Chapter 7 Algebraic Expression, Identities & Factorisation (Class 8 - Maths NCERT Exemplar Solutions)

To Find:

The term that is a like term to $24a^2bc$.

Solution:

Like terms are terms that have the same algebraic variables raised to the same powers. The numerical coefficients do not need to match.

The variable part of the given term is $a^2bc$. Let us simplify the options to find the one with the same variable part:

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

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

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

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

Clearly, option (a) has the same variable part $a^2bc$.

The correct option is (a).

Solution:

An algebraic identity is an equality that holds true regardless of the values chosen for its variables.

The standard identities are:

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

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

3. $a^2 - b^2 = (a + b)(a - b)$

Comparing these with the options:

• Option (a) is incorrect because the middle term $2pq$ is missing.
• Option (b) is incorrect because $p^2 - q^2 \neq p^2 - 2pq + q^2$.
• Option (d) matches the standard identity $(p + q)^2 = p^2 + 2pq + q^2$.

The correct option is (d).

To Find:

The irreducible factorisation of the expression $3a^3 + 6a$.

Solution:

To factorise the expression, we find the Greatest Common Factor (GCF) of the terms:

$3a^3 = 3 \times a \times a \times a$

$6a = 2 \times 3 \times a$

The common factors are $3$ and $a$. So, the common factor is $3a$.

Dividing each term by $3a$:

$3a^3 + 6a = 3a(a^2 + 2)$

Since $(a^2 + 2)$ cannot be factorised further over the set of real numbers, this is the irreducible factorisation.

The correct option is (a).

Solution:

The property $a(b + c) = ab + ac$ represents the multiplication being distributed over addition.

This is known as the Distributive Property of multiplication over addition.

The correct option is (b).

Solution:

When an algebraic expression is written as the product of two or more expressions, the process is called factorisation. Each of the expressions in the product is called a factor of the original expression.

Answer: factorisation

Solution:

This is one of the standard algebraic identities. Let us expand the Left Hand Side (LHS):

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

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

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

Comparing this with the given expression: $x^2 + (a + b)x + \text{ab}$.

Answer: ab

Solution:

By definition, an identity is an equation which remains true for any value that is substituted for its variables. This distinguishes it from a conditional equation, which is true only for specific values of the variable.

Answer: True (T)

Solution:

Let's find the factors of both terms:

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

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

The factors common to both are one $x$ and one $y$. Their product is $x \times y = xy$.

Thus, $xy$ is the common factor.

Answer: True (T)

Solution:

To divide the binomial by the monomial, we divide each term of the numerator by the denominator:

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

$= 1 + x$

The result is $1 + x$, not $3x^2$. Therefore, the statement is false.

Answer: False (F)

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

Using the distributive property, we multiply the monomial $-pqr$ with each term inside the trinomial:

(ii) Simplify: $(px + qy)(ax - by)$

Multiplying the two binomials using the distributive law:

(i) Expand: $(3x + 7y)(3x - 7y)$

Using the identity $(a + b)(a - b) = a^2 - b^2$:

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

Using the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$:

(i) Factorise: $21x^2y^3 + 27x^3y^2$

Find the common factors: $21x^2y^3 = 3 \times 7 \times x^2 \times y^3$ and $27x^3y^2 = 3 \times 9 \times x^3 \times y^2$.

The Greatest Common Factor (GCF) is $3x^2y^2$.

(ii) Factorise: $a^3 - 4a^2 + 12 - 3a$

Regrouping the terms:

(iii) Factorise: $4x^2 - 20x + 25$

Using the identity $a^2 - 2ab + b^2 = (a - b)^2$:

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

Using the identity $a^2 - b^2 = (a + b)(a - b)$:

(v) Factorise: $x^4 - 256$

(i) Evaluate: $(48)^2$

(ii) Evaluate: $181^2 - 19^2$

(iii) Evaluate: $497 \times 505$

(iv) Evaluate: $2.07 \times 1.93$

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

Expanding the LHS (Left Hand Side):

Since LHS = RHS, the expression is verified.

To Prove:

$(11pq + 4q)^2 - (11pq - 4q)^2 = 176pq^2$

Proof:

Let us consider the Left Hand Side (LHS) of the equation:

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

This expression is in the form $a^2 - b^2$, where $a = (11pq + 4q)$ and $b = (11pq - 4q)$.

Using the identity $a^2 - b^2 = (a + b)(a - b)$:

Simplifying the terms within the brackets:

Since LHS is equal to the Right Hand Side (RHS), the equation 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:

We know that the area of a rectangle is the product of its length and breadth.

First, let's factorise the area expression $x^2 + 12xy + 27y^2$ by splitting the middle term:

Now, substitute the values to find the breadth:

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

To Solve:

Divide $15(y + 3)(y^2 - 16)$ by $5(y^2 - y - 12)$.

Solution:

First, factorise the numerator and the denominator separately.

Factorising the numerator:

So, Numerator $= 15(y + 3)(y + 4)(y - 4)$.

Factorising the denominator:

So, Denominator $= 5(y - 4)(y + 3)$.

Performing the division:

Cancelling the common terms $(y + 3)$, $(y - 4)$, and dividing 15 by 5:

The result is $3(y + 4)$ or $3y + 12$.

Given:

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

To Find:

The value of $x^2 + \frac{1}{x^2}$

Solution:

Consider the identity $(a + b)^2 = a^2 + b^2 + 2ab$.

Let's square both sides of the given equation:

Expanding the LHS:

Transposing 2 to the right side:

The value is 23.

To Evaluate:

$\frac{38^2 - 22^2}{16}$

Solution:

Using the identity $a^2 - b^2 = (a + b)(a - b)$ in the numerator:

Now, substitute this back into the original expression:

Cancelling 16 from the numerator and denominator:

The value is 60.

Given:

$10000x = (9982)^2 - (18)^2$

To Find:

The value of $x$.

Solution:

Using the identity $a^2 - b^2 = (a + b)(a - b)$ on the RHS:

Divide both sides by 10000:

The value of $x$ is 9964.

A monomial is an expression containing only one term, and a binomial is an expression containing two terms.

According to the distributive law, when we multiply a monomial by a binomial, the monomial is multiplied by each term of the binomial.

Example:

Let the monomial be $x$ and the binomial be $(y + z)$.

Product $= x \times (y + z) = (x \times y) + (x \times z) = xy + xz$

The resulting expression $xy + xz$ has two terms, which is a binomial.

Correct Option: (b) binomial

By definition, a polynomial is an algebraic expression in which the variables involved have only non-negative integer powers (also known as whole numbers).

This means the exponents can be $0, 1, 2, 3, \dots$ and so on. Exponents cannot be negative or in fractional form.

Correct Option: (c) non-negative integers

We can check the identity by multiplying $(a - b)$ by itself:

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

$= a(a - b) - b(a - b)$

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

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

This matches option (b). The other options provide incorrect expansion of the algebraic identities.

Correct Option: (b) $(a - b)^2 = a^2 - 2ab + b^2$

To find the sum of like terms, we add their numerical coefficients while keeping the literal (variable) part the same.

Sum $= -7pq + 2pq$

$= (-7 + 2) \times pq$

$= -5pq$

Correct Option: (d) $-5pq$

The expression can be written as:

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

When we subtract a negative term, it becomes addition:

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

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

$= 4x^2y^2$

Correct Option: (d) $4x^2y^2$

Like terms are terms that have the same variables with the same corresponding exponents. The numerical coefficient does not need to be the same.

For the term $4m^3n^2$, the variable part is $m^3n^2$.

(a) $4m^2n^2$ $\rightarrow$ powers of $m$ are different.

(b) $-6m^3n^2$ $\rightarrow$ variable part $m^3n^2$ is exactly the same. (Like term)

(c) $6pm^3n^2$ $\rightarrow$ contains an extra variable $p$.

(d) $4m^3n$ $\rightarrow$ power of $n$ is different.

Correct Option: (b) $-6m^3n^2$

A binomial is an algebraic expression containing exactly two terms.

(a) $7 \times a + a = 7a + a = 8a$ (One term $\rightarrow$ Monomial)

(b) $6a^2 + 7b + 2c$ (Three terms $\rightarrow$ Trinomial)

(c) $4a \times 3b \times 2c = 24abc$ (One term $\rightarrow$ Monomial)

(d) $6(a^2 + b) = 6a^2 + 6b$ (Two terms $\rightarrow$ Binomial)

Correct Option: (d) $6(a^2 + b)$

To find the sum, we combine the given expressions and group the like terms together:

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

Rearranging the terms:

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

$= 0 + 0 + 2c + ab - bc - ac$

$= 2c + ab - bc - ac$

This can also be written as $2c + ab - ac - bc$.

Correct Option: (a) $2c + ab - ac - bc$

To find the product of monomials, we multiply their numerical coefficients and then multiply their variable parts using the laws of exponents.

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

$= [4 \times (-7) \times (-7)] \times [p \times q^3 \times p \times q]$

$= [196] \times [p^{(1+1)} \times q^{(3+1)}]$

$= 196 p^2 q^4$

Correct Option: (a) $196 p^2 q^4$

We know that the formula for the area of a rectangle is:

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

Given:

Length $= 4ab$

Breadth $= 6b^2$

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

$= (4 \times 6) \times (a \times b \times b^2)$

$= 24 \times a \times b^{(1+2)}$

$= 24ab^3$

Correct Option: (b) $24ab^3$

The volume of a rectangular box (cuboid) is given by the product of its length, breadth, and height.

Volume $= \text{Length} \times \text{Breadth} \times \text{Height}$

Given:

Length $= 2ab$

Breadth $= 3ac$

Height $= 2ac$

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

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

$= 12 \times a^{(1+1+1)} \times b \times c^{(1+1)}$

$= 12 a^3 b c^2$

Correct Option: (a) $12a^3bc^2$

To find the product of a trinomial and a monomial, we use the distributive property:

$m(x + y + z) = mx + my + mz$

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

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

$= 12 a^{(1+2)} b - 14 a b^{(1+1)} + 10 a^{(1+1)} b^{(1+1)}$

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

Correct Option: (b) $12a^3b - 14ab^2 + 10a^2b^2$

We use the standard identity for the square of a binomial difference:

$(A - B)^2 = A^2 - 2AB + B^2$

Here, $A = 3x$ and $B = 4y$.

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

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

This can be rearranged as $9x^2 + 16y^2 - 24xy$.

Correct Option: (d) $9x^2 + 16y^2 - 24xy$

Like terms are terms that have exactly the same literal (variable) factors with the same exponents.

(a) Variable parts are $xyz^2$ and $xy^2z$. Powers of $y$ and $z$ are different.

(b) Variable parts are $xyz^2$ and $xyz^2$. These are exactly the same. (Like terms)

(c) Variable parts are $xyz^2$ and $x^2yz$. Powers of $x$ and $z$ are different.

(d) Variable parts are $xyz^2$ and $x^2y^2z^2$. Powers of all variables are different.

Correct Option: (b) $-5xyz^2$ and $7xyz^2$

The coefficient of a variable in a term is the numerical factor multiplied by that variable.

The term $\frac{-y}{3}$ can be rewritten as:

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

Here, the numerical factor multiplied by $y$ is $\frac{-1}{3}$.

Correct Option: (c) $\frac{-1}{3}$

This is a standard algebraic identity representing the difference of two squares.

The identity states that:

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

Verification:

$(a + b)(a - b) = a(a - b) + b(a - b)$

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

$= a^2 - b^2$

Correct Option: (c) $(a + b)(a - b)$

To find the common factor, we first express each term in its irreducible factor form:

$17abc = 17 \times a \times b \times c$

$34ab^2 = 2 \times 17 \times a \times b \times b$

$51a^2b = 3 \times 17 \times a \times a \times b$

The common factors in all three terms are $17$, $a$, and $b$.

$\text{Common factor} = 17 \times a \times b = 17ab$

Correct Option: (b) 17ab

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$:

Here, $a = 9x$ and $b = 7xy$.

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

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

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

Correct Option: (c) 81x² + 49x²y² – 126x²y

We use the grouping method to factorise the given expression:

$23xy - 46x + 54y - 108$

$= (23xy - 46x) + (54y - 108)$

Factor out the common terms from each group:

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

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

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

Correct Option: (a) (23x + 54) (y – 2)

We use the splitting the middle term method. We need two numbers whose sum is $-10$ and product is $21$.

The numbers are $-7$ and $-3$ ($ -7 \times -3 = 21$ and $-7 + (-3) = -10$).

$r^2 - 10r + 21 = r^2 - 7r - 3r + 21$

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

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

Correct Option: (b) (r – 7) (r – 3)

We split the middle term. We need two numbers whose sum is $-17$ and product is $-38$.

The numbers are $-19$ and $+2$ ($-19 \times 2 = -38$ and $-19 + 2 = -17$).

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

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

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

Correct Option: (a) (p – 19) (p + 2)

We write the division as a fraction:

$\frac{57p^2qr}{114pq} = \frac{\cancel{57}^{1}}{\cancel{114}_{2}} \times \frac{p^2}{p} \times \frac{q}{q} \times r$

Using laws of exponents for the variables:

$= \frac{1}{2} \times p^{(2-1)} \times 1 \times r$

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

Correct Option: (c) $\frac{1}{2}$ pr

First, factorise the expression in the numerator:

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

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

$= 4p(p - 2)(p + 2)$

Now, perform the division:

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

Cancelling common factors:

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

Correct Option: (c) p + 2

The irreducible factors are:

$3ab = 3 \times a \times b \times 1$

$2cd = 2 \times c \times d \times 1$

There are no common numerical factors (other than 1) and no common variable factors between the two terms.

