frac{x^3y}{9}$ - $\frac{xy^3}{16}$ = $xy\left(\frac{x}{3} + \frac{y}{4}\right)\left(\frac{x}{3} - \frac{y}{4}\right)$.

(xvi) Factorise $1331x^3y – 11y^3x$.

First, factor out the GCF of the terms $1331x^3y$ and $11y^3x$.

GCF(1331, 11). $1331 = 11^3$. GCF is 11.

GCF($x^3$, $x$) = $x$.

GCF($y$, $y^3$) = $y$.

GCF = $11xy$.

$1331x^3y – 11y^3x = 11xy(121x^2 - y^2)$

Now, factorise the expression in the parenthesis using the identity:

$121x^2 - y^2 = (11x)^2 - y^2$

Using the identity with $A=11x$ and $B=y$:

$(11x)^2 - y^2 = (11x + y)(11x - y)$.

So, $1331x^3y – 11y^3x = 11xy(11x + y)(11x - y)$.

(xvii) Factorise $\frac{1}{36}$ a²b² - $\frac{16}{49}$ b²c².

First, factor out the GCF of the terms. The common variable factor is $b^2$.

$\frac{1}{36} a^2b^2 - \frac{16}{49} b^2c^2 = b^2\left(\frac{1}{36}a^2 - \frac{16}{49}c^2\right)$

Now, factorise the expression in the parenthesis using the identity:

$\frac{1}{36}a^2 - \frac{16}{49}c^2 = \left(\frac{1}{6}a\right)^2 - \left(\frac{4}{7}c\right)^2$

Using the identity with $A=\frac{1}{6}a$ and $B=\frac{4}{7}c$:

$\left(\frac{1}{6}a\right)^2 - \left(\frac{4}{7}c\right)^2 = \left(\frac{1}{6}a + \frac{4}{7}c\right)\left(\frac{1}{6}a - \frac{4}{7}c\right)$.

So, $\frac{1}{36}$ a²b² - $\frac{16}{49}$ b²c² = $b^2\left(\frac{1}{6}a + \frac{4}{7}c\right)\left(\frac{1}{6}a - \frac{4}{7}c\right)$.

(xviii) Factorise $a^4 – (a – b)^4$.

Write the expression as a difference of squares:

$a^4 – (a – b)^4 = (a^2)^2 - ((a - b)^2)^2$

Using the identity with $A=a^2$ and $B=(a-b)^2$:

$(a^2)^2 - ((a - b)^2)^2 = (a^2 - (a - b)^2)(a^2 + (a - b)^2)$

Now, factorise each factor separately.

The first factor is a difference of squares again: $a^2 - (a - b)^2$.

Using the identity with $A=a$ and $B=(a-b)$:

$a^2 - (a - b)^2 = (a - (a - b))(a + (a - b))$

$= (a - a + b)(a + a - b)$

$= (b)(2a - b)$

The second factor is a sum of squares: $a^2 + (a - b)^2$. Expand the square:

$a^2 + (a^2 - 2ab + b^2) = a^2 + a^2 - 2ab + b^2 = 2a^2 - 2ab + b^2$. This does not factor further over real numbers.

So, $a^4 – (a – b)^4 = (b)(2a - b)(a^2 + (a - b)^2) = b(2a - b)(2a^2 - 2ab + b^2)$.

(xix) Factorise $x^4 – 1$.

$x^4 - 1 = (x^2)^2 - 1^2$

Using the identity with $A=x^2$ and $B=1$:

$(x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)$.

The first factor is a difference of squares again: $x^2 - 1 = x^2 - 1^2$.

Using the identity with $A=x$ and $B=1$:

$x^2 - 1^2 = (x + 1)(x - 1)$.

So, $x^4 – 1 = (x + 1)(x - 1)(x^2 + 1)$.

(xx) Factorise $y^4 – 625$.

$y^4 - 625 = (y^2)^2 - 25^2$

Using the identity with $A=y^2$ and $B=25$:

$(y^2)^2 - 25^2 = (y^2 - 25)(y^2 + 25)$.

The first factor is a difference of squares again: $y^2 - 25 = y^2 - 5^2$.

Using the identity with $A=y$ and $B=5$:

$y^2 - 5^2 = (y + 5)(y - 5)$.

So, $y^4 – 625 = (y + 5)(y - 5)(y^2 + 25)$.

(xxi) Factorise $p^5 – 16p$.

First, factor out the GCF, which is $p$:

$p^5 - 16p = p(p^4 - 16)$

Now, factorise the expression in the parenthesis using the identity:

$p^4 - 16 = (p^2)^2 - 4^2$

Using the identity with $A=p^2$ and $B=4$:

$(p^2)^2 - 4^2 = (p^2 - 4)(p^2 + 4)$.

The first factor is a difference of squares again: $p^2 - 4 = p^2 - 2^2$.

Using the identity with $A=p$ and $B=2$:

$p^2 - 2^2 = (p + 2)(p - 2)$.

So, $p^5 – 16p = p(p + 2)(p - 2)(p^2 + 4)$.

(xxii) Factorise $16x^4 – 81$.

$16x^4 - 81 = (4x^2)^2 - 9^2$

Using the identity with $A=4x^2$ and $B=9$:

$(4x^2)^2 - 9^2 = (4x^2 - 9)(4x^2 + 9)$.

The first factor is a difference of squares again: $4x^2 - 9 = (2x)^2 - 3^2$.

Using the identity with $A=2x$ and $B=3$:

$(2x)^2 - 3^2 = (2x + 3)(2x - 3)$.

So, $16x^4 – 81 = (2x + 3)(2x - 3)(4x^2 + 9)$.

(xxiii) Factorise $x^4 – y^4$.

$x^4 – y^4 = (x^2)^2 - (y^2)^2$

Using the identity with $A=x^2$ and $B=y^2$:

$(x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)$.

The first factor is a difference of squares again: $x^2 - y^2$.

Using the identity with $A=x$ and $B=y$:

$x^2 - y^2 = (x + y)(x - y)$.

So, $x^4 – y^4 = (x + y)(x - y)(x^2 + y^2)$.

(xxiv) Factorise $y^4 – 81$.

$y^4 - 81 = (y^2)^2 - 9^2$

Using the identity with $A=y^2$ and $B=9$:

$(y^2)^2 - 9^2 = (y^2 - 9)(y^2 + 9)$.

The first factor is a difference of squares again: $y^2 - 9 = y^2 - 3^2$.

Using the identity with $A=y$ and $B=3$:

$y^2 - 3^2 = (y + 3)(y - 3)$.

So, $y^4 – 81 = (y + 3)(y - 3)(y^2 + 9)$.

(xxv) Factorise $16x^4 – 625y^4$.

$16x^4 – 625y^4 = (4x^2)^2 - (25y^2)^2$

Using the identity with $A=4x^2$ and $B=25y^2$:

$(4x^2)^2 - (25y^2)^2 = (4x^2 - 25y^2)(4x^2 + 25y^2)$.

The first factor is a difference of squares again: $4x^2 - 25y^2 = (2x)^2 - (5y)^2$.

Using the identity with $A=2x$ and $B=5y$:

$(2x)^2 - (5y)^2 = (2x + 5y)(2x - 5y)$.

So, $16x^4 – 625y^4 = (2x + 5y)(2x - 5y)(4x^2 + 25y^2)$.

(xxvi) Factorise $(a – b)^2 – (b – c)^2$.

This is in the form $A^2 - B^2$, where $A = (a - b)$ and $B = (b - c)$.

Using the identity $(A - B)(A + B)$:

$(a – b)^2 – (b – c)^2 = ((a - b) - (b - c))((a - b) + (b - c))$

$= (a - b - b + c)(a - b + b - c)$

$= (a - 2b + c)(a - c)$.

(xxvii) Factorise $(x + y)^4 – (x – y)^4$.

Write the expression as a difference of squares:

$(x + y)^4 – (x – y)^4 = ((x + y)^2)^2 - ((x - y)^2)^2$

Using the identity with $A=(x+y)^2$ and $B=(x-y)^2$:

$((x + y)^2 - (x - y)^2)((x + y)^2 + (x - y)^2)$

Factorise the first factor using the identity $A^2 - B^2 = (A-B)(A+B)$, or specifically $(A+B)^2 - (A-B)^2 = 4AB$.

$(x + y)^2 - (x - y)^2 = 4(x)(y) = 4xy$.

Factorise the second factor using the identity $(A+B)^2 + (A-B)^2 = 2(A^2 + B^2)$.

$(x + y)^2 + (x - y)^2 = 2(x^2 + y^2)$.

So, $(x + y)^4 – (x – y)^4 = (4xy)(2(x^2 + y^2))$

$= 8xy(x^2 + y^2)$.

(xxviii) Factorise $x^4 – y^4 + x^2 – y^2$.

Group the terms:

$(x^4 – y^4) + (x^2 – y^2)$

Factorise the first group as a difference of squares:

$x^4 – y^4 = (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)$.

Substitute this back into the expression:

$(x^2 - y^2)(x^2 + y^2) + (x^2 - y^2)$

Notice the common binomial factor $(x^2 - y^2)$. Factor it out:

$(x^2 - y^2)((x^2 + y^2) + 1)$

Factorise the first factor as a difference of squares again:

$x^2 - y^2 = (x - y)(x + y)$.

So, $x^4 – y^4 + x^2 – y^2 = (x - y)(x + y)(x^2 + y^2 + 1)$.

(xxix) Factorise $8a^3 – 2a$.

First, factor out the GCF of the terms $8a^3$ and $2a$.

GCF(8, 2) = 2.

GCF($a^3$, $a$) = $a$.

GCF = $2a$.

$8a^3 – 2a = 2a(4a^2 - 1)$

Now, factorise the expression in the parenthesis using the identity:

$4a^2 - 1 = (2a)^2 - 1^2$

Using the identity with $A=2a$ and $B=1$:

$(2a)^2 - 1^2 = (2a + 1)(2a - 1)$.

So, $8a^3 – 2a = 2a(2a + 1)(2a - 1)$.

(xxx) Factorise $x^2 - \frac{y^2}{100}$.

$x^2 - \frac{y^2}{100} = x^2 - \left(\frac{y}{10}\right)^2$

Using the identity with $A=x$ and $B=\frac{y}{10}$:

$x^2 - \left(\frac{y}{10}\right)^2 = \left(x + \frac{y}{10}\right)\left(x - \frac{y}{10}\right)$.

(xxxi) Factorise $9x^2 – (3y + z)^2$.

Write the expression as a difference of squares:

$9x^2 – (3y + z)^2 = (3x)^2 - (3y + z)^2$

Using the identity with $A=3x$ and $B=(3y + z)$:

$(3x)^2 - (3y + z)^2 = ((3x) - (3y + z))((3x) + (3y + z))$

$= (3x - 3y - z)(3x + 3y + z)$.

The area of a rectangle is given by the product of its length and breadth. We factorise the given quadratic expressions to find the possible factors, which represent the length and breadth.

For a quadratic trinomial of the form $x^2 + bx + c$, we look for two numbers $m$ and $n$ such that $m \times n = c$ and $m + n = b$. The factorised form is $(x+m)(x+n)$.

(i) Area = $x^2 – 6x + 8$.

We need two numbers that multiply to 8 and add up to -6.

The numbers are -2 and -4, since $(-2) \times (-4) = 8$ and $(-2) + (-4) = -6$.

So, $x^2 – 6x + 8 = (x - 2)(x - 4)$.

Possible lengths and breadths are $(x - 2)$ and $(x - 4)$.

(ii) Area = $x^2 – 3x + 2$.

We need two numbers that multiply to 2 and add up to -3.

The numbers are -1 and -2, since $(-1) \times (-2) = 2$ and $(-1) + (-2) = -3$.

So, $x^2 – 3x + 2 = (x - 1)(x - 2)$.

Possible lengths and breadths are $(x - 1)$ and $(x - 2)$.

(iii) Area = $x^2 – 7x + 10$.

We need two numbers that multiply to 10 and add up to -7.

The numbers are -2 and -5, since $(-2) \times (-5) = 10$ and $(-2) + (-5) = -7$.

So, $x^2 – 7x + 10 = (x - 2)(x - 5)$.

Possible lengths and breadths are $(x - 2)$ and $(x - 5)$.

(iv) Area = $x^2 + 19x – 20$.

We need two numbers that multiply to -20 and add up to 19.

The numbers are -1 and 20, since $(-1) \times 20 = -20$ and $(-1) + 20 = 19$.

So, $x^2 + 19x – 20 = (x - 1)(x + 20)$.

Possible lengths and breadths are $(x - 1)$ and $(x + 20)$.

(v) Area = $x^2 + 9x + 20$.

We need two numbers that multiply to 20 and add up to 9.

The numbers are 4 and 5, since $4 \times 5 = 20$ and $4 + 5 = 9$.

So, $x^2 + 9x + 20 = (x + 4)(x + 5)$.

Possible lengths and breadths are $(x + 4)$ and $(x + 5)$.

(i) Divide $51x^3y^2z$ by $17xyz$.

$\frac{51x^3y^2z}{17xyz} = \frac{51}{17} \times \frac{x^3}{x^1} \times \frac{y^2}{y^1} \times \frac{z^1}{z^1}$

$= 3 \times x^{3-1} \times y^{2-1} \times z^{1-1}$

$= 3 \times x^2 \times y^1 \times z^0$

Since $z^0 = 1$, the result is $3x^2y$.

(ii) Divide $76x^3yz^3$ by $19x^2y^2$.

$\frac{76x^3yz^3}{19x^2y^2} = \frac{76}{19} \times \frac{x^3}{x^2} \times \frac{y^1}{y^2} \times \frac{z^3}{z^0}$

$= 4 \times x^{3-2} \times y^{1-2} \times z^{3-0}$

$= 4 \times x^1 \times y^{-1} \times z^3$

$= 4xz^3y^{-1} = \frac{4xz^3}{y}$.

(iii) Divide $17ab^2c^3$ by $(–abc^2)$.

$\frac{17ab^2c^3}{-abc^2} = \frac{17}{-1} \times \frac{a^1}{a^1} \times \frac{b^2}{b^1} \times \frac{c^3}{c^2}$

$= -17 \times a^{1-1} \times b^{2-1} \times c^{3-2}$

$= -17 \times a^0 \times b^1 \times c^1$

$= -17 \times 1 \times b \times c$

$= -17bc$.

(iv) Divide –121p³q³r³ by (–11xy²z³).

$\frac{-121p^3q^3r^3}{-11xy^2z^3} = \frac{-121}{-11} \times \frac{p^3}{x^1} \times \frac{q^3}{y^2} \times \frac{r^3}{z^3}$

$= 11 \times \frac{p^3}{x} \times \frac{q^3}{y^2} \times \frac{r^3}{z^3}$

Note that the variables in the numerator ($p, q, r$) are different from the variables in the denominator ($x, y, z$). They cannot be simplified by subtracting exponents.

The result is $\frac{11p^3q^3r^3}{xy^2z^3}$.

To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial.

(i) Divide $(3pqr – 6p^2q^2r^2)$ by $3pq$.

$\frac{3pqr – 6p^2q^2r^2}{3pq} = \frac{3pqr}{3pq} - \frac{6p^2q^2r^2}{3pq}$

Divide each term separately:

$\frac{3pqr}{3pq} = \frac{\cancel{3pq}r}{\cancel{3pq}} = r$

$\frac{6p^2q^2r^2}{3pq} = \frac{6}{3} \times \frac{p^2}{p} \times \frac{q^2}{q} \times \frac{r^2}{1} = 2 \times p^{2-1} \times q^{2-1} \times r^2 = 2pr^2$

So, $(3pqr – 6p^2q^2r^2) \div 3pq = r - 2pr^2$.

(ii) Divide $(ax^3 – bx^2 + cx)$ by $(– dx)$.

$\frac{ax^3 – bx^2 + cx}{-dx} = \frac{ax^3}{-dx} - \frac{bx^2}{-dx} + \frac{cx}{-dx}$

Divide each term separately:

$\frac{ax^3}{-dx} = \frac{a}{-d} \times \frac{x^3}{x} = -\frac{a}{d} x^{3-1} = -\frac{a}{d} x^2$

$-\frac{bx^2}{-dx} = \left(-\frac{b}{-d}\right) \times \frac{x^2}{x} = \frac{b}{d} x^{2-1} = \frac{b}{d} x$

$\frac{cx}{-dx} = \frac{c}{-d} \times \frac{x}{x} = -\frac{c}{d} x^{1-1} = -\frac{c}{d} x^0 = -\frac{c}{d}$

So, $(ax^3 – bx^2 + cx) \div (– dx) = -\frac{a}{d}x^2 + \frac{b}{d}x - \frac{c}{d}$.

(iii) Divide $(x^3y^3 + x^2y^3 – xy^4 + xy)$ by $xy$.

$\frac{x^3y^3 + x^2y^3 – xy^4 + xy}{xy} = \frac{x^3y^3}{xy} + \frac{x^2y^3}{xy} - \frac{xy^4}{xy} + \frac{xy}{xy}$

Divide each term separately:

$\frac{x^3y^3}{xy} = x^{3-1}y^{3-1} = x^2y^2$

$\frac{x^2y^3}{xy} = x^{2-1}y^{3-1} = xy^2$

$\frac{xy^4}{xy} = x^{1-1}y^{4-1} = x^0y^3 = y^3$

$\frac{xy}{xy} = 1$

So, $(x^3y^3 + x^2y^3 – xy^4 + xy) \div xy = x^2y^2 + xy^2 - y^3 + 1$.

(iv) Divide $(– qrxy + pryz – rxyz)$ by $(– xyz)$.

$\frac{-qrxy + pryz – rxyz}{-xyz} = \frac{-qrxy}{-xyz} + \frac{pryz}{-xyz} - \frac{rxyz}{-xyz}$

Divide each term separately:

$\frac{-qrxy}{-xyz} = \left(\frac{-1}{-1}\right) \times \left(\frac{r}{r}\right) \times \left(\frac{x}{x}\right) \times \left(\frac{y}{y}\right) \times q = 1 \times 1 \times 1 \times 1 \times q = q$ (Variable $z$ in the denominator cancels nothing in the numerator).

Let's rewrite properly:

$\frac{-qrxy}{-xyz} = \frac{\cancel{-}q\cancel{r}\cancel{x}\cancel{y}}{\cancel{-}\cancel{x}\cancel{y}z} = \frac{qr}{z}$. Oh, wait. The variables in the numerator are $q, r, x, y$. The variables in the denominator are $x, y, z$. The common variables are $x$ and $y$. Let's re-do the division carefully.

$\frac{-qrxy}{-xyz} = \frac{-1}{-1} \times q \times r \times \frac{x}{x} \times \frac{y}{y} \times \frac{1}{z} = 1 \times q \times r \times 1 \times 1 \times \frac{1}{z} = \frac{qr}{z}$.

$\frac{pryz}{-xyz} = \frac{p \times r \times y \times z}{-1 \times x \times y \times z} = \frac{1}{-1} \times p \times r \times \frac{y}{y} \times \frac{z}{z} \times \frac{1}{x} = -1 \times p \times r \times 1 \times 1 \times \frac{1}{x} = -\frac{pr}{x}$.

$-\frac{rxyz}{-xyz} = -\left(\frac{rxyz}{xyz}\right) \times \left(\frac{1}{-1}\right) = -1 \times \left(\frac{\cancel{r}\cancel{x}\cancel{y}\cancel{z}}{\cancel{x}\cancel{y}\cancel{z}}\right) \times (-1) = -1 \times r \times (-1) = r$.