Therefore, the highest common factor is $1$.

Correct Option: (a) 1

An irreducible factor is a factor that cannot be expressed further as a product of other factors. Let us look at the prime factorisation of the given term:

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

(a) $x^2$ is reducible because it can be written as $x \times x$.

(b) $y^2$ is reducible because it can be written as $y \times y$.

(c) $x$ is irreducible because it cannot be factored further.

(d) $24x$ is reducible because it can be written as $2 \times 2 \times 2 \times 3 \times x$.

Correct Option: (c) x

In algebraic expressions, factors of a term are the parts that are multiplied together to obtain that term.

The expression $(a + b)^2$ can be written as:

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

Here, the factor $(a + b)$ is repeated twice. Therefore, the number of factors is $2$.

Correct Option: (c) 2

To factorise $3x - 24$, we need to find the Highest Common Factor (HCF) of the terms $3x$ and $24$.

$3x = 3 \times x$

$24 = 3 \times 8$

The HCF is $3$. Taking $3$ as a common factor out of the brackets:

$3x - 24 = 3(x - 8)$

Correct Option: (b) 3 (x – 8)

We can write $4$ as $2^2$. The expression then becomes a difference of two squares.

$x^2 - 4 = x^2 - 2^2$

Using the identity $a^2 - b^2 = (a + b)(a - b)$, where $a = x$ and $b = 2$:

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

Therefore, the factors are $(x + 2)$ and $(x - 2)$.

Correct Option: (b) (x + 2), (x – 2)

To divide the monomials, we express them as a fraction:

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

Separating the numerical and literal parts:

$\frac{\cancel{-27}^{3}}{\cancel{-9}_{1}} \times \frac{x^2}{x} \times \frac{y}{y}$

$= 3 \times x^{(2-1)} \times 1$

$= 3x$

Correct Option: (d) 3x

To divide $(2x^2 + 4)$ by $2$, we first factorise the numerator by taking the common factor out:

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

Now, perform the division:

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

Correct Option: (b) x² + 2

We divide each term of the trinomial by the monomial $3x$:

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

Simplify each term separately:

$\frac{3x^3}{3x} = x^{(3-1)} = x^2$

$\frac{9x^2}{3x} = 3 \times x^{(2-1)} = 3x$

$\frac{27x}{3x} = 9 \times 1 = 9$

The result is $x^2 + 3x + 9$.

Correct Option: (d) x² + 3x + 9

Using the standard algebraic identities:

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

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

Adding the two expressions:

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

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

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

Correct Option: (c) 2a² + 2b²

We use the standard algebraic identities to expand both parts of the expression:

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

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

Now, subtract the second expression from the first:

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

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

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

$= 0 + 0 + 4ab$

$= 4ab$

Correct Option: (a) 4ab

According to the rules of multiplication for signs:

$(+) \times (+) = (+)$

$(-) \times (-) = (+)$

In both cases, when the signs are the same (like signs), the resulting product is always positive.

Answer: positive

According to the rules of multiplication for signs:

$(+) \times (-) = (-)$

$(-) \times (+) = (-)$

When the signs are different (unlike signs), the resulting product is always negative.

Answer: negative

Using the Distributive Law of multiplication over addition:

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

Comparing this with the given blank format $a \times \text{____} + a \times \text{____}$:

The blanks should be filled with $b$ and $c$.

Answer: b, c

The expression $a^2 - 2ab + b^2$ is the result of multiplying $(a - b)$ by itself.

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

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

Answer: $^2$ (or square)

This is the standard algebraic identity for the difference of two squares.

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

Verification: $(a + b)(a - b) = a(a - b) + b(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2$.

Answer: (a – b)

Let the missing term be $X$.

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

Expanding the identity:

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

To find $X$, rearrange the equation:

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

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

$X = 2ab - 2b^2$

Answer: $2ab - 2b^2$

Expand the squared binomial using the identity $(a + b)^2 = a^2 + 2ab + b^2$:

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

Combine the like terms:

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

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

$= a^2 + b^2$

Answer: $a^2, b^2$

Solution:

According to the standard algebraic identity for the product of two binomials with a common term:

By comparing the given expression with the identity, the blank should be filled with $ab$.

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

Solution:

In algebra, when we multiply two polynomials, the resulting expression is always another polynomial. This is because the multiplication involves the distributive property, and the powers of the variables remain non-negative integers.

Hence, the product of two polynomials is a polynomial.

Solution:

Given: The expression is $ax^2 + bx$.

To find the common factor, we express each term as a product of its irreducible factors:

$ax^2 = a \times x \times x$

$bx = b \times x$

The common factor in both terms is $x$.

Therefore, we can write:

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

Hence, the common factor of $ax^2 + bx$ is $x$.

Solution:

Given: The expression is $18mn + 10mnp$.

We find the Highest Common Factor (HCF) of the two terms:

Factors of $18mn = 2 \times 3 \times 3 \times m \times n$

Factors of $10mnp = 2 \times 5 \times m \times n \times p$

Common factors are $2$, $m$, and $n$. So, the HCF is $2mn$.

Taking out the common factor $2mn$:

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

Hence, the factorised form is $2mn(9 + 5p)$.

Solution:

Given: $4y^2 - 12y + 9$

We can rewrite the terms as squares and a product:

$4y^2 = (2y)^2$

$9 = (3)^2$

$12y = 2(2y)(3)$

The expression is in the form $a^2 - 2ab + b^2$:

Here, $a = 2y$ and $b = 3$.

So, the factorised form is $(2y - 3)^2$ or $(2y - 3)(2y - 3)$.

Solution:

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

We can divide the numerical coefficients and the variables separately:

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

Using the laws of exponents $\frac{a^m}{a^n} = a^{m-n}$:

$2 \times x^{3-1} \times y^{2-2} \times z$

$2 \times x^2 \times y^0 \times z$

Hence, $38x^3y^2z \div 19xy^2 =$ $2x^2z$.

Solution:

Given:

Length $(l) = 2x$

Breadth $(b) = 3y$

Height $(h) = 4z$

The formula for the Volume $(V)$ of a rectangular box is:

Substituting the given values:

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

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

$V = 24xyz$

Hence, the volume is $24xyz$ cubic units.

Solution:

The expression is in the form of the identity $a^2 - b^2$.

Here, $a = 67$ and $b = 37$.

$67^2 - 37^2 = (67 - 37) \times (67 + 37)$

Calculating the values:

$(67 - 37) = 30$

$(67 + 37) = 104$

Product: $30 \times 104 = 3120$

Hence, $67^2 - 37^2 = (67 - 37) \times$ $(67 + 37)$ $=$ $3120$.

Solution:

Using the algebraic identity $a^2 - b^2 = (a + b)(a - b)$:

Calculating the values:

$(103 + 102) = 205$

$(103 - 102) = 1$

Product: $205 \times 1 = 205$

Hence, $103^2 - 102^2 =$ $(103 + 102)$ $\times (103 - 102) =$ $205$.

Given:

Side 1 (Length, $l$) = $4x^2$

Side 2 (Breadth, $b$) = $3y^2$

To Find: Area of the rectangular plot.

Solution:

The area of a rectangle is calculated by the product of its adjacent sides.

Substituting the given values:

Area = $4x^2 \times 3y^2$

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

Area = $12x^2y^2$

Hence, the area of the rectangular plot is $12x^2y^2$.

Given:

Length ($l$) = $2x$

Breadth ($b$) = $2x$

Height ($h$) = $2x$

To Find: Volume of the rectangular box.

Solution:

The volume ($V$) of a rectangular box (cuboid) is given by:

Substituting the values in equation (i):

$V = 2x \times 2x \times 2x$

$V = (2 \times 2 \times 2) \times (x \times x \times x)$

$V = 8x^3$

Hence, the volume of the rectangular box is $8x^3$.

Solution:

In an algebraic term, the numerical factor is called its numerical coefficient or simply the coefficient.

The given term is $-37abc$.

Here, the numerical factor is $-37$.

Hence, the coefficient in $-37abc$ is $-37$.

Solution:

Terms are added or subtracted to form an expression. Parts of an expression which are formed by products of variables and constants are single terms.

The given expression is $a^2 + bc \times d$.

We can simplify the second part: $bc \times d = bcd$.

So, the expression becomes $a^2 + bcd$.

The terms are:

1. $a^2$

2. $bcd$

Since there are two distinct parts separated by a plus sign, there are 2 terms.

Hence, the number of terms is $2$.

Solution:

The area of a square is given by $(\text{side})^2$.

Area of the first square with side $4a = (4a)^2 = 16a^2$.

Area of the second square with side $4b = (4b)^2 = 16b^2$.

Sum of the areas = $16a^2 + 16b^2$.

We can take $16$ as a common factor:

Sum = $16(a^2 + b^2)$.

Hence, the sum of areas is $16a^2 + 16b^2$ or $16(a^2 + b^2)$.

Solution:

In the common factor method, we look for factors that are common to every term in the expression and then use the property $a(b + c) = ab + ac$ in reverse.

This property is known as the Distributive property of multiplication over addition.

Hence, the method is based on the distributive property.

Given: Area of the square = $9y^2$

Solution:

We know that the area of a square is given by:

To find the side, we take the square root of the area:

$\text{Side} = \sqrt{\text{Area}}$

$\text{Side} = \sqrt{9y^2}$

$\text{Side} = \sqrt{3 \times 3 \times y \times y}$

$\text{Side} = 3y$

Hence, the side of the square is $3y$.

Solution:

The given expression is $\frac{3x + 3}{3}$.

First, we factorise the numerator by taking the common factor $3$ out:

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

Now, substituting this back into the expression:

$\frac{3(x + 1)}{3}$

Cancelling the common factor $3$ from the numerator and denominator:

$\frac{\cancel{3}^1(x + 1)}{\cancel{3}_1} = x + 1$

Hence, the simplified form is $x + 1$.

Solution:

To factorise $2x + 4y$, we identify the greatest common factor of the terms $2x$ and $4y$.

Factors of $2x = 2 \times x$

Factors of $4y = 2 \times 2 \times y$

The common factor is $2$.

Taking $2$ out as a common factor:

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

Hence, the factorisation is $2(x + 2y)$.

Solution:

The given statement is False.

According to the standard algebraic identity, the square of a binomial is given by:

The term $2ab$ is missing in the given statement, therefore it is incorrect.

Solution:

The given statement is False.

The expansion of $(a - b)^2$ is:

The expression $a^2 - b^2$ represents the difference of squares, which is actually equal to $(a - b)(a + b)$.

Solution:

The given statement is True.

This is a standard algebraic identity called the Difference of Squares identity.

Since the LHS equals the RHS, the statement is correct.

Solution:

The given statement is False.

According to the rules of multiplication for integers and algebraic terms, the product of two negative values is always a positive value.

Example: $(-2x) \times (-3y) = 6xy$, which is a positive term.

Solution:

The given statement is True.

According to the rules of signs, when a positive term is multiplied by a negative term, the resulting product is always negative.

Example: $(5a) \times (-2b) = -10ab$.

Solution:

The given statement is True.

In an algebraic term, the numerical part (including its sign) is known as the numerical coefficient.

In the term $-6x^2y^2$, the numerical factor multiplied with the variables is $-6$.

Solution:

The given statement is False.

A binomial is an algebraic expression containing exactly two terms. Let's analyze the given expression:

$p^2q + q^2r + r^2q$

This expression consists of three distinct terms: $p^2q$, $q^2r$, and $r^2q$. Since there are three terms, it is a trinomial, not a binomial.

Solution:

The given statement is False.

Using the identity for the square of a difference:

Expanding the square, we get:

The factors are $(a - b)$ and $(a - b)$. The factors $(a + b)$ and $(a + b)$ would result in $a^2 + 2ab + b^2$.

Solution:

The given statement is False.

An expression is a factor of another if it divides the latter completely. In the expression $2\pi(h + r)$, the individual factors are $2$, $\pi$, and $(h + r)$.

While $h$ is a term within the factor $(h + r)$, it is not a factor of the entire expression because we cannot divide $2\pi(h + r)$ by $h$ to get a polynomial result (unless $r$ is also a multiple of $h$).

For example, if we expand: $2\pi h + 2\pi r$. $h$ is present in the first term but not in the second term as a factor.

Solution:

The given statement is True.

To find the factors, let us factorise the given expression:

Given expression: $\frac{n^2}{2} + \frac{n}{2}$

We can take out the common factor $\frac{1}{2}$ from both terms:

$\frac{1}{2}(n^2 + n)$

Now, within the bracket, we can see that $n$ is a common factor for $n^2$ and $n$:

$\frac{1}{2} \cdot n \cdot (n + 1)$

As we can see, the expression is a product of $\frac{1}{2}$, $n$, and $(n + 1)$. Therefore, they are all factors of the given expression.

Solution:

The given statement is False.