So, $(– qrxy + pryz – rxyz) \div (– xyz) = \frac{qr}{z} - \frac{pr}{x} + r$.

Let's check the second term again: $\frac{pryz}{-xyz} = \frac{p \times r \times y \times z}{-1 \times x \times y \times z} = \frac{p \times r \times \cancel{y} \times \cancel{z}}{-1 \times x \times \cancel{y} \times \cancel{z}} = \frac{pr}{-x} = -\frac{pr}{x}$. This is correct.

Let's check the third term again: $-\frac{rxyz}{-xyz} = \frac{rxyz}{xyz} = r$. This is correct.

So, $(– qrxy + pryz – rxyz) \div (– xyz) = \frac{qr}{z} - \frac{pr}{x} + r$.

(i) Factorise $x^2 – 22x + 117$ and divide by $(x – 13)$.

We need two numbers that multiply to 117 and add up to -22. Since the product is positive and the sum is negative, both numbers are negative.

Factors of 117 are (1, 117), (3, 39), (9, 13).

Negative factors are (-1, -117), (-3, -39), (-9, -13).

Sums are -118, -42, -22.

The numbers are -9 and -13.

$x^2 – 22x + 117 = (x - 9)(x - 13)$.

Now, divide by $(x - 13)$:

$\frac{(x - 9)(x - 13)}{(x - 13)}$

Assuming $x \neq 13$, we can cancel the $(x - 13)$ term.

The result is $x - 9$.

(ii) Factorise $x^3 + x^2 – 132x$ and divide by $x (x – 11)$.

First, factor out the GCF from the numerator $x^3 + x^2 – 132x$. The GCF is $x$.

$x^3 + x^2 – 132x = x(x^2 + x - 132)$.

Now, factorise the trinomial $x^2 + x - 132$. We need two numbers that multiply to -132 and add up to 1. Since the product is negative, one number is positive and one is negative. Since the sum is positive, the positive number has a larger absolute value.

Factors of 132: (1, 132), (2, 66), (3, 44), (4, 33), (6, 22), (11, 12).

Look for a pair with a difference of 1. (11, 12).

The numbers are -11 and 12, since $(-11) \times 12 = -132$ and $(-11) + 12 = 1$.

$x^2 + x - 132 = (x - 11)(x + 12)$.

So, the numerator is $x(x - 11)(x + 12)$.

Now, divide by the denominator $x(x - 11)$:

$\frac{x(x - 11)(x + 12)}{x(x - 11)}$

Assuming $x \neq 0$ and $x \neq 11$, we can cancel the $x$ and $(x - 11)$ terms.

The result is $x + 12$.

(iii) Factorise $2x^3 – 12x^2 + 16x$ and divide by $(x – 2) (x – 4)$.

First, factor out the GCF from the numerator $2x^3 – 12x^2 + 16x$. The GCF of the coefficients (2, 12, 16) is 2. The GCF of the variable parts ($x^3, x^2, x$) is $x$. So, the GCF is $2x$.

$2x^3 – 12x^2 + 16x = 2x(x^2 - 6x + 8)$.

Now, factorise the trinomial $x^2 - 6x + 8$. We need two numbers that multiply to 8 and add up to -6.

The numbers are -2 and -4, since $(-2) \times (-4) = 8$ and $(-2) + (-4) = -6$.

$x^2 - 6x + 8 = (x - 2)(x - 4)$.

So, the numerator is $2x(x - 2)(x - 4)$.

Now, divide by the denominator $(x - 2)(x - 4)$:

$\frac{2x(x - 2)(x - 4)}{(x - 2)(x - 4)}$

Assuming $x \neq 2$ and $x \neq 4$, we can cancel the $(x - 2)$ and $(x - 4)$ terms.

The result is $2x$.

(iv) Factorise $9x^2 – 4$ and divide by $(3x + 2)$.

Factorise the numerator $9x^2 – 4$ using the difference of squares identity $A^2 - B^2 = (A+B)(A-B)$.

$9x^2 - 4 = (3x)^2 - 2^2$

Using the identity with $A=3x$ and $B=2$:

$(3x)^2 - 2^2 = (3x + 2)(3x - 2)$.

Now, divide by the denominator $(3x + 2)$:

$\frac{(3x + 2)(3x - 2)}{(3x + 2)}$

Assuming $3x + 2 \neq 0$, we can cancel the $(3x + 2)$ term.

The result is $3x - 2$.

(v) Factorise $3x^2 – 48$ and divide by $(x – 4)$.

First, factor out the GCF from the numerator $3x^2 – 48$. The GCF of the coefficients (3, 48) is 3.

$3x^2 – 48 = 3(x^2 - 16)$.

Now, factorise the expression in the parenthesis using the difference of squares identity:

$x^2 - 16 = x^2 - 4^2 = (x + 4)(x - 4)$.

So, the numerator is $3(x + 4)(x - 4)$.

Now, divide by the denominator $(x – 4)$:

$\frac{3(x + 4)(x - 4)}{(x - 4)}$

Assuming $x \neq 4$, we can cancel the $(x - 4)$ term.

The result is $3(x + 4) = 3x + 12$.

(vi) Factorise $x^4 – 16$ and divide by $x^3 + 2x^2 + 4x + 8$.

Factorise the numerator $x^4 – 16$ using the difference of squares identity:

$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4)$.

The first factor $x^2 - 4$ is also a difference of squares: $x^2 - 2^2 = (x - 2)(x + 2)$.

So, the numerator is $(x - 2)(x + 2)(x^2 + 4)$.

Factorise the denominator $x^3 + 2x^2 + 4x + 8$ by grouping:

$(x^3 + 2x^2) + (4x + 8)$

$x^2(x + 2) + 4(x + 2)$

$(x + 2)(x^2 + 4)$.

So, the denominator is $(x + 2)(x^2 + 4)$.

Now, perform the division:

$\frac{(x - 2)(x + 2)(x^2 + 4)}{(x + 2)(x^2 + 4)}$

Assuming $(x+2) \neq 0$ and $(x^2+4) \neq 0$, we can cancel the $(x+2)$ and $(x^2+4)$ terms.

Note that $x^2 + 4$ is never zero for real values of $x$.

The result is $x - 2$.

(vii) Factorise $(3x^4 – 1875)$ and divide by $(3x^2 – 75)$.

Factorise the numerator $3x^4 – 1875$. Factor out the GCF, which is 3.

$3x^4 – 1875 = 3(x^4 - 625)$.

Factorise $x^4 - 625$ using the difference of squares identity:

$x^4 - 625 = (x^2)^2 - 25^2 = (x^2 - 25)(x^2 + 25)$.

The first factor $x^2 - 25$ is also a difference of squares: $x^2 - 5^2 = (x - 5)(x + 5)$.

So, the numerator is $3(x - 5)(x + 5)(x^2 + 25)$.

Factorise the denominator $3x^2 – 75$. Factor out the GCF, which is 3.

$3x^2 – 75 = 3(x^2 - 25)$.

Factorise $x^2 - 25$ using the difference of squares identity:

$x^2 - 25 = x^2 - 5^2 = (x - 5)(x + 5)$.

So, the denominator is $3(x - 5)(x + 5)$.

Now, perform the division:

$\frac{3(x - 5)(x + 5)(x^2 + 25)}{3(x - 5)(x + 5)}$

Assuming $3 \neq 0$, $(x-5) \neq 0$, and $(x+5) \neq 0$, we can cancel the $3$, $(x-5)$, and $(x+5)$ terms.

The result is $x^2 + 25$.

The area of a square is given by the formula $Area = (side)^2$.

The given area is $4x^2 + 12xy + 9y^2$.

To find the side of the square, we need to find the expression whose square is $4x^2 + 12xy + 9y^2$. This means we need to factorise the given trinomial and check if it is a perfect square.

We look for the pattern of a perfect square trinomial: $A^2 + 2AB + B^2 = (A+B)^2$.

The first term of the expression is $4x^2$. This can be written as $(2x)^2$. So, let $A = 2x$.

The last term of the expression is $9y^2$. This can be written as $(3y)^2$. So, let $B = 3y$.

Now, let's check if the middle term of the expression, $12xy$, matches the middle term of the perfect square identity, $2AB$.

$2AB = 2(2x)(3y) = 2 \times 2 \times 3 \times xy = 12xy$.

Since the middle term $12xy$ matches $2AB$, the trinomial $4x^2 + 12xy + 9y^2$ is a perfect square trinomial.

$4x^2 + 12xy + 9y^2 = (2x)^2 + 2(2x)(3y) + (3y)^2 = (2x + 3y)^2$.

So, the area of the square is $(2x + 3y)^2$.

Since $Area = (side)^2$, we have $(side)^2 = (2x + 3y)^2$.

Taking the square root of both sides, the side of the square is $(2x + 3y)$.

The side of the square is $\mathbf{2x + 3y}$.

The area of a square is given by the formula $Area = (side)^2$.

The given area is $9x^2 + 24xy + 16y^2$.

To find the side of the square, we need to find the expression whose square is $9x^2 + 24xy + 16y^2$. This means we need to factorise the given trinomial and check if it is a perfect square.

We look for the pattern of a perfect square trinomial: $A^2 + 2AB + B^2 = (A+B)^2$.

The first term of the expression is $9x^2$. This can be written as $(3x)^2$. So, let $A = 3x$.

The last term of the expression is $16y^2$. This can be written as $(4y)^2$. So, let $B = 4y$.

Now, let's check if the middle term of the expression, $24xy$, matches the middle term of the perfect square identity, $2AB$.

$2AB = 2(3x)(4y) = 2 \times 3 \times 4 \times xy = 24xy$.

Since the middle term $24xy$ matches $2AB$, the trinomial $9x^2 + 24xy + 16y^2$ is a perfect square trinomial.

$9x^2 + 24xy + 16y^2 = (3x)^2 + 2(3x)(4y) + (4y)^2 = (3x + 4y)^2$.

So, the area of the square is $(3x + 4y)^2$.

Since $Area = (side)^2$, we have $(side)^2 = (3x + 4y)^2$.

Taking the square root of both sides (and considering the side length must be non-negative), the side of the square is $(3x + 4y)$.

The side of the square is $\mathbf{3x + 4y}$.

The area of a rectangle is given by the formula $Area = Length \times Breadth$.

We are given the Area = $x^2 + 7x + 12$ and Breadth = $(x + 3)$.

We need to find the Length.

From the formula, $Length = \frac{Area}{Breadth}$.

So, $Length = \frac{x^2 + 7x + 12}{x + 3}$.

To perform this division, we can factorise the numerator $x^2 + 7x + 12$.

We need two numbers that multiply to 12 and add up to 7.

The factors of 12 are (1, 12), (2, 6), (3, 4).

The sums are 13, 8, 7.

The numbers are 3 and 4.

So, $x^2 + 7x + 12 = (x + 3)(x + 4)$.

Now, substitute the factored form of the numerator back into the expression for Length:

$Length = \frac{(x + 3)(x + 4)}{x + 3}$

Assuming $x + 3 \neq 0$, we can cancel the $(x + 3)$ term in the numerator and the denominator.

$Length = x + 4$.

Alternatively, we can use polynomial long division to divide $x^2 + 7x + 12$ by $x + 3$.

$\begin{array}{r} x+4\phantom{)} \\ x+3{\overline{\smash{\big)}\,x^2+7x+12}} \\ \underline{-~\phantom{(}(x^2+3x)\phantom{+12)}} \\ 0+4x+12\phantom{)} \\ \underline{-~\phantom{()}(4x+12)} \\ 0+0\phantom{)} \end{array}$

The quotient is $x + 4$, with a remainder of 0.

So, the length of the rectangle is $\mathbf{x + 4}$.

Given:

Curved Surface Area (C.S.A.) of the cylinder = $2\pi (y^2 – 7y + 12)$

Radius ($r$) of the cylinder = $(y – 3)$

Formula for C.S.A. of a cylinder = $2\pi rh$, where $h$ is the height.

We have the equation:

To find the height ($h$), we can rearrange the formula:

$h = \frac{\text{C.S.A.}}{2\pi r}$

Substitute the given values for C.S.A. and $r$ into the formula for $h$:

$h = \frac{2\pi (y^2 – 7y + 12)}{2\pi (y – 3)}$

We can cancel the $2\pi$ term from the numerator and the denominator (assuming $2\pi \neq 0$):

$h = \frac{y^2 – 7y + 12}{y – 3}$

Now, we need to perform the division. We can factorise the quadratic expression in the numerator, $y^2 – 7y + 12$. We look for two numbers that multiply to 12 and add up to -7.

Let the two numbers be $m$ and $n$. We need $m \times n = 12$ and $m + n = -7$.

Since the product is positive and the sum is negative, both numbers must be negative.

Let's consider the negative factors of 12:

• $-1$ and $-12$: $(-1) + (-12) = -13$
• $-2$ and $-6$: $(-2) + (-6) = -8$
• $-3$ and $-4$: $(-3) + (-4) = -7$

The numbers are $-3$ and $-4$.

So, the numerator can be factorised as $y^2 – 7y + 12 = (y - 3)(y - 4)$.

Substitute this factored form back into the expression for $h$:

$h = \frac{(y - 3)(y - 4)}{y - 3}$

Assuming the radius $(y - 3)$ is not zero, we can cancel the common factor $(y - 3)$ from the numerator and the denominator:

$h = y - 4$

Thus, the height of the cylinder is $(y - 4)$.

For the height and radius to be physically meaningful, $y-3 > 0$ and $y-4 \ge 0$, which implies $y > 4$.

The height of the cylinder is $\mathbf{y - 4}$.

Given:

The area of a circle is given by the expression $\pi x^2 + 6\pi x + 9\pi$.

To Find:

The radius of the circle.

Solution:

The formula for the area of a circle with radius $r$ is:

Area $= \pi r^2$

We are given that the area of the circle is $\pi x^2 + 6\pi x + 9\pi$.

Equating the given expression for the area with the formula, we have:

$\pi r^2 = \pi x^2 + 6\pi x + 9\pi$

To find the radius $r$, we can divide both sides of the equation by $\pi$:

$r^2 = \frac{\pi x^2 + 6\pi x + 9\pi}{\pi}$

$r^2 = x^2 + 6x + 9$

Now, we observe that the expression on the right side, $x^2 + 6x + 9$, is a perfect square trinomial. It can be factored as $(x+3)^2$ because $(x+3)^2 = x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9$.

Substitute this factorization back into the equation for $r^2$:

$r^2 = (x+3)^2$

Taking the square root of both sides to solve for $r$:

$\sqrt{r^2} = \sqrt{(x+3)^2}$

$r = \pm (x+3)$

Since the radius of a circle must be a non-negative value, we take the positive square root.

$r = x+3$

We assume that the context of the problem implies that $x+3 \geq 0$, so $x \geq -3$. If $x+3$ could be negative, the radius would be $|x+3|$. However, typically in such problems, the expression represents the radius directly.

Thus, the radius of the circle is $x+3$.

Given:

The expression for the sum of the first $n$ natural numbers is $\frac{n^2}{2}+\frac{n}{2}$.

To Factorise:

Factorise the given expression $\frac{n^2}{2}+\frac{n}{2}$.

Solution:

The given expression is:

$\frac{n^2}{2}+\frac{n}{2}$

We can observe that both terms in the expression have a common denominator of $2$. We can also see that both terms have a common factor of $n$ in the numerator.

So, the common factor for both terms is $\frac{n}{2}$.

We can factor out the common factor $\frac{n}{2}$ from both terms:

$\frac{n}{2} \left( \frac{n^2/2}{n/2} + \frac{n/2}{n/2} \right)$

Simplify the terms inside the parenthesis:

$\frac{n^2/2}{n/2} = \frac{n^2}{2} \times \frac{2}{n} = \frac{n^2}{n} = n$

$\frac{n/2}{n/2} = 1$

Substitute the simplified terms back into the expression:

$\frac{n}{2} (n + 1)$

Thus, the factored form of the expression $\frac{n^2}{2}+\frac{n}{2}$ is $\frac{n(n+1)}{2}$.

Given:

Number of observations = $x+5$

Sum of observations = $x^4 - 625$

To Find:

The mean of the observations.

Solution:

The mean of a set of observations is calculated by dividing the sum of the observations by the number of observations.

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}}$

Substitute the given values into the formula:

Mean = $\frac{x^4 - 625}{x+5}$

Now, we need to simplify the expression by factorising the numerator. The numerator $x^4 - 625$ is a difference of squares, as $x^4 = (x^2)^2$ and $625 = 25^2$.

$x^4 - 625 = (x^2)^2 - (25)^2 = (x^2 - 25)(x^2 + 25)$

The factor $x^2 - 25$ is also a difference of squares, as $x^2 = x^2$ and $25 = 5^2$.

$x^2 - 25 = x^2 - 5^2 = (x-5)(x+5)$

Substitute this back into the factorization of the numerator:

$x^4 - 625 = (x-5)(x+5)(x^2 + 25)$

Now substitute this factored form into the expression for the mean:

Mean = $\frac{(x-5)(x+5)(x^2 + 25)}{x+5}$

Assuming $x+5 \neq 0$ (i.e., $x \neq -5$), we can cancel the common factor $(x+5)$ from the numerator and the denominator.

Mean = $\frac{(x-5)\cancel{(x+5)}(x^2 + 25)}{\cancel{x+5}}$

Mean = $(x-5)(x^2 + 25)$

This can also be expanded by multiplying the terms:

Mean = $x(x^2 + 25) - 5(x^2 + 25)$

Mean = $x^3 + 25x - 5x^2 - 125$

Mean = $x^3 - 5x^2 + 25x - 125$

Both $(x-5)(x^2 + 25)$ and $x^3 - 5x^2 + 25x - 125$ are valid forms for the mean.

Given:

Height of the triangle $= x^4 + y^4$

Base of the triangle $= 14xy$

To Find:

The area of the triangle.

Solution:

The formula for the area of a triangle is:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

Substitute the given values into the formula:

Area $= \frac{1}{2} \times (14xy) \times (x^4 + y^4)$

Multiply the numerical and algebraic terms:

Area $= \left(\frac{1}{2} \times 14\right) xy (x^4 + y^4)$

Area $= 7xy (x^4 + y^4)$

We can also distribute $7xy$ across the terms in the parenthesis:

Area $= 7xy \cdot x^4 + 7xy \cdot y^4$

Area $= 7x^{1+4}y + 7xy^{1+4}$

Area $= 7x^5y + 7xy^5$

The area of the triangle is $7xy(x^4 + y^4)$ or $7x^5y + 7xy^5$.

Given:

Cost of one chocolate = $\textsf{₹} (x+y)$

Number of chocolates bought = $(x+y)$

To Find:

The total amount paid in terms of $x$.

The amount paid when $x = 10$.

Solution:

The total amount paid is calculated by multiplying the cost of one chocolate by the number of chocolates bought.

Total Amount = (Cost of one chocolate) $\times$ (Number of chocolates)

Total Amount $= (x+y) \times (x+y)$

Total Amount $= (x+y)^2$

Expanding this expression using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$, where $a=x$ and $b=y$:

Total Amount $= x^2 + 2(x)(y) + y^2$

Total Amount $= x^2 + 2xy + y^2$

This is the total amount paid in terms of $x$ and $y$. The question asks for the amount in terms of $x$. Without a specific relationship between $x$ and $y$, the expression remains in terms of both variables. Assuming the question implies expressing the general form using $x$, the total amount paid is $x^2 + 2xy + y^2$.