In algebra, we distinguish between an Identity and an Equation:

1. An Identity is an equality that is true for all values of the variables. For example: $(a+b)^2 = a^2 + 2ab + b^2$.

2. An Equation is an equality that is true only for certain specific values of the variables. For example: $2x + 3 = 7$ is only true when $x = 2$.

Therefore, saying an equation is true for all values is incorrect.

Solution:

The given statement is False.

Let us check the Left Hand Side (LHS) and Right Hand Side (RHS):

Now let's expand the RHS:

RHS: $(a + b)(x + ab) = ax + a^2b + bx + ab^2$

Clearly, $(x + a)(x + b) \neq (a + b)(x + ab)$. Hence the statement is false.

Solution:

The given statement is False.

Let's find the Highest Common Factor (HCF) of the three terms:

1. $11pq^2 = 11 \times p \times q \times q$

2. $121p^2q^3 = 11 \times 11 \times p \times p \times q \times q \times q$

3. $1331p^2q = 11 \times 11 \times 11 \times p \times p \times q$

Common numerical factor: $11$

Common variable factors: $p$ (lowest power is $p^1$) and $q$ (lowest power is $q^1$).

HCF = $11 \times p \times q = 11pq$.

The statement claims the common factor is $11p^2q^2$, which is incorrect.

Solution:

The given statement is True.

Let's find the common factors of the three terms in the expression:

Term 1: $12a^2b^2 = 4 \times 3 \times a^2 \times b^2$

Term 2: $4ab^2 = 4 \times 1 \times a \times b^2$

Term 3: $-32 = 4 \times (-8)$

As we can see, the number $4$ is the greatest numerical factor common to all three terms. Since the third term $(-32)$ has no variables, there are no variable common factors.

Thus, the common factor is indeed $4$.

Solution:

The given statement is False.

Let's factorise the expression $-3a^2 + 3ab + 3ac$:

Taking $3a$ as a common factor from each term:

$-3a^2 = 3a \times (-a)$

$3ab = 3a \times (b)$

$3ac = 3a \times (c)$

So, $-3a^2 + 3ab + 3ac = 3a(-a + b + c)$.

The statement says $3a(-a - b - c)$, which has incorrect signs for the $b$ and $c$ terms.

Solution:

The given statement is False.

To check this, let us multiply the factors provided in the statement:

$(p + 18)(p - 12) = p(p - 12) + 18(p - 12)$

$= p^2 - 12p + 18p - 216$

$= p^2 + 6p - 216$

This does not match the original expression $p^2 + 30p + 216$.

Note: The correct factorisation of $p^2 + 30p + 216$ is $(p + 18)(p + 12)$ because $18 + 12 = 30$ and $18 \times 12 = 216$.

Solution:

The given statement is True.

Let the two consecutive numbers be $n$ and $n+1$.

Difference of their squares:

Using identity $a^2 - b^2 = (a+b)(a-b)$:

$[(n+1) + n] \times [(n+1) - n]$

$[2n + 1] \times [1]$

$2n + 1$

Now, let's find their sum:

Since (i) = (ii), the difference of squares of two consecutive numbers is equal to their sum.

Solution:

The given statement is True.

An algebraic expression is called a monomial if it contains only one term. In the given expression $abc + bca + cab$, let us observe the terms:

By the commutative property of multiplication, we know that $abc = bca = cab$.

Therefore, we can simplify the expression as:

$abc + abc + abc = 3abc$

Since the simplified expression $3abc$ consists of only one single term, it is a monomial.

Solution:

The given statement is False.

To find the quotient, we perform the division operation:

$\frac{p}{3} \div \frac{3}{p}$

To divide by a fraction, we multiply by its reciprocal:

$\frac{p}{3} \times \frac{p}{3}$

$\frac{p \times p}{3 \times 3} = \frac{p^2}{9}$

The quotient is $\frac{p^2}{9}$, not $9$.

Solution:

The given statement is True.

Given: $51^2 - 49^2 = 100p$

Using the identity $a^2 - b^2 = (a - b)(a + b)$ for the left-hand side (LHS):

$(51 - 49)(51 + 49) = 100p$

$(2)(100) = 100p$

$200 = 100p$

$p = \frac{200}{100}$

$p = 2$

The value of $p$ is indeed $2$.

Solution:

The given statement is False.

Let us perform the division:

$\frac{9x - 51}{9}$

Distributing the divisor over the numerator:

$\frac{9x}{9} - \frac{51}{9}$

$x - \frac{17}{3}$

As we can see, $x - \frac{17}{3} \neq x - 51$. The division must apply to every term in the expression.

Solution:

The given statement is True.

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

First, we multiply $(a + 1)(a - 1)$:

Now, we multiply the result from (i) with $(a^2 + 1)$:

$(a^2 - 1)(a^2 + 1)$

Again using the same identity where $x = a^2$ and $y = 1$:

$(a^2)^2 - (1)^2 = a^4 - 1$

Thus, $(a + 1)(a - 1)(a^2 + 1) = a^4 - 1$.

Solution (i):

We need to add the like terms together:

$(7a^2bc + 3a^2bc) + (-3abc^2 + 2abc^2)$

$\begin{array}{cc} & 7a^2bc & -3abc^2 \\ + & 3a^2bc & +2abc^2 \\ \hline & 10a^2bc & -abc^2 \\ \hline \end{array}$

The sum is $10a^2bc - abc^2$.

Solution (ii):

Summing the given expressions by grouping like terms:

$(9ax + ax) + (3by - 5by) + (-cz + 3cz)$

$\begin{array}{cc} & 9ax & +3by & -cz \\ + & ax & -5by & +3cz \\ \hline & 10ax & -2by & +2cz \\ \hline \end{array}$

The sum is $10ax - 2by + 2cz$.

Solution (iii):

Group the like terms:

$(xy^2z^2 + 3xy^2z^2) + (3x^2y^2z - 9x^2y^2z) + (-4x^2yz^2 + x^2yz^2)$

$\begin{array}{cc} & xy^2z^2 & +3x^2y^2z & -4x^2yz^2 \\ + & 3xy^2z^2 & -9x^2y^2z & +x^2yz^2 \\ \hline & 4xy^2z^2 & -6x^2y^2z & -3x^2yz^2 \\ \hline \end{array}$

The sum is $4xy^2z^2 - 6x^2y^2z - 3x^2yz^2$.

Solution (iv):

Arrange the terms in order and add:

$(5x^2 - 2x^2) + (-3xy + 5xy) + (4y^2 + 7y^2) + (-9 + 13)$

$\begin{array}{cc} & 5x^2 & -3xy & +4y^2 & -9 \\ + & -2x^2 & +5xy & +7y^2 & +13 \\ \hline & 3x^2 & +2xy & +11y^2 & +4 \\ \hline \end{array}$

The sum is $3x^2 + 2xy + 11y^2 + 4$.

Solution (v):

Add the coefficients of the terms with the same powers of $p$:

$\begin{array}{cc} & 2p^4 & -3p^3 & +p^2 & -5p & +7 \\ + & -3p^4 & -7p^3 & -3p^2 & -p & -12 \\ \hline & -p^4 & -10p^3 & -2p^2 & -6p & -5 \\ \hline \end{array}$

The sum is $-p^4 - 10p^3 - 2p^2 - 6p - 5$.

Solution (vi):

First, expand each expression:

$3a(a - b + c) = 3a^2 - 3ab + 3ac$

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

Now, add them together:

$3a^2 - 3ab + 3ac + 2ab - 2b^2 + 2bc$

$3a^2 + (-3ab + 2ab) + 3ac - 2b^2 + 2bc$

$3a^2 - ab + 3ac - 2b^2 + 2bc$

Solution (vii):

First, expand each expression:

$3a(2b + 5c) = 6ab + 15ac$

$3c(2a + 2b) = 6ac + 6bc$

Now, add them together:

$6ab + 15ac + 6ac + 6bc$

$6ab + (15ac + 6ac) + 6bc$

$6ab + 21ac + 6bc$

Solution (i):

We need to subtract $5a^2b^2c^2$ from $-7a^2b^2c^2$.

$\begin{array}{cc} & -7a^2b^2c^2 \\ - & 5a^2b^2c^2 \\ \hline & -12a^2b^2c^2 \\ \hline \end{array}$

The result is $-12a^2b^2c^2$.

Solution (ii):

Rearranging terms to align like variables:

Subtract $(6x^2 - 4xy + 5y^2)$ from $(-3x^2 + 6xy + 8y^2)$.

$\begin{array}{cc} & -3x^2 & +6xy & +8y^2 \\ - & (6x^2 & -4xy & +5y^2) \\ \hline & -9x^2 & +10xy & +3y^2 \\ \hline \end{array}$

The result is $-9x^2 + 10xy + 3y^2$.

Solution (iii):

Aligning like terms based on variables and powers:

$\begin{array}{cc} & -10a^2b^2c & +4ab^2c^2 & +2a^2bc^2 \\ - & (4a^2b^2c & +2ab^2c^2 & -5a^2bc^2) \\ \hline & -14a^2b^2c & +2ab^2c^2 & +7a^2bc^2 \\ \hline \end{array}$

The result is $-14a^2b^2c + 2ab^2c^2 + 7a^2bc^2$.

Solution (iv):

Subtracting term by term:

$\begin{array}{cc} & -4t^4 & +8t^3 & -4t^2 & -2t & +11 \\ - & (3t^4 & -4t^3 & +2t^2 & -6t & +6) \\ \hline & -7t^4 & +12t^3 & -6t^2 & +4t & +5 \\ \hline \end{array}$

The result is $-7t^4 + 12t^3 - 6t^2 + 4t + 5$.

Solution (v):

$\begin{array}{cc} & 5ab & -2bc & -2ac & +10abc \\ - & (2ab & +5bc & -7ac & +0) \\ \hline & 3ab & -7bc & +5ac & +10abc \\ \hline \end{array}$

The result is $3ab - 7bc + 5ac + 10abc$.

Solution (vi):

First, expand both expressions:

$7p(3q + 7p) = 21pq + 49p^2$

$8p(2p - 7q) = 16p^2 - 56pq$

Subtracting: $(16p^2 - 56pq) - (49p^2 + 21pq)$

$\begin{array}{cc} & 16p^2 & -56pq \\ - & 49p^2 & +21pq \\ \hline & -33p^2 & -77pq \\ \hline \end{array}$

The result is $-33p^2 - 77pq$.

Solution (vii):

Expand the target expression: $3p(-p - a - r) = -3p^2 - 3pa - 3pr$

Subtracting: $(-3p^2 - 3pa - 3pr) - (-3p^2 + 3pq + 3px)$

$= -3p^2 - 3pa - 3pr + 3p^2 - 3pq - 3px$

$= -3pa - 3pr - 3pq - 3px$

The result is $-3pa - 3pr - 3pq - 3px$.

(i) $-7pq^2r^3$ and $-13p^3q^2r$

To multiply these monomials, we multiply their numerical coefficients and their literal factors separately.

Product $= (-7 \times -13) \times (p \times p^3) \times (q^2 \times q^2) \times (r^3 \times r)$

Using the law of exponents $a^m \times a^n = a^{m+n}$:

Product $= 91 \times p^{1+3} \times q^{2+2} \times r^{3+1}$

Final Result: $91p^4q^4r^4$

(ii) $3x^2y^2z^2$ and $17xyz$

Product $= (3 \times 17) \times (x^2 \times x) \times (y^2 \times y) \times (z^2 \times z)$

Product $= 51 \times x^{2+1} \times y^{2+1} \times z^{2+1}$

Final Result: $51x^3y^3z^3$

(iii) $15xy^2$ and $17yz^2$

Product $= (15 \times 17) \times x \times (y^2 \times y) \times z^2$

Product $= 255 \times x \times y^{2+1} \times z^2$

Final Result: $255xy^3z^2$

(iv) $-5a^2bc, 11ab$ and $13abc^2$

Product $= (-5 \times 11 \times 13) \times (a^2 \times a \times a) \times (b \times b \times b) \times (c \times c^2)$

Product $= -715 \times a^{2+1+1} \times b^{1+1+1} \times c^{1+2}$

Final Result: $-715a^4b^3c^3$

(v) $-3x^2y$ and $(5y - xy)$

Using the Distributive Property: $a(b - c) = ab - ac$

Product $= (-3x^2y \times 5y) - (-3x^2y \times xy)$

Product $= -15x^2y^2 - (-3x^3y^2)$

Final Result: $-15x^2y^2 + 3x^3y^2$

(vi) $abc$ and $(bc + ca)$

Product $= (abc \times bc) + (abc \times ca)$

Product $= ab^{1+1}c^{1+1} + a^{1+1}bc^{1+1}$

Final Result: $ab^2c^2 + a^2bc^2$

(vii) $7pqr$ and $(p - q + r)$

Product $= (7pqr \times p) - (7pqr \times q) + (7pqr \times r)$

Final Result: $7p^2qr - 7pq^2r + 7pqr^2$

(viii) $x^2y^2z^2$ and $(xy - yz + zx)$

Product $= (x^2y^2z^2 \times xy) - (x^2y^2z^2 \times yz) + (x^2y^2z^2 \times zx)$

Final Result: $x^3y^3z^2 - x^2y^3z^3 + x^3y^2z^3$

(ix) $(p + 6)$ and $(q - 7)$

Using binomial multiplication: $(a+b)(c+d) = a(c+d) + b(c+d)$

Product $= p(q - 7) + 6(q - 7)$

Product $= pq - 7p + 6q - 42$

Final Result: $pq - 7p + 6q - 42$

(x) $6mn$ and $0mn$

Any term multiplied by zero results in zero.