Now, we need to find the amount paid when $x = 10$. Substitute $x = 10$ into the expression for the total amount:

Amount (when $x=10$) $= (10)^2 + 2(10)y + y^2$

Amount (when $x=10$) $= 100 + 20y + y^2$

Since the value of $y$ is not given, the amount paid when $x=10$ is $\textsf{₹} (100 + 20y + y^2)$.

Given:

Base of the parallelogram $= (2x+3)$ units

Height of the parallelogram $= (2x-3)$ units

To Find:

The area of the parallelogram in terms of $x$.

The area of the parallelogram when $x=30$ units.

Solution:

The formula for the area of a parallelogram is:

Area $= \text{base} \times \text{height}$

Substitute the given values for base and height:

Area $= (2x+3) \times (2x-3)$

This expression is in the form of a difference of squares, $(a+b)(a-b) = a^2 - b^2$, where $a=2x$ and $b=3$.

Using this identity:

Area $= (2x)^2 - (3)^2$

Area $= 4x^2 - 9$

So, the area of the parallelogram in terms of $x$ is $(4x^2 - 9)$ square units.

Now, we need to find the area when $x = 30$ units.

Substitute $x=30$ into the area expression $4x^2 - 9$:

Area $= 4(30)^2 - 9$

First, calculate the square of 30:

$(30)^2 = 30 \times 30 = 900$

Now substitute this value back into the area expression:

Area $= 4(900) - 9$

Perform the multiplication:

$4 \times 900 = 3600$

Finally, perform the subtraction:

Area $= 3600 - 9$

Area $= 3591$

The area of the parallelogram when $x=30$ units is 3591 square units.

Given:

Radius of the circle, $r = 7ab - 7bc - 14ac$

Value of $\pi = \frac{22}{7}$

To Find:

The circumference of the circle.

Solution:

The formula for the circumference of a circle is:

Circumference, $C = 2\pi r$

Substitute the given values of $\pi$ and $r$ into the formula:

$C = 2 \times \frac{22}{7} \times (7ab - 7bc - 14ac)$

First, multiply $2$ by $\frac{22}{7}$:

$2 \times \frac{22}{7} = \frac{44}{7}$

So, the expression for the circumference becomes:

$C = \frac{44}{7} (7ab - 7bc - 14ac)$

Observe that the terms inside the parenthesis have a common factor of $7$. Factor out $7$ from the expression for the radius:

$7ab - 7bc - 14ac = 7(ab - bc - 2ac)$

Substitute this factored form back into the circumference expression:

$C = \frac{44}{7} \times 7(ab - bc - 2ac)$

Now, we can cancel the $7$ in the denominator with the $7$ outside the parenthesis:

$C = \frac{44}{\cancel{7}} \times \cancel{7}(ab - bc - 2ac)$

$C = 44(ab - bc - 2ac)$

Distribute $44$ across the terms inside the parenthesis:

$C = 44 \times ab - 44 \times bc - 44 \times 2ac$

$C = 44ab - 44bc - 88ac$

The circumference of the circle is $44(ab - bc - 2ac)$ or $44ab - 44bc - 88ac$.

Given:

$p+q = 12$

$pq = 22$

To Find:

The value of $p^2 + q^2$.

Solution:

We know the algebraic identity for the square of a sum of two terms:

$(p+q)^2 = p^2 + 2pq + q^2$

We can rearrange this identity to find an expression for $p^2 + q^2$:

$p^2 + q^2 = (p+q)^2 - 2pq$

We are given the values of $(p+q)$ and $pq$. Substitute these values into the rearranged identity:

$p^2 + q^2 = (12)^2 - 2(22)$

Calculate the square of 12:

$(12)^2 = 12 \times 12 = 144$

Calculate the product of 2 and 22:

$2 \times 22 = 44$

Substitute these results back into the expression for $p^2 + q^2$:

$p^2 + q^2 = 144 - 44$

Perform the subtraction:

$144 - 44 = 100$

Therefore, the value of $p^2 + q^2$ is 100.

Given:

$a+b = 25$

$a^2 + b^2 = 225$

To Find:

The value of $ab$.

Solution:

We use the algebraic identity for the square of the sum of two terms:

$(a+b)^2 = a^2 + 2ab + b^2$

We can rearrange this identity to solve for $2ab$:

$2ab = (a+b)^2 - (a^2 + b^2)$

We are given the values of $(a+b)$ and $(a^2 + b^2)$. Substitute these values into the rearranged identity:

$2ab = (25)^2 - (225)$

Calculate the square of 25:

$(25)^2 = 25 \times 25 = 625$

Substitute this value back into the equation:

$2ab = 625 - 225$

Perform the subtraction:

$2ab = 400$

Now, to find $ab$, divide both sides of the equation by 2:

$ab = \frac{400}{2}$

$ab = 200$

Therefore, the value of $ab$ is 200.

Given:

$x - y = 13$

$xy = 28$

To Find:

The value of $x^2 + y^2$.

Solution:

We know the algebraic identity for the square of the difference of two terms:

$(x-y)^2 = x^2 - 2xy + y^2$

We can rearrange this identity to find an expression for $x^2 + y^2$:

$x^2 + y^2 = (x-y)^2 + 2xy$

We are given the values of $(x-y)$ and $xy$. Substitute these values into the rearranged identity:

$x^2 + y^2 = (13)^2 + 2(28)$

Calculate the square of 13:

$(13)^2 = 13 \times 13 = 169$

Calculate the product of 2 and 28:

$2 \times 28 = 56$

Substitute these results back into the expression for $x^2 + y^2$:

$x^2 + y^2 = 169 + 56$

Perform the addition:

$169 + 56 = 225$

Therefore, the value of $x^2 + y^2$ is 225.

Given:

$m - n = 16$

$m^2 + n^2 = 400$

To Find:

The value of $mn$.

Solution:

We use the algebraic identity for the square of the difference of two terms:

$(m-n)^2 = m^2 - 2mn + n^2$

We can rearrange this identity to solve for $-2mn$ or $2mn$:

$(m-n)^2 = (m^2 + n^2) - 2mn$

Now, substitute the given values of $(m-n)$ and $(m^2 + n^2)$ into the equation:

$(16)^2 = (400) - 2mn$

Calculate the square of 16:

$(16)^2 = 16 \times 16 = 256$

Substitute this value back into the equation:

$256 = 400 - 2mn$

Now, we need to isolate the term $2mn$. Add $2mn$ to both sides of the equation and subtract $256$ from both sides:

$2mn = 400 - 256$

Perform the subtraction:

$400 - 256 = 144$

So, we have:

$2mn = 144$

To find $mn$, divide both sides of the equation by 2:

$mn = \frac{144}{2}$

$mn = 72$

Therefore, the value of $mn$ is 72.

Given:

$a^2 + b^2 = 74$

$ab = 35$

To Find:

The value of $a+b$.

Solution:

We know the algebraic identity for the square of the sum of two terms:

$(a+b)^2 = a^2 + 2ab + b^2$

We can rearrange the right side of the identity as:

$(a+b)^2 = (a^2 + b^2) + 2ab$

We are given the values of $(a^2 + b^2)$ and $ab$. Substitute these values into the identity:

$(a+b)^2 = (74) + 2(35)$

Calculate the product of 2 and 35:

$2 \times 35 = 70$

Substitute this value back into the equation:

$(a+b)^2 = 74 + 70$

Perform the addition:

$(a+b)^2 = 144$

Now, to find $a+b$, take the square root of both sides of the equation:

$\sqrt{(a+b)^2} = \sqrt{144}$

$a+b = \pm 12$

Therefore, the value of $a+b$ can be $+12$ or $-12$.

To Verify:

(i) $(ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0$

We use the identity $(A+B)(A-B) = A^2 - B^2$.

LHS: $(ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab)$

Apply the identity to each term:

$(ab)^2 - (bc)^2 + (bc)^2 - (ca)^2 + (ca)^2 - (ab)^2$

$= a^2b^2 - b^2c^2 + b^2c^2 - c^2a^2 + c^2a^2 - a^2b^2$

Combine like terms:

$= (a^2b^2 - a^2b^2) + (-b^2c^2 + b^2c^2) + (-c^2a^2 + c^2a^2)$

$= 0 + 0 + 0 = 0$

RHS: $0$

Since LHS = RHS, the identity is verified.

(ii) $(a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) = a^3 + b^3 + c^3 – 3abc$

LHS: $(a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca)$

Expand the product:

$= a(a^2 + b^2 + c^2 – ab – bc – ca) + b(a^2 + b^2 + c^2 – ab – bc – ca) + c(a^2 + b^2 + c^2 – ab – bc – ca)$

$= (a^3 + ab^2 + ac^2 – a^2b – abc – a^2c) + (a^2b + b^3 + bc^2 – ab^2 – b^2c – abc) + (a^2c + b^2c + c^3 – abc – bc^2 – ac^2)$

Combine like terms:

$= a^3 + b^3 + c^3 + (ab^2 - ab^2) + (ac^2 - ac^2) + (-a^2b + a^2b) + (-abc - abc - abc) + (bc^2 - bc^2) + (-b^2c + b^2c) + (-a^2c + a^2c)$

$= a^3 + b^3 + c^3 + 0 + 0 + 0 - 3abc + 0 + 0 + 0$

$= a^3 + b^3 + c^3 - 3abc$

RHS: $a^3 + b^3 + c^3 – 3abc$

Since LHS = RHS, the identity is verified.

(iii) $(p – q) (p^2 + pq + q^2) = p^3 – q^3$

LHS: $(p – q) (p^2 + pq + q^2)$

Expand the product:

$= p(p^2 + pq + q^2) - q(p^2 + pq + q^2)$

$= p^3 + p^2q + pq^2 - qp^2 - pq^2 - q^3$

Combine like terms ($p^2q = qp^2$):

$= p^3 + (p^2q - p^2q) + (pq^2 - pq^2) - q^3$

$= p^3 + 0 + 0 - q^3$

$= p^3 - q^3$

RHS: $p^3 – q^3$

Since LHS = RHS, the identity is verified.

(iv) $(m + n) (m^2 – mn + n^2) = m^3 + n^3$

LHS: $(m + n) (m^2 – mn + n^2)$

Expand the product:

$= m(m^2 – mn + n^2) + n(m^2 – mn + n^2)$

$= m^3 - m^2n + mn^2 + nm^2 - mn^2 + n^3$

Combine like terms ($m^2n = nm^2$):

$= m^3 + (-m^2n + m^2n) + (mn^2 - mn^2) + n^3$

$= m^3 + 0 + 0 + n^3$

$= m^3 + n^3$

RHS: $m^3 + n^3$

Since LHS = RHS, the identity is verified.

(v) $(a + b) (a + b) (a + b) = a^3 + 3a^2b + 3ab^2 + b^3$

LHS: $(a + b) (a + b) (a + b) = (a+b)^3$

We know that $(a+b)^2 = a^2 + 2ab + b^2$.

So, $(a+b)^3 = (a+b)^2 (a+b)$

$= (a^2 + 2ab + b^2)(a+b)$

Expand the product:

$= a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$

$= a^3 + 2a^2b + ab^2 + ba^2 + 2ab^2 + b^3$

Combine like terms ($ba^2 = a^2b$):

$= a^3 + (2a^2b + a^2b) + (ab^2 + 2ab^2) + b^3$

$= a^3 + 3a^2b + 3ab^2 + b^3$

RHS: $a^3 + 3a^2b + 3ab^2 + b^3$

Since LHS = RHS, the identity is verified.

(vi) $(a – b) (a – b) (a – b) = a^3 – 3a^2b + 3ab^2 – b^3$

LHS: $(a – b) (a – b) (a – b) = (a-b)^3$

We know that $(a-b)^2 = a^2 - 2ab + b^2$.

So, $(a-b)^3 = (a-b)^2 (a-b)$

$= (a^2 - 2ab + b^2)(a-b)$

Expand the product:

$= a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$

$= a^3 - 2a^2b + ab^2 - ba^2 + 2ab^2 - b^3$

Combine like terms ($ba^2 = a^2b$):

$= a^3 + (-2a^2b - a^2b) + (ab^2 + 2ab^2) - b^3$

$= a^3 - 3a^2b + 3ab^2 - b^3$

RHS: $a^3 – 3a^2b + 3ab^2 – b^3$

Since LHS = RHS, the identity is verified.

(vii) $(a^2 – b^2) (a^2 + b^2) + (b^2 – c^2) (b^2 + c^2) + (c^2 – a^2) + (c^2 + a^2) = 0$

There appears to be a typo in the third term on the LHS. Assuming the third term is $(c^2 – a^2) (c^2 + a^2)$, we proceed with the verification.

We use the identity $(A-B)(A+B) = A^2 - B^2$.

LHS: $(a^2 – b^2) (a^2 + b^2) + (b^2 – c^2) (b^2 + c^2) + (c^2 – a^2) (c^2 + a^2)$

Apply the identity to each term:

$(a^2)^2 - (b^2)^2 + (b^2)^2 - (c^2)^2 + (c^2)^2 - (a^2)^2$

$= a^4 - b^4 + b^4 - c^4 + c^4 - a^4$

Combine like terms:

$= (a^4 - a^4) + (-b^4 + b^4) + (-c^4 + c^4)$

$= 0 + 0 + 0 = 0$

RHS: $0$

Since LHS = RHS (assuming the typo correction), the identity is verified.

(viii) $(5x + 8)^2 – 160x = (5x – 8)^2$

We will expand the LHS and show it equals the RHS.

LHS: $(5x + 8)^2 – 160x$

Expand $(5x + 8)^2$ using $(A+B)^2 = A^2 + 2AB + B^2$:

$(5x + 8)^2 = (5x)^2 + 2(5x)(8) + 8^2 = 25x^2 + 80x + 64$

Substitute this back into the LHS expression:

LHS $= (25x^2 + 80x + 64) - 160x$

Combine like terms:

LHS $= 25x^2 + (80x - 160x) + 64$

LHS $= 25x^2 - 80x + 64$

Now, expand the RHS:

RHS: $(5x – 8)^2$

Expand $(5x – 8)^2$ using $(A-B)^2 = A^2 - 2AB + B^2$:

RHS $= (5x)^2 - 2(5x)(8) + 8^2 = 25x^2 - 80x + 64$

Since LHS = RHS, the identity is verified.

(ix) $(7p – 13q)^2 + 364pq = (7p + 13q)^2$

We will expand the LHS and show it equals the RHS.

LHS: $(7p – 13q)^2 + 364pq$

Expand $(7p – 13q)^2$ using $(A-B)^2 = A^2 - 2AB + B^2$:

$(7p – 13q)^2 = (7p)^2 - 2(7p)(13q) + (13q)^2 = 49p^2 - 182pq + 169q^2$

Substitute this back into the LHS expression:

LHS $= (49p^2 - 182pq + 169q^2) + 364pq$

Combine like terms:

LHS $= 49p^2 + (-182pq + 364pq) + 169q^2$

LHS $= 49p^2 + 182pq + 169q^2$

Now, expand the RHS:

RHS: $(7p + 13q)^2$

Expand $(7p + 13q)^2$ using $(A+B)^2 = A^2 + 2AB + B^2$:

RHS $= (7p)^2 + 2(7p)(13q) + (13q)^2 = 49p^2 + 182pq + 169q^2$

Since LHS = RHS, the identity is verified.

(x) $\left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$ - $\left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$ = 2

There appears to be a typo in the question. The left-hand side is a term subtracted from itself, which results in 0.

LHS: $\left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$ - $\left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$

Let $A = \left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$. The expression is $A - A = 0$.

So, LHS = 0.

RHS: $2$

Since $0 \neq 2$, the given equation is not true as stated.

Assuming the question intended to be $\left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$ - $\left( \frac{3p}{7}-\frac{7}{6p} \right)^{2} = 2$, we verify that instead.

Let $A = \frac{3p}{7}$ and $B = \frac{7}{6p}$. The assumed equation is $(A+B)^2 - (A-B)^2 = 2$.

We know the identity $(A+B)^2 - (A-B)^2 = 4AB$.

LHS (Assumed): $\left( \frac{3p}{7}+\frac{7}{6p} \right)^{2}$ - $\left( \frac{3p}{7}-\frac{7}{6p} \right)^{2}$

Using the identity, this equals $4 \times \frac{3p}{7} \times \frac{7}{6p}$

$= 4 \times \frac{3\cancel{p}}{\cancel{7}} \times \frac{\cancel{7}}{6\cancel{p}}$

$= 4 \times \frac{3}{6}$

$= 4 \times \frac{1}{2}$

$= \frac{4}{2} = 2$

RHS (Assumed): $2$

Assuming the typo, LHS = RHS, and the identity is verified for the corrected expression.

To Find: The value of $a$ in each equation.

(i) $8a = 35^2 – 27^2$

We use the identity $A^2 - B^2 = (A-B)(A+B)$. Here, $A=35$ and $B=27$.

$35^2 - 27^2 = (35 - 27)(35 + 27)$

$= (8)(62)$

$= 496$

So, the equation becomes:

$8a = 496$

To find $a$, divide both sides by 8:

$a = \frac{496}{8}$

$a = 62$

(ii) $9a = 76^2 – 67^2$

We use the identity $A^2 - B^2 = (A-B)(A+B)$. Here, $A=76$ and $B=67$.

$76^2 - 67^2 = (76 - 67)(76 + 67)$

$= (9)(143)$

$= 1287$

So, the equation becomes:

$9a = 1287$

To find $a$, divide both sides by 9:

$a = \frac{1287}{9}$

We can perform the division:

$a = 143$

(iii) $pqa = (3p + q)^2 – (3p – q)^2$

We use the identity $(A+B)^2 - (A-B)^2 = 4AB$. Here, $A=3p$ and $B=q$.

$(3p + q)^2 – (3p – q)^2 = 4(3p)(q)$

$= 12pq$

So, the equation becomes:

$pqa = 12pq$

Assuming $pq \neq 0$, we can divide both sides by $pq$:

$a = \frac{12pq}{pq}$

$a = 12$

(iv) $pq^2a = (4pq + 3q)^2 – (4pq – 3q)^2$

We use the identity $(A+B)^2 - (A-B)^2 = 4AB$. Here, $A=4pq$ and $B=3q$.

$(4pq + 3q)^2 – (4pq – 3q)^2 = 4(4pq)(3q)$

$= 4 \times 4pq \times 3q$

$= (4 \times 4 \times 3) \times (p \times q \times q)$

$= 48pq^2$

So, the equation becomes:

$pq^2a = 48pq^2$

Assuming $pq^2 \neq 0$, we can divide both sides by $pq^2$:

$a = \frac{48pq^2}{pq^2}$

$a = 48$

Given:

First expression: $4c(-a+b+c)$

Second expression: $3a(a+b+c) - 2b(a-b+c)$

To Find:

The expression that should be added to the first expression to obtain the second expression.