Product $= (6 \times 0) \times (m \times m) \times (n \times n)$

Final Result: $0$

(xi) $a, a^5$ and $a^6$

Product $= a^1 \times a^5 \times a^6 = a^{1+5+6}$

Final Result: $a^{12}$

(xii) $-7st, -1$ and $-13st^2$

Product $= (-7 \times -1 \times -13) \times (s \times s) \times (t \times t^2)$

Product $= -91 \times s^{1+1} \times t^{1+2}$

Final Result: $-91s^2t^3$

(xiii) $b^3, 3b^2$ and $7ab^5$

Product $= (1 \times 3 \times 7) \times a \times (b^3 \times b^2 \times b^5)$

Product $= 21 \times a \times b^{3+2+5}$

Final Result: $21ab^{10}$

(xiv) $-\frac{100}{9}rs$ and $\frac{3}{4}r^3s^2$

Product $= \left( - \frac{100}{9} \times \frac{3}{4} \right) \times (r \times r^3) \times (s \times s^2)$

Simplifying the fraction:

Product $= \left( - \frac{\cancel{100}^{25}}{\cancel{9}_3} \times \frac{\cancel{3}^1}{\cancel{4}_1} \right) \times r^4 \times s^3$

Final Result: $ - \frac{25}{3}r^4s^3$

(xv) $(a^2 - b^2)$ and $(a^2 + b^2)$

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

Here, $x = a^2$ and $y = b^2$.

Product $= (a^2)^2 - (b^2)^2$

Final Result: $a^4 - b^4$

(xvi) $(ab + c)$ and $(ab + c)$

This is $(ab + c)^2$. Using the identity: $(x + y)^2 = x^2 + 2xy + y^2$

Product $= (ab)^2 + 2(ab)(c) + c^2$

Final Result: $a^2b^2 + 2abc + c^2$

(xvii) $(pq - 2r)$ and $(pq - 2r)$

This is $(pq - 2r)^2$. Using the identity: $(x - y)^2 = x^2 - 2xy + y^2$

Product $= (pq)^2 - 2(pq)(2r) + (2r)^2$

Final Result: $p^2q^2 - 4pqr + 4r^2$

(xviii) $\left( \frac{3}{4}x - \frac{4}{3}y \right)$ and $\left( \frac{2}{3}x + \frac{3}{2}y \right)$

Multiplying term by term:

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

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

$= \frac{1}{2}x^2 + \frac{9}{8}xy - \frac{8}{9}xy - 2y^2$

To combine $xy$ terms, take LCM of $8$ and $9$ which is $72$:

$= \frac{1}{2}x^2 + \left( \frac{81 - 64}{72} \right)xy - 2y^2$

Final Result: $\frac{1}{2}x^2 + \frac{17}{72}xy - 2y^2$

(xix) $\left( \frac{3}{2}p^2 + \frac{2}{3}q^2 \right)$ and $(2p^2 - 3q^2)$

$= \frac{3}{2}p^2(2p^2 - 3q^2) + \frac{2}{3}q^2(2p^2 - 3q^2)$

$= \left( \frac{3}{2} \times 2 \right)p^4 - \left( \frac{3}{2} \times 3 \right)p^2q^2 + \left( \frac{2}{3} \times 2 \right)q^2p^2 - \left( \frac{2}{3} \times 3 \right)q^4$

$= 3p^4 - \frac{9}{2}p^2q^2 + \frac{4}{3}p^2q^2 - 2q^4$

Combining $p^2q^2$ terms with LCM $6$:

$= 3p^4 + \left( \frac{-27 + 8}{6} \right)p^2q^2 - 2q^4$

Final Result: $3p^4 - \frac{19}{6}p^2q^2 - 2q^4$

(xx) $(x^2 - 5x + 6)$ and $(2x + 7)$

$= 2x(x^2 - 5x + 6) + 7(x^2 - 5x + 6)$

$= (2x^3 - 10x^2 + 12x) + (7x^2 - 35x + 42)$

Grouping like terms:

$= 2x^3 + (-10x^2 + 7x^2) + (12x - 35x) + 42$

Final Result: $2x^3 - 3x^2 - 23x + 42$

(xxi) $(3x^2 + 4x - 8)$ and $(2x^2 - 4x + 3)$

$= 3x^2(2x^2 - 4x + 3) + 4x(2x^2 - 4x + 3) - 8(2x^2 - 4x + 3)$

$= (6x^4 - 12x^3 + 9x^2) + (8x^3 - 16x^2 + 12x) - (16x^2 - 32x + 24)$

$= 6x^4 - 12x^3 + 8x^3 + 9x^2 - 16x^2 - 16x^2 + 12x + 32x - 24$

Final Result: $6x^4 - 4x^3 - 23x^2 + 44x - 24$

(xxii) $(2x - 2y - 3)$ and $(x + y + 5)$

$= 2x(x + y + 5) - 2y(x + y + 5) - 3(x + y + 5)$

$= (2x^2 + 2xy + 10x) - (2xy + 2y^2 + 10y) - (3x + 3y + 15)$

$= 2x^2 + 2xy - 2xy - 2y^2 + 10x - 3x - 10y - 3y - 15$

Final Result: $2x^2 - 2y^2 + 7x - 13y - 15$

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

To simplify this expression, we use the algebraic identities for $(a + b)^2$ and $(a - b)^2$.

Expanding the given terms:

$(3x + 2y)^2 = (3x)^2 + 2(3x)(2y) + (2y)^2 = 9x^2 + 12xy + 4y^2$

$(3x - 2y)^2 = (3x)^2 - 2(3x)(2y) + (2y)^2 = 9x^2 - 12xy + 4y^2$

Adding both expressions:

$(9x^2 + 12xy + 4y^2) + (9x^2 - 12xy + 4y^2)$

$= 9x^2 + 9x^2 + 12xy - 12xy + 4y^2 + 4y^2$

$= 18x^2 + 8y^2$

Alternative Method:

$= 2[(3x)^2 + (2y)^2] = 2(9x^2 + 4y^2) = 18x^2 + 8y^2$

Final Result: $18x^2 + 8y^2$

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

We can use the identity for the difference of squares of sum and difference:

Here, $a = 3x$ and $b = 2y$.

$= 4(3x)(2y)$

$= 24xy$

Final Result: $24xy$

(iii) $\left( \frac{7}{9}a + \frac{9}{7}b \right)^2 - ab$

Expanding the squared term using $(x + y)^2 = x^2 + 2xy + y^2$:

$= \left( \frac{7}{9}a \right)^2 + 2 \left( \frac{7}{9}a \right) \left( \frac{9}{7}b \right) + \left( \frac{9}{7}b \right)^2 - ab$

$= \frac{49}{81}a^2 + 2 \left( \frac{\cancel{7}}{\cancel{9}} \cdot \frac{\cancel{9}}{\cancel{7}} \right) ab + \frac{81}{49}b^2 - ab$

$= \frac{49}{81}a^2 + 2ab + \frac{81}{49}b^2 - ab$

Combining like terms ($2ab - ab$):

$= \frac{49}{81}a^2 + ab + \frac{81}{49}b^2$

Final Result: $\frac{49}{81}a^2 + ab + \frac{81}{49}b^2$

(iv) $\left( \frac{3}{4}x - \frac{4}{3}y \right)^2 + 2xy$

Expanding the squared term using $(a - b)^2 = a^2 - 2ab + b^2$:

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

$= \frac{9}{16}x^2 - 2 \left( \frac{\cancel{3}}{\cancel{4}} \cdot \frac{\cancel{4}}{\cancel{3}} \right) xy + \frac{16}{9}y^2 + 2xy$

$= \frac{9}{16}x^2 - 2xy + \frac{16}{9}y^2 + 2xy$

The $-2xy$ and $+2xy$ cancel each other out.

Final Result: $\frac{9}{16}x^2 + \frac{16}{9}y^2$

(v) $(1.5p + 1.2q)^2 - (1.5p - 1.2q)^2$

Using the identity $(a + b)^2 - (a - b)^2 = 4ab$:

Here, $a = 1.5p$ and $b = 1.2q$.

$= 4(1.5p)(1.2q)$

$= 4 \times 1.5 \times 1.2 \times p \times q$

$= 6.0 \times 1.2 \times pq$

$= 7.2pq$

Final Result: $7.2pq$

(vi) $(2.5m + 1.5q)^2 + (2.5m - 1.5q)^2$

Using the identity $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$:

Here, $a = 2.5m$ and $b = 1.5q$.

$= 2[(2.5m)^2 + (1.5q)^2]$

$= 2(6.25m^2 + 2.25q^2)$

$= 12.5m^2 + 4.5q^2$

Final Result: $12.5m^2 + 4.5q^2$

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

Removing the parentheses and combining like terms:

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

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

$= 2x^2 + 0 + 16$

Final Result: $2x^2 + 16$

(viii) $(ab - c)^2 + 2abc$

Expanding using $(x - y)^2 = x^2 - 2xy + y^2$:

$= (ab)^2 - 2(ab)(c) + c^2 + 2abc$

$= a^2b^2 - 2abc + c^2 + 2abc$

The $-2abc$ and $+2abc$ terms cancel out.

Final Result: $a^2b^2 + c^2$

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

We recognize these as the standard factorisations for the difference and sum of cubes:

Substituting these into the expression:

$= (a^3 - b^3) - (a^3 + b^3)$

$= a^3 - b^3 - a^3 - b^3$

$= -2b^3$

Final Result: $-2b^3$

(x) $(b^2 - 49)(b + 7) + 343$

Expanding the product using distributive property:

$= b^2(b + 7) - 49(b + 7) + 343$

$= b^3 + 7b^2 - 49b - 343 + 343$

$= b^3 + 7b^2 - 49b$

Final Result: $b^3 + 7b^2 - 49b$

(xi) $(4.5a + 1.5b)^2 + (4.5b + 1.5a)^2$

Expanding both squares:

$(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$

Adding them together:

$= (20.25a^2 + 2.25a^2) + (13.5ab + 13.5ab) + (2.25b^2 + 20.25b^2)$

$= 22.5a^2 + 27ab + 22.5b^2$

Final Result: $22.5a^2 + 27ab + 22.5b^2$

(xii) $(pq - qr)^2 + 4pq^2r$

Expanding the squared term:

$= (pq)^2 - 2(pq)(qr) + (qr)^2 + 4pq^2r$

$= p^2q^2 - 2pq^2r + q^2r^2 + 4pq^2r$

Combining like terms ($-2pq^2r + 4pq^2r$):

$= p^2q^2 + 2pq^2r + q^2r^2$

This is now in the form $x^2 + 2xy + y^2$, which is $(pq + qr)^2$.

Final Result: $(pq + qr)^2$ or $p^2q^2 + 2pq^2r + q^2r^2$

(xiii) $(s^2t + tq^2)^2 - (2stq)^2$

Expanding both squares:

$(s^2t + tq^2)^2 = (s^2t)^2 + 2(s^2t)(tq^2) + (tq^2)^2 = s^4t^2 + 2s^2t^2q^2 + t^2q^4$

$(2stq)^2 = 4s^2t^2q^2$

Subtracting them:

$= s^4t^2 + 2s^2t^2q^2 + t^2q^4 - 4s^2t^2q^2$

$= s^4t^2 - 2s^2t^2q^2 + t^2q^4$

This matches the identity for $(a - b)^2 = a^2 - 2ab + b^2$.

Final Result: $(s^2t - tq^2)^2$

(i) $(xy + yz)^2$

Using the identity for the square of a sum:

Here, $a = xy$ and $b = yz$.

$(xy + yz)^2 = (xy)^2 + 2(xy)(yz) + (yz)^2$

$= x^2y^2 + 2xy^2z + y^2z^2$

Final Result: $x^2y^2 + 2xy^2z + y^2z^2$

(ii) $(x^2y - xy^2)^2$

Using the identity for the square of a difference:

Here, $a = x^2y$ and $b = xy^2$.

$(x^2y - xy^2)^2 = (x^2y)^2 - 2(x^2y)(xy^2) + (xy^2)^2$

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

Final Result: $x^4y^2 - 2x^3y^3 + x^2y^4$

(iii) $\left( \frac{4}{5}a + \frac{5}{4}b \right)^2$

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

$\left( \frac{4}{5}a + \frac{5}{4}b \right)^2 = \left( \frac{4}{5}a \right)^2 + 2\left( \frac{4}{5}a \right)\left( \frac{5}{4}b \right) + \left( \frac{5}{4}b \right)^2$

$= \frac{16}{25}a^2 + 2\left( \frac{\cancel{4}}{\cancel{5}} \cdot \frac{\cancel{5}}{\cancel{4}} \right)ab + \frac{25}{16}b^2$

$= \frac{16}{25}a^2 + 2ab + \frac{25}{16}b^2$

Final Result: $\frac{16}{25}a^2 + 2ab + \frac{25}{16}b^2$

(iv) $\left( \frac{2}{3}x - \frac{3}{2}y \right)^2$

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

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

$= \frac{4}{9}x^2 - 2\left( \frac{\cancel{2}}{\cancel{3}} \cdot \frac{\cancel{3}}{\cancel{2}} \right)xy + \frac{9}{4}y^2$

$= \frac{4}{9}x^2 - 2xy + \frac{9}{4}y^2$

Final Result: $\frac{4}{9}x^2 - 2xy + \frac{9}{4}y^2$

(v) $\left( \frac{4}{5}p + \frac{5}{3}q \right)^2$

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

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

$= \frac{16}{25}p^2 + 2\left( \frac{4}{\cancel{5}} \cdot \frac{\cancel{5}}{3} \right)pq + \frac{25}{9}q^2$

$= \frac{16}{25}p^2 + \frac{8}{3}pq + \frac{25}{9}q^2$

Final Result: $\frac{16}{25}p^2 + \frac{8}{3}pq + \frac{25}{9}q^2$

(vi) $(x + 3)(x + 7)$

Using the identity for product of binomials with a common term:

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

$(x + 3)(x + 7) = x^2 + (3 + 7)x + (3 \times 7)$

$= x^2 + 10x + 21$

Final Result: $x^2 + 10x + 21$

(vii) $(2x + 9)(2x - 7)$

Using the identity $(X + a)(X + b) = X^2 + (a + b)X + ab$:

Here, $X = 2x$, $a = 9$, and $b = -7$.

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

$= 4x^2 + (2)(2x) - 63$

$= 4x^2 + 4x - 63$

Final Result: $4x^2 + 4x - 63$

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

Using $(X + a)(X + b) = X^2 + (a + b)X + ab$:

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

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

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

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

Final Result: $\frac{16x^2}{25} + \frac{4xy}{5} + \frac{3y^2}{16}$

(ix) $\left( \frac{2x}{3} - \frac{2}{3} \right) \left( \frac{2x}{3} + \frac{2a}{3} \right)$

Using $(X + a)(X + b) = X^2 + (a + b)X + ab$:

Here, $X = \frac{2x}{3}$, $a = -\frac{2}{3}$, and $b = \frac{2a}{3}$.

$= \left( \frac{2x}{3} \right)^2 + \left( -\frac{2}{3} + \frac{2a}{3} \right)\left( \frac{2x}{3} \right) + \left( -\frac{2}{3} \cdot \frac{2a}{3} \right)$

$= \frac{4x^2}{9} + \left( \frac{2a - 2}{3} \right)\frac{2x}{3} - \frac{4a}{9}$

$= \frac{4x^2}{9} + \frac{4x(a - 1)}{9} - \frac{4a}{9}$

Final Result: $\frac{4x^2}{9} + \frac{4ax - 4x}{9} - \frac{4a}{9}$

(x) $(2x - 5y)(2x - 5y)$

This is $(2x - 5y)^2$. Using $(a - b)^2 = a^2 - 2ab + b^2$:

$= (2x)^2 - 2(2x)(5y) + (5y)^2$

$= 4x^2 - 20xy + 25y^2$

Final Result: $4x^2 - 20xy + 25y^2$

(xi) $\left( \frac{2a}{3} + \frac{b}{3} \right) \left( \frac{2a}{3} - \frac{b}{3} \right)$

Using the identity for difference of squares:

Here, $x = \frac{2a}{3}$ and $y = \frac{b}{3}$.

$= \left( \frac{2a}{3} \right)^2 - \left( \frac{b}{3} \right)^2$

$= \frac{4a^2}{9} - \frac{b^2}{9}$

Final Result: $\frac{4a^2}{9} - \frac{b^2}{9}$

(xii) $(x^2 + y^2)(x^2 - y^2)$

Using $(a + b)(a - b) = a^2 - b^2$ where $a = x^2$ and $b = y^2$:

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

$= x^4 - y^4$

Final Result: $x^4 - y^4$

(xiii) $(a^2 + b^2)^2$

Using $(x + y)^2 = x^2 + 2xy + y^2$ where $x = a^2$ and $y = b^2$:

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

$= a^4 + 2a^2b^2 + b^4$

Final Result: $a^4 + 2a^2b^2 + b^4$

(xiv) $(7x + 5)^2$

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

$= (7x)^2 + 2(7x)(5) + (5)^2$

$= 49x^2 + 70x + 25$

Final Result: $49x^2 + 70x + 25$

(xv) $(0.9p - 0.5q)^2$

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

$= (0.9p)^2 - 2(0.9p)(0.5q) + (0.5q)^2$

$= 0.81p^2 - 0.90pq + 0.25q^2$

Final Result: $0.81p^2 - 0.9pq + 0.25q^2$

(xvi) $x^2y^2 = (xy)^2$

According to the Power of a Product Rule in exponents:

Applying this to $x^2y^2$:

$x^2y^2 = (x \cdot y)^2 = (xy)^2$

The statement effectively shows that the product of the squares is equal to the square of the product.

(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 $b=271$:

$(729)^2 – (271)^2 = (729 - 271)(729 + 271)$

$= (458)(1000)$

$= 458000$

To find the Greatest Common Factor (GCF) or Highest Common Factor (HCF) of algebraic terms, we find the HCF of the numerical coefficients and multiply it by the common literal variables raised to their lowest powers present in all terms.

(i) $-18a^2, 108a$

First, let's find the prime factorisation of the numerical coefficients:

Numerical factors: $18 = 2 \times 3^2$ and $108 = 2^2 \times 3^3$.

The HCF of 18 and 108 is $2 \times 3^2 = 18$.

Common literal factor: The lowest power of $a$ is $a^1$.

Hence, the Greatest Common Factor is $18a$.

(ii) $3x^2y, 18xy^2, -6xy$

Numerical coefficients are 3, 18, and 6.

The HCF of (3, 18, 6) is 3.

Common variables with lowest powers:

Variable $x$: $x^2, x^1, x^1 \implies$ lowest power is $x$.

Variable $y$: $y^1, y^2, y^1 \implies$ lowest power is $y$.

Hence, the Greatest Common Factor is $3xy$.

(iii) $2xy, -y^2, 2x^2y$

Numerical coefficients are 2, -1, and 2. The HCF of the numbers is 1.

Common variables with lowest powers:

Variable $x$: Present in first and third term only, so it is not common.

Variable $y$: $y^1, y^2, y^1 \implies$ lowest power is $y$.

Hence, the Greatest Common Factor is $y$.

(iv) $l^2m^2n, lm^2n^2, l^2mn^2$

The numerical coefficient in all terms is 1.

Common variables with lowest powers:

For $l$: $l^2, l^1, l^2 \implies$ lowest is $l$.

For $m$: $m^2, m^2, m^1 \implies$ lowest is $m$.

For $n$: $n^1, n^2, n^2 \implies$ lowest is $n$.

Hence, the Greatest Common Factor is $lmn$.

(v) $21pqr, -7p^2q^2r^2, 49p^2qr$

Numerical coefficients are 21, 7, and 49. The HCF of (21, 7, 49) is 7.

Common variables with lowest powers:

For $p$: $p^1, p^2, p^2 \implies$ lowest is $p$.

For $q$: $q^1, q^2, q^1 \implies$ lowest is $q$.

For $r$: $r^1, r^2, r^1 \implies$ lowest is $r$.

Hence, the Greatest Common Factor is $7pqr$.

(vi) $qrxy, pryz, rxyz$

Common variables with lowest powers:

Variable $r$ is present in all terms: lowest power is $r$.

Variable $y$ is present in all terms: lowest power is $y$.

Variables $q, p, x, z$ are not present in all terms.

Hence, the Greatest Common Factor is $ry$.

(vii) $3x^3y^2z, -6xy^3z^2, 12x^2yz^3$

Numerical coefficients are 3, 6, and 12. The HCF of (3, 6, 12) is 3.

Common variables with lowest powers:

For $x$: $x^3, x^1, x^2 \implies$ lowest is $x$.

For $y$: $y^2, y^3, y^1 \implies$ lowest is $y$.

For $z$: $z^1, z^2, z^3 \implies$ lowest is $z$.

Hence, the Greatest Common Factor is $3xyz$.

(viii) $63p^2a^2r^2s, -9pq^2r^2s^2, 15p^2qr^2s^2, -60p^2a^2rs^2$

First, find the HCF of 63, 9, 15, and 60.

$63 = 3 \times 3 \times 7$

$9 = 3 \times 3$

$15 = 3 \times 5$

$60 = 2 \times 2 \times 3 \times 5$

The common numerical factor is 3.

Common variables with lowest powers:

For $p$: $p^2, p^1, p^2, p^2 \implies$ lowest is $p$.

For $r$: $r^2, r^2, r^2, r^1 \implies$ lowest is $r$.

For $s$: $s^1, s^2, s^2, s^2 \implies$ lowest is $s$.

Variable $a$ is only in 1st and 4th term. Variable $q$ is only in 2nd and 3rd term.

Hence, the Greatest Common Factor is $3prs$.

(ix) $13x^2y, 169xy$

Numerical coefficients are 13 and 169. Since $169 = 13 \times 13$, the HCF is 13.

Common variables with lowest powers:

For $x$: $x^2, x^1 \implies$ lowest is $x$.

For $y$: $y^1, y^1 \implies$ lowest is $y$.

Hence, the Greatest Common Factor is $13xy$.

(x) $11x^2, 12y^2$

Numerical coefficients are 11 and 12. 11 is prime and 12 is $2 \times 2 \times 3$. They have no common factor other than 1.

Common variables: $x$ is in the first term, $y$ is in the second term. There are no common variables.

Hence, the Greatest Common Factor is $1$.

Factorisation involves writing an algebraic expression as a product of its factors. We use methods such as taking out common factors, grouping terms, or using algebraic identities.

Solution (i): $6ab + 12bc$

First, we identify the Greatest Common Factor (GCF) of the two terms.

Factors of $6ab = 2 \times 3 \times a \times b$

Factors of $12bc = 2 \times 2 \times 3 \times b \times c$

The GCF is $2 \times 3 \times b = 6b$. Taking this common:

$6ab + 12bc = 6b(a + 2c)$

Hence, the factorised form is $6b(a + 2c)$.

Solution (ii): $-xy - ay$

Both terms have $y$ as a common variable. We can also take out the negative sign.

$-xy - ay = -y(x + a)$

Hence, the factorised form is $-y(x + a)$.

Solution (iii): $ax^3 - bx^2 + cx$

All three terms contain the variable $x$. The lowest power of $x$ present is $x^1$.

$ax^3 - bx^2 + cx = x(ax^2 - bx + c)$

Hence, the factorised form is $x(ax^2 - bx + c)$.

Solution (iv): $l^2m^2n - lm^2n^2 - l^2mn^2$

The common literal factors in all three terms are $l$, $m$, and $n$. The lowest power for each is $l^1$, $m^1$, and $n^1$.

Taking $lmn$ common:

$l^2m^2n - lm^2n^2 - l^2mn^2 = lmn(lm - mn - ln)$

Hence, the factorised form is $lmn(lm - mn - ln)$.

Solution (v): $3pqr - 6p^2q^2r^2 - 15r^2$

The numerical GCF of 3, 6, and 15 is 3. The common variable in all terms is $r$ with the lowest power $r^1$.

Taking $3r$ common:

$3pqr - 6p^2q^2r^2 - 15r^2 = 3r(pq - 2p^2q^2r - 5r)$

Hence, the factorised form is $3r(pq - 2p^2q^2r - 5r)$.

Solution (vi): $x^3y^2 + x^2y^3 - xy^4 + xy$

The common factor for all terms is $xy$.

$x^3y^2 + x^2y^3 - xy^4 + xy = xy(x^2y + xy^2 - y^3 + 1)$

Hence, the factorised form is $xy(x^2y + xy^2 - y^3 + 1)$.

Solution (vii): $4xy^2 - 10x^2y + 16x^2y^2 + 2xy$

The numerical HCF of 4, 10, 16, and 2 is 2. The variable HCF is $xy$.

$4xy^2 - 10x^2y + 16x^2y^2 + 2xy = 2xy(2y - 5x + 8xy + 1)$

Hence, the factorised form is $2xy(2y - 5x + 8xy + 1)$.

Solution (viii): $2a^3 - 3a^2b + 5ab^2 - ab$

Taking the common variable $a$ out from all terms:

$2a^3 - 3a^2b + 5ab^2 - ab = a(2a^2 - 3ab + 5b^2 - b)$

Hence, the factorised form is $a(2a^2 - 3ab + 5b^2 - b)$.

Solution (ix): $63p^2q^2r^2s - 9pq^2r^2s^2 + 15p^2qr^2s^2 - 60p^2q^2rs^2$

Numerical HCF of 63, 9, 15, and 60 is 3. Variable HCF is $pqrs$.

$= 3pqrs(21pqr - 3qrs + 5prs - 20pqs)$

Hence, the factorised form is $3pqrs(21pqr - 3qrs + 5prs - 20pqs)$.

Solution (x): $24x^2yz^3 - 6xy^3z^2 + 15x^2y^2z - 5xyz$

The numerical factors have no common factor other than 1 for all four terms simultaneously (due to the coefficient 5). However, $xyz$ is common to all terms.

$24x^2yz^3 - 6xy^3z^2 + 15x^2y^2z - 5xyz = xyz(24xz^2 - 6y^2z + 15xy - 5)$

Hence, the factorised form is $xyz(24xz^2 - 6y^2z + 15xy - 5)$.