Solution:

Let the expression to be added be $X$. According to the problem, we have:

$4c(-a+b+c) + X = 3a(a+b+c) - 2b(a-b+c)$

To find $X$, we can rearrange the equation:

$X = \left[ 3a(a+b+c) - 2b(a-b+c) \right] - 4c(-a+b+c)$

First, let's simplify the terms:

Simplify the first part of the second expression, $3a(a+b+c)$:

$3a(a+b+c) = 3a \times a + 3a \times b + 3a \times c = 3a^2 + 3ab + 3ac$

Simplify the second part of the second expression, $-2b(a-b+c)$:

$-2b(a-b+c) = -2b \times a -2b \times (-b) -2b \times c = -2ab + 2b^2 - 2bc$

So, the second expression is:

$(3a^2 + 3ab + 3ac) + (-2ab + 2b^2 - 2bc) = 3a^2 + 3ab + 3ac - 2ab + 2b^2 - 2bc$

Combine like terms in the second expression:

$3a^2 + (3ab - 2ab) + 2b^2 + 3ac - 2bc = 3a^2 + ab + 2b^2 + 3ac - 2bc$

Now, simplify the first expression, $4c(-a+b+c)$:

$4c(-a+b+c) = 4c \times (-a) + 4c \times b + 4c \times c = -4ac + 4bc + 4c^2$

Now, subtract the first expression from the simplified second expression to find $X$:

$X = (3a^2 + ab + 2b^2 + 3ac - 2bc) - (-4ac + 4bc + 4c^2)$

$X = 3a^2 + ab + 2b^2 + 3ac - 2bc + 4ac - 4bc - 4c^2$

Combine like terms:

$X = 3a^2 + 2b^2 - 4c^2 + ab + (3ac + 4ac) + (-2bc - 4bc)$

$X = 3a^2 + 2b^2 - 4c^2 + ab + 7ac - 6bc$

The expression that should be added is $3a^2 + 2b^2 - 4c^2 + ab + 7ac - 6bc$.

Given:

First expression: $b (b^2 + b – 7) + 5$

Second expression: $3b^2 – 8$

Value of $b = -3$

To Find:

The expression obtained by subtracting the first expression from the second expression.

The value of the obtained expression for $b = -3$.

Solution:

First, let's simplify the first expression:

$b (b^2 + b – 7) + 5 = b \cdot b^2 + b \cdot b + b \cdot (-7) + 5$

$= b^3 + b^2 - 7b + 5$

Now, we need to subtract this expression from the second expression, $3b^2 - 8$.

Expression obtained $= (3b^2 - 8) - (b^3 + b^2 - 7b + 5)$

Remove the parenthesis, changing the sign of each term inside the second parenthesis:

$= 3b^2 - 8 - b^3 - b^2 + 7b - 5$

Combine like terms:

$= -b^3 + (3b^2 - b^2) + 7b + (-8 - 5)$

$= -b^3 + 2b^2 + 7b - 13$

The obtained expression is $-b^3 + 2b^2 + 7b - 13$.

Now, we need to find the value of this expression when $b = -3$.

Substitute $b = -3$ into the expression:

Value $= -(-3)^3 + 2(-3)^2 + 7(-3) - 13$

Calculate the powers and products:

$(-3)^3 = (-3) \times (-3) \times (-3) = 9 \times (-3) = -27$

$-(-3)^3 = -(-27) = 27$

$(-3)^2 = (-3) \times (-3) = 9$

$2(-3)^2 = 2(9) = 18$

$7(-3) = -21$

Substitute these values back into the expression:

Value $= 27 + 18 - 21 - 13$

Perform the addition and subtraction:

Value $= 45 - 21 - 13$

Value $= 24 - 13$

Value $= 11$

The value of the expression when $b = -3$ is 11.

Given:

$x - \frac{1}{x} = 7$

To Find:

The value of $x^2 + \frac{1}{x^2}$.

Solution:

We are given the equation:

$x - \frac{1}{x} = 7$

To find the value of $x^2 + \frac{1}{x^2}$, we can square both sides of the given equation.

$\left(x - \frac{1}{x}\right)^2 = 7^2$

We use the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$. Here, $a=x$ and $b=\frac{1}{x}$.

Expanding the left side:

$x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 49$

Simplify the middle term:

$2(x)\left(\frac{1}{x}\right) = 2 \times \frac{x}{x} = 2 \times 1 = 2$

And $\left(\frac{1}{x}\right)^2 = \frac{1^2}{x^2} = \frac{1}{x^2}$.

Substitute these back into the equation:

$x^2 - 2 + \frac{1}{x^2} = 49$

Now, isolate $x^2 + \frac{1}{x^2}$ by adding 2 to both sides of the equation:

$x^2 + \frac{1}{x^2} = 49 + 2$

$x^2 + \frac{1}{x^2} = 51$

Therefore, the value of $x^2 + \frac{1}{x^2}$ is 51.

Given:

The expression to factorise is $x^2 + \frac{1}{x^2} + 2 - 3x - \frac{3}{x}$.

To Factorise:

Factorise the given expression.

Solution:

The given expression is:

$x^2 + \frac{1}{x^2} + 2 - 3x - \frac{3}{x}$

Rearrange the terms to group similar parts:

$(x^2 + \frac{1}{x^2} + 2) - (3x + \frac{3}{x})$

Observe the first group of terms, $x^2 + \frac{1}{x^2} + 2$. This is the expansion of $(x + \frac{1}{x})^2$, since $(a+b)^2 = a^2 + 2ab + b^2$.

Here, $a=x$ and $b=\frac{1}{x}$. So, $(x + \frac{1}{x})^2 = x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$.

So, we can write the first group as $(x + \frac{1}{x})^2$.

Now consider the second group of terms, $3x + \frac{3}{x}$. We can factor out a common factor of 3 from these terms:

$3x + \frac{3}{x} = 3(x + \frac{1}{x})$

Substitute these back into the expression:

$(x + \frac{1}{x})^2 - 3(x + \frac{1}{x})$

Now, we can see that $(x + \frac{1}{x})$ is a common factor in both terms.

Factor out $(x + \frac{1}{x})$:

$(x + \frac{1}{x}) \left[ (x + \frac{1}{x}) - 3 \right]$

The factored form of the expression is $\left(x + \frac{1}{x}\right)\left(x + \frac{1}{x} - 3\right)$.

Given:

The expression to factorise is $p^4 + q^4 + p^2q^2$.

To Factorise:

Factorise the expression $p^4 + q^4 + p^2q^2$.

Solution:

The given expression is:

$p^4 + q^4 + p^2q^2$

We can rearrange the terms and add and subtract a term to complete the square for $(p^2+q^2)^2$.

Recall the identity $(a+b)^2 = a^2 + 2ab + b^2$. If we let $a=p^2$ and $b=q^2$, then $(p^2+q^2)^2 = (p^2)^2 + 2(p^2)(q^2) + (q^2)^2 = p^4 + 2p^2q^2 + q^4$.

The given expression $p^4 + q^4 + p^2q^2$ can be written as:

$(p^4 + 2p^2q^2 + q^4) - p^2q^2$

The term $(p^4 + 2p^2q^2 + q^4)$ is equal to $(p^2 + q^2)^2$. So the expression becomes:

$(p^2 + q^2)^2 - p^2q^2$

We can write $p^2q^2$ as $(pq)^2$. So we have:

$(p^2 + q^2)^2 - (pq)^2$

This is now in the form of a difference of squares, $A^2 - B^2$, where $A = (p^2 + q^2)$ and $B = pq$.

Using the identity $A^2 - B^2 = (A-B)(A+B)$, we substitute $A$ and $B$:

$\left[ (p^2 + q^2) - pq \right] \left[ (p^2 + q^2) + pq \right]$

Rearranging the terms within the parentheses gives the standard form:

$(p^2 - pq + q^2)(p^2 + pq + q^2)$

Thus, the factored form of $p^4 + q^4 + p^2q^2$ is $(p^2 - pq + q^2)(p^2 + pq + q^2)$.

To Find:

The value of the given expressions.

Solution:

We will use the algebraic identity: $a^2 - b^2 = (a-b)(a+b)$.

(i) $\frac{6.25 \;×\; 6.25 \;–\; 1.75 \;×\; 1.75}{4.5}$

The numerator is in the form $a^2 - b^2$, where $a = 6.25$ and $b = 1.75$.

Numerator $= (6.25)^2 - (1.75)^2$

Using the identity, $a^2 - b^2 = (a-b)(a+b)$:

Numerator $= (6.25 - 1.75)(6.25 + 1.75)$

Calculate the terms in the parentheses:

$6.25 - 1.75 = 4.50$

$6.25 + 1.75 = 8.00$

So, the numerator is $(4.50)(8.00) = 4.5 \times 8$.

The expression becomes:

$\frac{4.5 \times 8}{4.5}$

Cancel the common term $4.5$ in the numerator and denominator:

$\frac{\cancel{4.5} \times 8}{\cancel{4.5}} = 8$

The value of the expression is 8.

(ii) $\frac{198 \;×\; 198 \;–\; 102 \;×\; 102}{96}$

The numerator is in the form $a^2 - b^2$, where $a = 198$ and $b = 102$.

Numerator $= (198)^2 - (102)^2$

Using the identity, $a^2 - b^2 = (a-b)(a+b)$:

Numerator $= (198 - 102)(198 + 102)$

Calculate the terms in the parentheses:

$198 - 102 = 96$

$198 + 102 = 300$

So, the numerator is $(96)(300)$.

The expression becomes:

$\frac{96 \times 300}{96}$

Cancel the common term $96$ in the numerator and denominator:

$\frac{\cancel{96} \times 300}{\cancel{96}} = 300$

The value of the expression is 300.

Given:

Product of two expressions $= x^5 + x^3 + x$

One expression $= x^2 + x + 1$

To Find:

The other expression.

Solution:

Let the other expression be $E$.

We are given that:

$(x^2 + x + 1) \times E = x^5 + x^3 + x$

To find the other expression $E$, we need to divide the product by the given expression:

$E = \frac{x^5 + x^3 + x}{x^2 + x + 1}$

We can perform polynomial long division to find the quotient.