Solution (xi): $a^3 + a^2 + a + 1$

We use the method of Factorisation by Grouping.

Group the first two terms and the last two terms:

$(a^3 + a^2) + (a + 1)$

Take $a^2$ common from the first group and $1$ from the second:

$a^2(a + 1) + 1(a + 1)$

Now, $(a + 1)$ is common to both terms:

$(a + 1)(a^2 + 1)$

Hence, the factorised form is $(a + 1)(a^2 + 1)$.

Solution (xii): $lx + my + mx + ly$

Rearrange the terms to group them by common variables:

$(lx + mx) + (ly + my)$

Take $x$ common from the first group and $y$ from the second:

$x(l + m) + y(l + m)$

Take $(l + m)$ common:

$(l + m)(x + y)$

Hence, the factorised form is $(l + m)(x + y)$.

Solution (xiii): $a^3x - x^4 + a^2x^2 - ax^3$

First, take $x$ common from all terms:

$x(a^3 - x^3 + a^2x - ax^2)$

Now group the terms inside the bracket:

$x[(a^3 + a^2x) - (ax^2 + x^3)]$

$x[a^2(a + x) - x^2(a + x)]$

$x(a + x)(a^2 - x^2)$

Using identity $(a^2 - x^2) = (a + x)(a - x)$:

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

Hence, the factorised form is $x(a + x)^2(a - x)$.

Solution (xiv): $2x^2 - 2y + 4xy - x$

Rearrange the terms for easier grouping:

$(2x^2 + 4xy) - (x + 2y)$

$2x(x + 2y) - 1(x + 2y)$

Take $(x + 2y)$ common:

$(x + 2y)(2x - 1)$

Hence, the factorised form is $(x + 2y)(2x - 1)$.

Solution (xv): $y^2 + 8zx - 2xy - 4yz$

Rearrange the terms:

$(y^2 - 2xy) - (4yz - 8zx)$

$y(y - 2x) - 4z(y - 2x)$

Take $(y - 2x)$ common:

$(y - 2x)(y - 4z)$

Hence, the factorised form is $(y - 2x)(y - 4z)$.

Solution (xvi): $ax^2y - bxyz - ax^2z + bxy^2$

Rearrange the terms:

$(ax^2y - ax^2z) + (bxy^2 - bxyz)$

$ax^2(y - z) + bxy(y - z)$

Take $(y - z)$ common:

$(y - z)(ax^2 + bxy)$

We can further take $x$ common from the second bracket:

$x(y - z)(ax + by)$

Hence, the factorised form is $x(y - z)(ax + by)$.

Solution (xvii): $a^2b + a^2c + ab + ac + b^2c + c^2b$

Group the terms in pairs:

$a^2(b + c) + a(b + c) + bc(b + c)$

Now, $(b + c)$ is common to all three terms:

$(b + c)(a^2 + a + bc)$

Hence, the factorised form is $(b + c)(a^2 + a + bc)$.

Solution (xviii): $2ax^2 + 4axy + 3bx^2 + 2ay^2 + 6bxy + 3by^2$

Rearrange and group the terms containing '$a$' and '$b$' separately:

$(2ax^2 + 4axy + 2ay^2) + (3bx^2 + 6bxy + 3by^2)$

Take $2a$ common from the first group and $3b$ from the second:

$2a(x^2 + 2xy + y^2) + 3b(x^2 + 2xy + y^2)$

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

$2a(x + y)^2 + 3b(x + y)^2$

Take $(x + y)^2$ common:

$(2a + 3b)(x + y)^2$

Hence, the factorised form is $(2a + 3b)(x + y)^2$.

(i) $x^2 + 6x + 9$

To factorise this, we identify the square terms:

$x^2$ is the square of $x$ and $9$ is the square of $3$.

Hence, the factorised form is $(x + 3)^2$.

(ii) $x^2 + 12x + 36$

Identifying the square terms:

$x^2 = (x)^2$ and $36 = (6)^2$.

Hence, the factorised form is $(x + 6)^2$.

(iii) $x^2 + 14x + 49$

Identifying the square terms:

$x^2 = (x)^2$ and $49 = (7)^2$.

Hence, the factorised form is $(x + 7)^2$.

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

Identifying the square terms:

$x^2 = (x)^2$ and $1 = (1)^2$.

Hence, the factorised form is $(x + 1)^2$.

(v) $4x^2 + 4x + 1$

Identifying the square terms:

$4x^2 = (2x)^2$ and $1 = (1)^2$.

Hence, the factorised form is $(2x + 1)^2$.

(vi) $a^2x^2 + 2ax + 1$

Identifying the square terms:

$a^2x^2 = (ax)^2$ and $1 = (1)^2$.

Hence, the factorised form is $(ax + 1)^2$.

(vii) $a^2x^2 + 2abx + b^2$

Identifying the square terms:

$a^2x^2 = (ax)^2$ and $b^2 = (b)^2$.

Hence, the factorised form is $(ax + b)^2$.

(viii) $a^2x^2 + 2abxy + b^2y^2$

Identifying the square terms:

$a^2x^2 = (ax)^2$ and $b^2y^2 = (by)^2$.

Hence, the factorised form is $(ax + by)^2$.

(ix) $4x^2 + 12x + 9$

Identifying the square terms:

$4x^2 = (2x)^2$ and $9 = (3)^2$.

Hence, the factorised form is $(2x + 3)^2$.

(x) $16x^2 + 40x + 25$

Identifying the square terms:

$16x^2 = (4x)^2$ and $25 = (5)^2$.

Hence, the factorised form is $(4x + 5)^2$.

(xi) $9x^2 + 24x + 16$

Identifying the square terms:

$9x^2 = (3x)^2$ and $16 = (4)^2$.

Hence, the factorised form is $(3x + 4)^2$.

(xii) $9x^2 + 30x + 25$

Identifying the square terms:

$9x^2 = (3x)^2$ and $25 = (5)^2$.

Hence, the factorised form is $(3x + 5)^2$.

(xiii) $2x^3 + 24x^2 + 72x$

First, we take out the common factor $2x$:

$2x(x^2 + 12x + 36)$

Now, factorising the term inside the bracket:

Substituting this back into the expression:

$2x(x + 6)^2$

Hence, the factorised form is $2x(x + 6)^2$.

(xiv) $a^2x^3 + 2abx^2 + b^2x$

First, we take out the common factor $x$:

$x(a^2x^2 + 2abx + b^2)$

Now, factorising the term inside the bracket:

Hence, the factorised form is $x(ax + b)^2$.

(xv) $4x^4 + 12x^3 + 9x^2$

First, we take out the common factor $x^2$:

$x^2(4x^2 + 12x + 9)$

Now, factorising the term inside the bracket:

Hence, the factorised form is $x^2(2x + 3)^2$.

(xvi) $\frac{x^2}{4} + 2x + 4$

Identifying the square terms:

$\frac{x^2}{4} = \left(\frac{x}{2}\right)^2$ and $4 = (2)^2$.

The middle term is $2\left(\frac{x}{2}\right)(2) = 2x$.

Hence, the factorised form is $\left(\frac{x}{2} + 2\right)^2$.

(xvii) $9x^2 + 2xy + \frac{y^2}{9}$

Identifying the square terms:

$9x^2 = (3x)^2$ and $\frac{y^2}{9} = \left(\frac{y}{3}\right)^2$.

The middle term is $2(3x)\left(\frac{y}{3}\right) = 2xy$.

Hence, the factorised form is $\left(3x + \frac{y}{3}\right)^2$.

Solution (i): $x^2 - 8x + 16$

To factorise this expression, we first identify the square terms and the middle term.

The middle term is $8x = 2 \times x \times 4$.

The expression can be written as:

Here, $a = x$ and $b = 4$.

Hence, the factorised form is $(x - 4)^2$.

Solution (ii): $x^2 - 10x + 25$

Identifying the square terms:

$x^2 = (x)^2$ and $25 = (5)^2$.

The middle term is $10x = 2 \times x \times 5$.

The expression is:

Hence, the factorised form is $(x - 5)^2$.

Solution (iii): $y^2 - 14y + 49$

Identifying the square terms:

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

The middle term is $14y = 2 \times y \times 7$.

The expression is:

Hence, the factorised form is $(y - 7)^2$.

Solution (iv): $p^2 - 2p + 1$

Identifying the square terms:

$p^2 = (p)^2$ and $1 = (1)^2$.

The middle term is $2p = 2 \times p \times 1$.

The expression is:

Hence, the factorised form is $(p - 1)^2$.

Solution (v): $4a^2 - 4ab + b^2$

Identifying the square terms:

$4a^2 = (2a)^2$ and $b^2 = (b)^2$.

The middle term is $4ab = 2 \times (2a) \times b$.

The expression is:

Hence, the factorised form is $(2a - b)^2$.

Solution (vi): $p^2y^2 - 2py + 1$

Identifying the square terms:

$p^2y^2 = (py)^2$ and $1 = (1)^2$.

The middle term is $2py = 2 \times (py) \times 1$.

The expression is:

Hence, the factorised form is $(py - 1)^2$.

Solution (vii): $a^2y^2 - 2aby + b^2$

Identifying the square terms:

$a^2y^2 = (ay)^2$ and $b^2 = (b)^2$.

The middle term is $2aby = 2 \times (ay) \times b$.

The expression is:

Hence, the factorised form is $(ay - b)^2$.

Solution (viii): $9x^2 - 12x + 4$

Identifying the square terms:

$9x^2 = (3x)^2$ and $4 = (2)^2$.

The middle term is $12x = 2 \times (3x) \times 2$.

The expression is:

Hence, the factorised form is $(3x - 2)^2$.

Solution (ix): $4y^2 - 12y + 9$

Identifying the square terms:

$4y^2 = (2y)^2$ and $9 = (3)^2$.

The middle term is $12y = 2 \times (2y) \times 3$.

The expression is:

Hence, the factorised form is $(2y - 3)^2$.

Solution (x): $\frac{x^2}{4} - 2x + 4$

Identifying the square terms:

$\frac{x^2}{4} = (\frac{x}{2})^2$ and $4 = (2)^2$.

The middle term is $2x = 2 \times (\frac{x}{2}) \times 2$.

The expression is:

Hence, the factorised form is $(\frac{x}{2} - 2)^2$.

Solution (xi): $a^2y^3 - 2aby^2 + b^2y$

Given: $a^2y^3 - 2aby^2 + b^2y$

First, we take out the common factor $y$ from all terms:

$y(a^2y^2 - 2aby + b^2)$

Now, factorising the expression inside the bracket:

The factorised bracket is $(ay - b)^2$.

Hence, the complete factorised form is $y(ay - b)^2$.

Solution (xii): $9y^2 - 4xy + \frac{4x^2}{9}$

Identifying the square terms:

$9y^2 = (3y)^2$ and $\frac{4x^2}{9} = (\frac{2x}{3})^2$.

The middle term is $4xy = 2 \times (3y) \times (\frac{2x}{3})$.

The expression is:

Hence, the factorised form is $(3y - \frac{2x}{3})^2$.

To factorise these trinomials of the form $x^2 + px + q$, we use the method of splitting the middle term. We find two numbers whose product is equal to the constant term ($q$) and whose sum is equal to the coefficient of the middle term ($p$).

Solution (i): $x^2 + 15x + 26$

We need to find two numbers whose product is $26$ and sum is $15$. These numbers are $13$ and $2$.

$x^2 + 13x + 2x + 26$

$x(x + 13) + 2(x + 13)$

Hence, the factorised form is $(x + 13)(x + 2)$.

Solution (ii): $x^2 + 9x + 20$

Find two numbers with product $20$ and sum $9$. These numbers are $5$ and $4$.

$x^2 + 5x + 4x + 20$

$x(x + 5) + 4(x + 5)$

Hence, the factorised form is $(x + 5)(x + 4)$.

Solution (iii): $y^2 + 18y + 65$

Find two numbers with product $65$ and sum $18$. These numbers are $13$ and $5$.

$y^2 + 13y + 5y + 65$

$y(y + 13) + 5(y + 13)$

Hence, the factorised form is $(y + 13)(y + 5)$.

Solution (iv): $p^2 + 14p + 13$

Find two numbers with product $13$ and sum $14$. These numbers are $13$ and $1$.

$p^2 + 13p + p + 13$

$p(p + 13) + 1(p + 13)$

Hence, the factorised form is $(p + 13)(p + 1)$.

Solution (v): $y^2 + 4y - 21$

Find two numbers with product $-21$ and sum $4$. These numbers are $7$ and $-3$.

$y^2 + 7y - 3y - 21$

$y(y + 7) - 3(y + 7)$

Hence, the factorised form is $(y + 7)(y - 3)$.

Solution (vi): $y^2 - 2y - 15$

Find two numbers with product $-15$ and sum $-2$. These numbers are $-5$ and $3$.

$y^2 - 5y + 3y - 15$

$y(y - 5) + 3(y - 5)$

Hence, the factorised form is $(y - 5)(y + 3)$.

Solution (vii): $18 + 11x + x^2$

Rearranging in descending powers: $x^2 + 11x + 18$.

Find two numbers with product $18$ and sum $11$. These numbers are $9$ and $2$.

$x^2 + 9x + 2x + 18$

$x(x + 9) + 2(x + 9)$

Hence, the factorised form is $(x + 9)(x + 2)$.

Solution (viii): $x^2 - 10x + 21$

Find two numbers with product $21$ and sum $-10$. These numbers are $-7$ and $-3$.

$x^2 - 7x - 3x + 21$

$x(x - 7) - 3(x - 7)$

Hence, the factorised form is $(x - 7)(x - 3)$.

Solution (ix): $x^2 - 17x + 60$

Find two numbers with product $60$ and sum $-17$. These numbers are $-12$ and $-5$.

$x^2 - 12x - 5x + 60$

$x(x - 12) - 5(x - 12)$

Hence, the factorised form is $(x - 12)(x - 5)$.

Solution (x): $x^2 + 4x - 77$

Find two numbers with product $-77$ and sum $4$. These numbers are $11$ and $-7$.

$x^2 + 11x - 7x - 77$

$x(x + 11) - 7(x + 11)$

Hence, the factorised form is $(x + 11)(x - 7)$.

Solution (xi): $y^2 + 7y + 12$

Find two numbers with product $12$ and sum $7$. These numbers are $4$ and $3$.

$y^2 + 4y + 3y + 12$

$y(y + 4) + 3(y + 4)$

Hence, the factorised form is $(y + 4)(y + 3)$.

Solution (xii): $p^2 - 13p - 30$

Find two numbers with product $-30$ and sum $-13$. These numbers are $-15$ and $2$.

$p^2 - 15p + 2p - 30$

$p(p - 15) + 2(p - 15)$

Hence, the factorised form is $(p - 15)(p + 2)$.

Solution (xiii): $a^2 - 16a - 80$

(Note: Assuming variable $p$ in input is a typo for $a$)

Find two numbers with product $-80$ and sum $-16$. These numbers are $-20$ and $4$.

$a^2 - 20a + 4a - 80$

$a(a - 20) + 4(a - 20)$

Hence, the factorised form is $(a - 20)(a + 4)$.

Solution (i): $x^2 - 9$

We can write the expression as a difference of two squares:

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

Applying the identity $a^2 - b^2 = (a + b)(a - b)$:

Final Result: $(x + 3)(x - 3)$

Solution (ii): $4x^2 - 25y^2$

Expressing as squares:

$4x^2 - 25y^2 = (2x)^2 - (5y)^2$

Applying the identity:

Final Result: $(2x + 5y)(2x - 5y)$

Solution (iii): $4x^2 - 49y^2$

Expressing as squares:

$4x^2 - 49y^2 = (2x)^2 - (7y)^2$

Applying the identity:

Final Result: $(2x + 7y)(2x - 7y)$

Solution (iv): $3a^2b^3 - 27a^4b$

First, we take out the greatest common factor $3a^2b$:

$3a^2b(b^2 - 9a^2)$

Now, factorise the expression inside the bracket as a difference of squares:

$3a^2b[(b)^2 - (3a)^2]$

Final Result: $3a^2b(b + 3a)(b - 3a)$

Solution (v): $28ay^2 - 175ax^2$

Taking out the common factor $7a$:

$7a(4y^2 - 25x^2)$

Factorising the bracket:

$7a[(2y)^2 - (5x)^2]$

Final Result: $7a(2y + 5x)(2y - 5x)$

Solution (vi): $9x^2 - 1$

$9x^2 - 1 = (3x)^2 - (1)^2$

Final Result: $(3x + 1)(3x - 1)$

Solution (vii): $25ax^2 - 25a$

Taking out common factor $25a$:

$25a(x^2 - 1)$

Factorising the bracket:

$25a(x + 1)(x - 1)$

Final Result: $25a(x + 1)(x - 1)$

Solution (viii): $\frac{x^2}{9} - \frac{y^2}{25}$

Expressing as squares:

$\left( \frac{x}{3} \right)^2 - \left( \frac{y}{5} \right)^2$

Final Result: $\left( \frac{x}{3} + \frac{y}{5} \right)\left( \frac{x}{3} - \frac{y}{5} \right)$

Solution (ix): $\frac{2p^2}{25} - 32q^2$

Taking out the common factor $2$:

$2 \left( \frac{p^2}{25} - 16q^2 \right)$

Factorising the bracket:

$2 \left[ \left( \frac{p}{5} \right)^2 - (4q)^2 \right]$

Final Result: $2 \left( \frac{p}{5} + 4q \right)\left( \frac{p}{5} - 4q \right)$

Solution (x): $49x^2 - 36y^2$

$(7x)^2 - (6y)^2$

Final Result: $(7x + 6y)(7x - 6y)$

Solution (xi): $y^3 - \frac{y}{9}$

Taking out the common factor $y$:

$y \left( y^2 - \frac{1}{9} \right)$

Factorising the bracket:

$y \left[ (y)^2 - \left( \frac{1}{3} \right)^2 \right]$

Final Result: $y \left( y + \frac{1}{3} \right)\left( y - \frac{1}{3} \right)$

Solution (xii): $\frac{x^2}{25} - 625$

$\left( \frac{x}{5} \right)^2 - (25)^2$

Final Result: $\left( \frac{x}{5} + 25 \right)\left( \frac{x}{5} - 25 \right)$

Solution (xiii): $\frac{x^2}{8} - \frac{y^2}{18}$

Taking out the common factor $\frac{1}{2}$:

$\frac{1}{2} \left( \frac{x^2}{4} - \frac{y^2}{9} \right)$

Factorising the bracket:

$\frac{1}{2} \left[ \left( \frac{x}{2} \right)^2 - \left( \frac{y}{3} \right)^2 \right]$

Final Result: $\frac{1}{2} \left( \frac{x}{2} + \frac{y}{3} \right)\left( \frac{x}{2} - \frac{y}{3} \right)$

Solution (xiv): $\frac{4x^2}{9} - \frac{9y^2}{16}$

$\left( \frac{2x}{3} \right)^2 - \left( \frac{3y}{4} \right)^2$

Final Result: $\left( \frac{2x}{3} + \frac{3y}{4} \right)\left( \frac{2x}{3} - \frac{3y}{4} \right)$

Solution (xv): $\frac{x^3y}{9} - \frac{xy^3}{16}$

Taking out the common factor $xy$:

$xy \left( \frac{x^2}{9} - \frac{y^2}{16} \right)$

Factorising the bracket:

$xy \left[ \left( \frac{x}{3} \right)^2 - \left( \frac{y}{4} \right)^2 \right]$

Final Result: $xy \left( \frac{x}{3} + \frac{y}{4} \right)\left( \frac{x}{3} - \frac{y}{4} \right)$

Solution (xvi): $1331x^3y - 11y^3x$

Taking out the common factor $11xy$:

$11xy(121x^2 - y^2)$

Factorising the bracket:

$11xy[(11x)^2 - (y)^2]$

Final Result: $11xy(11x + y)(11x - y)$

Solution (xvii): $\frac{1}{36}a^2b^2 - \frac{16}{49}b^2c^2$

Taking out common factor $b^2$:

$b^2 \left( \frac{a^2}{36} - \frac{16c^2}{49} \right)$

Factorising the bracket:

$b^2 \left[ \left( \frac{a}{6} \right)^2 - \left( \frac{4c}{7} \right)^2 \right]$

Final Result: $b^2 \left( \frac{a}{6} + \frac{4c}{7} \right)\left( \frac{a}{6} - \frac{4c}{7} \right)$

Solution (xviii): $a^4 - (a - b)^4$

$(a^2)^2 - [(a - b)^2]^2$

$[a^2 + (a - b)^2][a^2 - (a - b)^2]$

Apply the identity again to the second bracket:

$[a^2 + a^2 - 2ab + b^2][(a + (a - b))(a - (a - b))]$

$[2a^2 - 2ab + b^2][(2a - b)(b)]$

Final Result: $b(2a - b)(2a^2 - 2ab + b^2)$

Solution (xix): $x^4 - 1$

$(x^2)^2 - (1)^2$

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

Factorising the second bracket again:

Final Result: $(x^2 + 1)(x + 1)(x - 1)$

Solution (xx): $y^4 - 625$

$(y^2)^2 - (25)^2$

$(y^2 + 25)(y^2 - 25)$

Factorising the second bracket again:

Final Result: $(y^2 + 25)(y + 5)(y - 5)$

Solution (xxi): $p^5 - 16p$

Taking $p$ common: $p(p^4 - 16)$

$p[(p^2)^2 - (4)^2]$

$p(p^2 + 4)(p^2 - 4)$

Final Result: $p(p^2 + 4)(p + 2)(p - 2)$

Solution (xxii): $16x^4 - 81$

$(4x^2)^2 - (9)^2$

$(4x^2 + 9)(4x^2 - 9)$

Final Result: $(4x^2 + 9)(2x + 3)(2x - 3)$

Solution (xxiii): $x^4 - y^4$

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

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

Final Result: $(x^2 + y^2)(x + y)(x - y)$

Solution (xxiv): $y^4 - 81$

$(y^2)^2 - (9)^2$

$(y^2 + 9)(y^2 - 9)$

Final Result: $(y^2 + 9)(y + 3)(y - 3)$

Solution (xxv): $16x^4 - 625y^4$

$(4x^2)^2 - (25y^2)^2$

$(4x^2 + 25y^2)(4x^2 - 25y^2)$

Final Result: $(4x^2 + 25y^2)(2x + 5y)(2x - 5y)$

Solution (xxvi): $(a - b)^2 - (b - c)^2$

Let $X = (a - b)$ and $Y = (b - c)$. Using $X^2 - Y^2 = (X + Y)(X - Y)$:

$[(a - b) + (b - c)][(a - b) - (b - c)]$

$(a - c)(a - 2b + c)$

Final Result: $(a - c)(a - 2b + c)$

Solution (xxvii): $(x + y)^4 - (x - y)^4$

$[(x + y)^2]^2 - [(x - y)^2]^2$

$[(x + y)^2 + (x - y)^2][(x + y)^2 - (x - y)^2]$

Expand first bracket and factorise second:

$[x^2 + 2xy + y^2 + x^2 - 2xy + y^2][((x + y) + (x - y))((x + y) - (x - y))]$

$[2x^2 + 2y^2][(2x)(2y)]$

$2(x^2 + y^2)(4xy)$

Final Result: $8xy(x^2 + y^2)$

Solution (xxviii): $x^4 - y^4 + x^2 - y^2$

Grouping: $(x^4 - y^4) + (x^2 - y^2)$

$(x^2 + y^2)(x^2 - y^2) + 1(x^2 - y^2)$

Taking $(x^2 - y^2)$ common:

$(x^2 - y^2)(x^2 + y^2 + 1)$

Final Result: $(x + y)(x - y)(x^2 + y^2 + 1)$

Solution (xxix): $8a^3 - 2a$

Taking $2a$ common: $2a(4a^2 - 1)$

$2a[(2a)^2 - (1)^2]$

Final Result: $2a(2a + 1)(2a - 1)$

Solution (xxx): $x^2 - \frac{y^2}{100}$

$(x)^2 - \left( \frac{y}{10} \right)^2$

Final Result: $\left( x + \frac{y}{10} \right)\left( x - \frac{y}{10} \right)$

Solution (xxxi): $9x^2 - (3y + z)^2$

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

$[3x + (3y + z)][3x - (3y + z)]$

Final Result: $(3x + 3y + z)(3x - 3y - z)$

The area of a rectangle is the product of its length and breadth ($\text{Area} = \text{Length} \times \text{Breadth}$). To find the dimensions, we need to factorise the given quadratic expressions.

Solution (i):

Given: $\text{Area} = x^2 - 6x + 8$

We factorise by splitting the middle term. We need two numbers whose product is $8$ and sum is $-6$. These numbers are $-4$ and $-2$.

$x^2 - 4x - 2x + 8$

$x(x - 4) - 2(x - 4)$

$(x - 4)(x - 2)$

Possible dimensions: Length $= (x - 4)$ and Breadth $= (x - 2)$ (or vice versa).

Solution (ii):

Given: $\text{Area} = x^2 - 3x + 2$

We need two numbers whose product is $2$ and sum is $-3$. These are $-2$ and $-1$.

$x^2 - 2x - x + 2$

$x(x - 2) - 1(x - 2)$

$(x - 2)(x - 1)$

Possible dimensions: Length $= (x - 2)$ and Breadth $= (x - 1)$.

Solution (iii):

Given: $\text{Area} = x^2 - 7x + 10$

We need two numbers whose product is $10$ and sum is $-7$. These are $-5$ and $-2$.

$x^2 - 5x - 2x + 10$

$x(x - 5) - 2(x - 5)$

$(x - 5)(x - 2)$

Possible dimensions: Length $= (x - 5)$ and Breadth $= (x - 2)$.

Solution (iv):

Given: $\text{Area} = x^2 + 19x - 20$

We need two numbers whose product is $-20$ and sum is $19$. These are $20$ and $-1$.

$x^2 + 20x - x - 20$

$x(x + 20) - 1(x + 20)$

$(x + 20)(x - 1)$

Possible dimensions: Length $= (x + 20)$ and Breadth $= (x - 1)$.

Solution (v):

Given: $\text{Area} = x^2 + 9x + 20$

We need two numbers whose product is $20$ and sum is $9$. These are $5$ and $4$.