Divide $x^5 + x^3 + x$ by $x^2 + x + 1$. We can write the dividend as $x^5 + 0x^4 + x^3 + 0x^2 + x + 0$ for clarity in the long division process.

The quotient obtained from the division is $x^3 - x^2 + x$.

Since the remainder is 0, the other expression is exactly $x^3 - x^2 + x$.

Thus, the other expression is $x^3 - x^2 + x$.

Given:

Area of the square $= 625$ square units.

Side length of the square $= (3x-1)$ units (from the image).

To Find:

The length of the side of the square.

The value of $x$.

Solution:

The area of a square is given by the formula:

Area $= (\text{side length})^2$

We are given that the area is 625 square units. Let the side length be $s$.

$s^2 = 625$

To find the side length $s$, we take the square root of the area:

$s = \sqrt{625}$

We know that $25 \times 25 = 625$, so $\sqrt{625} = 25$.

Since the side length of a square must be positive, the length of the side of the square is 25 units.

We are also given that the side length of the square is represented by the expression $(3x-1)$ units.

So, we can equate the calculated side length to the given expression:

$3x - 1 = 25$

Now, we solve this linear equation for $x$. Add 1 to both sides of the equation:

$3x - 1 + 1 = 25 + 1$

$3x = 26$

Divide both sides by 3:

$x = \frac{26}{3}$

The value of $x$ is $\frac{26}{3}$.

The length of the side of the square is 25 units.

The value of $x$ is $\frac{26}{3}$.

Given:

Cards representing areas: Green (G) = $x \times x = x^2$, Red (R) = $x \times 1 = x$, Yellow (Y) = $1 \times 1 = 1$.

Expressions to factorise: (i) $2x^2 + 6x + 4$, (ii) $x^2 + 4x + 4$.

A figure showing an arrangement of cards.

To Factorise:

(i) $2x^2 + 6x + 4$ using the card method.

(ii) $x^2 + 4x + 4$ using the card method.

Factorise $2x^2 + 6x + 4$ using the figure (interpreting this as using the cards shown in the figure).

To Calculate:

The area of the figure shown.

Solution:

The card method involves representing the terms of an expression with the corresponding cards and arranging them to form a rectangle. The lengths of the sides of the rectangle are the factors of the expression.

(i) Factorise $2x^2 + 6x + 4$ using the card method:

The expression $2x^2 + 6x + 4$ requires:

• $2$ Green cards ($2x^2$)
• $6$ Red cards ($6x$)
• $4$ Yellow cards ($4$)

We arrange these cards to form a rectangle. We look for linear expressions for the dimensions (length and width) that multiply to give the expression.

We can factor this expression algebraically: $2x^2 + 6x + 4 = 2(x^2 + 3x + 2)$. Factoring the quadratic, $x^2 + 3x + 2 = (x+1)(x+2)$. So, the expression is $2(x+1)(x+2) = (x+1)(2x+4)$.

This suggests a rectangle with sides $(x+1)$ and $(2x+4)$. Let's verify the card count for this arrangement:

A rectangle with dimensions $(x+1)$ and $(2x+4)$ would have an area of $(x+1)(2x+4) = 2x^2 + 4x + 2x + 4 = 2x^2 + 6x + 4$.

To form this rectangle with cards:

• The $x \times 2x$ part requires two $x^2$ (Green) cards.
• The $x \times 4$ part requires four $x$ (Red) cards.
• The $1 \times 2x$ part requires two $x$ (Red) cards.
• The $1 \times 4$ part requires four $1$ (Yellow) cards.

Total: 2 Green, $4+2=6$ Red, 4 Yellow cards. This matches the required cards for $2x^2 + 6x + 4$.

Thus, the factorization is $(x+1)(2x+4)$.

(ii) Factorise $x^2 + 4x + 4$ using the card method:

The expression $x^2 + 4x + 4$ requires:

• $1$ Green card ($x^2$)
• $4$ Red cards ($4x$)
• $4$ Yellow cards ($4$)

We arrange these cards to form a rectangle. This expression is a perfect square trinomial: $x^2 + 4x + 4 = (x+2)^2$.

This suggests a square with side length $(x+2)$. Let's verify the card count for this arrangement:

A square with dimensions $(x+2)$ and $(x+2)$ would have an area of $(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4$.

To form this square with cards:

• The $x \times x$ part requires one $x^2$ (Green) card.
• The $x \times 2$ part requires two $x$ (Red) cards.
• The $2 \times x$ part requires two $x$ (Red) cards.
• The $2 \times 2$ part requires four $1$ (Yellow) cards.

Total: 1 Green, $2+2=4$ Red, 4 Yellow cards. This matches the required cards for $x^2 + 4x + 4$.

Thus, the factorization is $(x+2)(x+2)$ or $(x+2)^2$.

Factorise $2x^2 + 6x + 4$ by using the figure:

The figure displays an arrangement of cards: 2 Green ($2x^2$), 6 Red ($6x$), and 4 Yellow ($4$). The sum of the areas of these cards is $2x^2 + 6x + 4$. The instruction likely means to factorise the expression represented by the collection of cards shown in the figure.

As determined in part (i), the factorization of $2x^2 + 6x + 4$ using the card method results in the factors $(x+1)$ and $(2x+4)$.

The factors are $(x+1)$ and $(2x+4)$.

Calculate the area of figure:

The figure shows a rectangle formed by arranging the cards. We can find its area by determining the lengths of its sides from the dimensions of the cards forming the edges.

Looking at the left vertical side of the rectangle in the figure, it is formed by the $x$ side of a Green card followed by the $1$ side of two Red cards arranged vertically. The total length of this side is $x + 1 + 1 = x+2$ units.

Looking at the top horizontal side of the rectangle in the figure, it is formed by the $x$ sides of two Green cards followed by the $1$ sides of three Red cards arranged horizontally. The total length of this side is $x + x + 1 + 1 + 1 = 2x+3$ units.

The area of the rectangle is the product of its length and width.

Area $= (2x+3) \times (x+2)$

Expand the product:

Area $= 2x(x+2) + 3(x+2)$

Area $= 2x^2 + 4x + 3x + 6$

Combine the like terms ($4x$ and $3x$):

Area $= 2x^2 + 7x + 6$

The area of the figure shown is $2x^2 + 7x + 6$ square units.

Given:

Dimensions of the wall: Length $= (2x+10)$, Height $= (2x+5)$

Dimensions of the window: Length $= x$, Height $= x$

Dimensions of the door: Length $= x$, Height $= 2x$

To Find:

An algebraic expression for the area of the wall to be painted.

Solution:

The area of the wall to be painted is the total area of the wall minus the areas of the window and the door (which are not painted).

Area to be painted = Area of Wall - (Area of Window + Area of Door)

The area of a rectangle is given by the product of its length and height.

Calculate the Area of the Wall:

Area of Wall $= \text{Length} \times \text{Height}$

Area of Wall $= (2x+10)(2x+5)$

Expand the product:

Area of Wall $= 2x(2x+5) + 10(2x+5)$

Area of Wall $= 4x^2 + 10x + 20x + 50$

Area of Wall $= 4x^2 + 30x + 50$

Calculate the Area of the Window:

Area of Window $= \text{Length} \times \text{Height}$

Area of Window $= x \times x$

Area of Window $= x^2$

Calculate the Area of the Door:

Area of Door $= \text{Length} \times \text{Height}$

Area of Door $= x \times 2x$

Area of Door $= 2x^2$

Now, substitute these areas into the formula for the area to be painted:

Area to be painted $= (4x^2 + 30x + 50) - (x^2 + 2x^2)$

Area to be painted $= 4x^2 + 30x + 50 - (3x^2)$

Area to be painted $= 4x^2 + 30x + 50 - 3x^2$

Combine the like terms ($4x^2$ and $-3x^2$):

Area to be painted $= (4x^2 - 3x^2) + 30x + 50$

Area to be painted $= x^2 + 30x + 50$

The algebraic expression for the area of the wall to be painted is $x^2 + 30x + 50$ square units.

To match the expressions, we will expand the expressions in Column I using algebraic identities.

(1) $(21x + 13y)^2$

This is in the form of $(a+b)^2 = a^2 + 2ab + b^2$, where $a = 21x$ and $b = 13y$.

$(21x + 13y)^2 = (21x)^2 + 2(21x)(13y) + (13y)^2$

$= 21^2 x^2 + (2 \times 21 \times 13) xy + 13^2 y^2$

Calculate the squares and the product:

$21^2 = 441$

$13^2 = 169$

$2 \times 21 \times 13 = 42 \times 13 = 546$

So, $(21x + 13y)^2 = 441x^2 + 546xy + 169y^2$.

Rearranging the terms, we get $441x^2 + 169y^2 + 546xy$.

This matches with option (b) in Column II.

(2) $(21x – 13y)^2$

This is in the form of $(a-b)^2 = a^2 - 2ab + b^2$, where $a = 21x$ and $b = 13y$.

$(21x – 13y)^2 = (21x)^2 - 2(21x)(13y) + (13y)^2$

$= 21^2 x^2 - (2 \times 21 \times 13) xy + 13^2 y^2$

Using the values calculated above:

$= 441x^2 - 546xy + 169y^2$

Rearranging the terms, we get $441x^2 + 169y^2 - 546xy$.

This matches with option (c) in Column II.

(3) $(21x – 13y) (21x + 13y)$

This is in the form of $(a-b)(a+b) = a^2 - b^2$, where $a = 21x$ and $b = 13y$.

$(21x – 13y) (21x + 13y) = (21x)^2 - (13y)^2$

$= 21^2 x^2 - 13^2 y^2$

Using the values calculated above:

$= 441x^2 - 169y^2$

This matches with option (a) in Column II.

Summary of the matches:

(1) $\leftrightarrow$ (b)

(2) $\leftrightarrow$ (c)

(3) $\leftrightarrow$ (a)