$x^2 + 5x + 4x + 20$

$x(x + 5) + 4(x + 5)$

$(x + 5)(x + 4)$

Possible dimensions: Length $= (x + 5)$ and Breadth $= (x + 4)$.

Solution (i): $51x^3y^2z \div 17xyz$

$\frac{51x^3y^2z}{17xyz} = \left(\frac{51}{17}\right) \times x^{3-1} \times y^{2-1} \times z^{1-1}$

$= 3 \times x^2 \times y^1 \times z^0$

$= 3x^2y$

Final Result: $3x^2y$

Solution (ii): $76x^3yz^3 \div 19x^2y^2$

$\frac{76x^3yz^3}{19x^2y^2} = \left(\frac{76}{19}\right) \times x^{3-2} \times y^{1-2} \times z^3$

$= 4 \times x^1 \times y^{-1} \times z^3$

$= \frac{4xz^3}{y}$

Final Result: $\frac{4xz^3}{y}$

Solution (iii): $17ab^2c^3 \div (-abc^2)$

$\frac{17ab^2c^3}{-1abc^2} = \left(\frac{17}{-1}\right) \times a^{1-1} \times b^{2-1} \times c^{3-2}$

$= -17 \times a^0 \times b^1 \times c^1$

$= -17bc$

Final Result: $-17bc$

Solution (iv): $-121p^3q^3r^3 \div (-11xy^2z^3)$

$\frac{-121p^3q^3r^3}{-11xy^2z^3} = \left(\frac{-121}{-11}\right) \times \frac{p^3q^3r^3}{xy^2z^3}$

$= \frac{11p^3q^3r^3}{xy^2z^3}$

Final Result: $\frac{11p^3q^3r^3}{xy^2z^3}$

Solution (i): $(3pqr - 6p^2q^2r^2) \div 3pq$

$\frac{3pqr - 6p^2q^2r^2}{3pq} = \frac{3pqr}{3pq} - \frac{6p^2q^2r^2}{3pq}$

$= r - 2pqr^2$

Final Result: $r - 2pqr^2$

Solution (ii): $(ax^3 - bx^2 + cx) \div (-dx)$

$\frac{ax^3 - bx^2 + cx}{-dx} = \frac{ax^3}{-dx} - \frac{bx^2}{-dx} + \frac{cx}{-dx}$

$= -\frac{ax^2}{d} + \frac{bx}{d} - \frac{c}{d}$

Final Result: $\frac{-ax^2 + bx - c}{d}$

Solution (iii): $(x^3y^3 + x^2y^3 - xy^4 + xy) \div xy$

$\frac{x^3y^3 + x^2y^3 - xy^4 + xy}{xy} = \frac{x^3y^3}{xy} + \frac{x^2y^3}{xy} - \frac{xy^4}{xy} + \frac{xy}{xy}$

$= x^2y^2 + xy^2 - y^3 + 1$

Final Result: $x^2y^2 + xy^2 - y^3 + 1$

Solution (iv): $(-qrxy + pryz - rxyz) \div (-xyz)$

$\frac{-qrxy + pryz - rxyz}{-xyz} = \frac{-qrxy}{-xyz} + \frac{pryz}{-xyz} - \frac{rxyz}{-xyz}$

$= \frac{qr}{z} - \frac{pr}{x} + r$

Final Result: $\frac{qr}{z} - \frac{pr}{x} + r$

Solution (i): $(x^2 - 22x + 117) \div (x - 13)$

First, we factorise the dividend $x^2 - 22x + 117$ by splitting the middle term. We look for two numbers whose product is $117$ and whose sum is $-22$. These numbers are $-13$ and $-9$.

$x^2 - 13x - 9x + 117$

$x(x - 13) - 9(x - 13)$

$(x - 13)(x - 9)$

Now, we perform the division:

$\frac{(x - 13)(x - 9)}{(x - 13)}$

$\frac{\cancel{(x - 13)}^1(x - 9)}{\cancel{(x - 13)}_1} = x - 9$

Final Result: $x - 9$

Solution (ii): $(x^3 + x^2 - 132x) \div x(x - 11)$

First, factorise the dividend by taking out the common factor $x$:

$x(x^2 + x - 132)$

Now, factorise the quadratic expression $x^2 + x - 132$ by splitting the middle term. We need two numbers with product $-132$ and sum $1$. These are $12$ and $-11$.

$x[x^2 + 12x - 11x - 132]$

$x[x(x + 12) - 11(x + 12)]$

$x(x + 12)(x - 11)$

Now, perform the division:

$\frac{x(x + 12)(x - 11)}{x(x - 11)}$

$\frac{\cancel{x}^1(x + 12)\cancel{(x - 11)}^1}{\cancel{x}_1\cancel{(x - 11)}_1} = x + 12$

Final Result: $x + 12$

Solution (iii): $(2x^3 - 12x^2 + 16x) \div (x - 2)(x - 4)$

Factorise the dividend by taking $2x$ as a common factor:

$2x(x^2 - 6x + 8)$

Factorising the quadratic inside the bracket (product $= 8$, sum $= -6 \implies -4, -2$):

$2x[x^2 - 4x - 2x + 8]$

$2x[x(x - 4) - 2(x - 4)]$

$2x(x - 4)(x - 2)$

Now, perform the division:

$\frac{2x(x - 4)(x - 2)}{(x - 2)(x - 4)}$

$\frac{2x\cancel{(x - 4)}^1\cancel{(x - 2)}^1}{\cancel{(x - 2)}_1\cancel{(x - 4)}_1} = 2x$

Final Result: $2x$

Solution (iv): $(9x^2 - 4) \div (3x + 2)$

Factorise the dividend using the identity $a^2 - b^2 = (a - b)(a + b)$:

$(3x + 2)(3x - 2)$

Now, perform the division:

$\frac{(3x + 2)(3x - 2)}{(3x + 2)}$

$\frac{\cancel{(3x + 2)}^1(3x - 2)}{\cancel{(3x + 2)}_1} = 3x - 2$

Final Result: $3x - 2$

Solution (v): $(3x^2 - 48) \div (x - 4)$

First, take $3$ common from the dividend:

$3(x^2 - 16)$

Now, factorise $(x^2 - 16)$ as $(x^2 - 4^2)$:

$3(x + 4)(x - 4)$

Now, perform the division:

$\frac{3(x + 4)(x - 4)}{(x - 4)}$

$\frac{3(x + 4)\cancel{(x - 4)}^1}{\cancel{(x - 4)}_1} = 3(x + 4)$

Final Result: $3x + 12$

Solution (vi): $(x^4 - 16) \div (x^3 + 2x^2 + 4x + 8)$

Factorise the dividend $(x^4 - 16)$:

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

Further factorising $(x^2 - 4)$: $(x^2 + 4)(x + 2)(x - 2)$

Now, factorise the divisor $(x^3 + 2x^2 + 4x + 8)$ by grouping:

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

Performing the division:

$\frac{(x^2 + 4)(x + 2)(x - 2)}{(x + 2)(x^2 + 4)}$

$\frac{\cancel{(x^2 + 4)}^1\cancel{(x + 2)}^1(x - 2)}{\cancel{(x + 2)}_1\cancel{(x^2 + 4)}_1} = x - 2$

Final Result: $x - 2$

Solution (vii): $(3x^4 - 1875) \div (3x^2 - 75)$

First, take $3$ common from both dividend and divisor:

Dividend: $3(x^4 - 625) = 3((x^2)^2 - (25)^2) = 3(x^2 + 25)(x^2 - 25)$

Divisor: $3(x^2 - 25)$

Now, perform the division:

$\frac{3(x^2 + 25)(x^2 - 25)}{3(x^2 - 25)}$

$\frac{\cancel{3}^1(x^2 + 25)\cancel{(x^2 - 25)}^1}{\cancel{3}_1\cancel{(x^2 - 25)}_1} = x^2 + 25$

Final Result: $x^2 + 25$

Given:

Area of the square = $4x^2 + 12xy + 9y^2$

To Find:

The side of the square.

Solution:

We know that the area of a square is the square of its side.

We need to express the given expression $4x^2 + 12xy + 9y^2$ as a perfect square. We use the identity $a^2 + 2ab + b^2 = (a + b)^2$.

Here, we can write:

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

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

The middle term $12xy$ can be written as $2(2x)(3y)$.

So, the expression becomes:

Area = $(2x)^2 + 2(2x)(3y) + (3y)^2$

Area = $(2x + 3y)^2$

Comparing this with the area formula $(\text{Side})^2$:

$\text{Side} = \sqrt{(2x + 3y)^2}$

$\text{Side} = 2x + 3y$

Hence, the side of the square is $2x + 3y$ units.

Given:

Area of the square = $9x^2 + 24xy + 16y^2$

To Find:

The side of the square.

Solution:

The area of a square is defined as:

We factorise the given expression using the identity $a^2 + 2ab + b^2 = (a + b)^2$.

We have:

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

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

The middle term $24xy$ matches $2(3x)(4y)$.

Thus, Area = $(3x)^2 + 2(3x)(4y) + (4y)^2$

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

From equation (i), we have:

$(\text{Side})^2 = (3x + 4y)^2$

$\text{Side} = 3x + 4y$

Hence, the side of the square is $3x + 4y$ units.

Given:

Area of the rectangle = $x^2 + 7x + 12$

Breadth of the rectangle = $(x + 3)$

To Find:

The length of the rectangle.

Solution:

The area of a rectangle is the product of its length and breadth.

To find the length, we divide the area by the breadth:

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

Length = $\frac{x^2 + 7x + 12}{x + 3}$

First, we factorise the numerator $x^2 + 7x + 12$ by splitting the middle term ($7x$):

$x^2 + 4x + 3x + 12$

$x(x + 4) + 3(x + 4)$

$(x + 4)(x + 3)$

Now, substituting this back into the length formula:

Length = $\frac{(x + 4)\cancel{(x + 3)}^1}{\cancel{(x + 3)}_1}$

Length = $x + 4$

Hence, the length of the rectangle is $x + 4$ units.

Given:

Curved Surface Area (C.S.A.) = $2\pi(y^2 - 7y + 12)$

Radius ($r$) = $(y - 3)$

To Find:

The height ($h$) of the cylinder.

Solution:

The formula for the Curved Surface Area of a cylinder is:

Equating the given value with the formula:

$2\pi(y^2 - 7y + 12) = 2\pi(y - 3)h$

Dividing both sides by $2\pi$:

$y^2 - 7y + 12 = (y - 3)h$

To find $h$, we need to factorise the expression $y^2 - 7y + 12$. We look for two numbers whose product is $12$ and whose sum is $-7$. These numbers are $-4$ and $-3$.

$y^2 - 4y - 3y + 12 = (y - 3)h$

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

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

Dividing by $(y - 3)$:

$h = \frac{(y - 4)\cancel{(y - 3)}^1}{\cancel{(y - 3)}_1}$

$h = y - 4$

Hence, the height of the cylinder is $y - 4$ units.

Given:

Area of the circle ($A$) = $\pi x^2 + 6\pi x + 9\pi$

To Find:

The radius ($r$) of the circle.

Solution:

We know that the area of a circle is given by the formula:

The given expression for the area is:

$A = \pi x^2 + 6\pi x + 9\pi$

Taking $\pi$ as a common factor from all terms:

$A = \pi(x^2 + 6x + 9)$

Now, we factorise the expression inside the bracket using the identity $a^2 + 2ab + b^2 = (a + b)^2$:

$x^2 + 6x + 9 = (x)^2 + 2(x)(3) + (3)^2$

$x^2 + 6x + 9 = (x + 3)^2$

Substituting this back into the area expression:

Comparing equation (i) with the standard formula $A = \pi r^2$:

$\pi r^2 = \pi(x + 3)^2$

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

$r = x + 3$

Hence, the radius of the circle is $x + 3$ units.

Given:

Expression = $\frac{n^2}{2} + \frac{n}{2}$

Solution:

To factorise the given expression, we identify the common factors in both terms.

In the expression $\frac{n^2}{2} + \frac{n}{2}$, the common factor is $\frac{n}{2}$.

Taking $\frac{n}{2}$ out as a common factor:

$\frac{n^2}{2} + \frac{n}{2} = \frac{n}{2}(n) + \frac{n}{2}(1)$

$\frac{n^2}{2} + \frac{n}{2} = \frac{n}{2}(n + 1)$

Hence, the factorised form is $\frac{n(n + 1)}{2}$.

Given:

Sum of observations = $x^4 - 625$

Number of observations = $x + 5$

To Find:

Mean of the observations.

Solution:

The mean of observations is calculated using the formula:

First, let's factorise the sum $x^4 - 625$ using the identity $a^2 - b^2 = (a - b)(a + b)$:

$x^4 - 625 = (x^2)^2 - (25)^2$

$x^4 - 625 = (x^2 - 25)(x^2 + 25)$

Now, factorise $(x^2 - 25)$ further:

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

So, the sum of observations becomes:

Sum = $(x - 5)(x + 5)(x^2 + 25)$

Now, calculating the Mean:

$\text{Mean} = \frac{(x - 5)(x + 5)(x^2 + 25)}{x + 5}$

Cancelling the common factor $(x + 5)$ from the numerator and denominator:

$\text{Mean} = \frac{(x - 5)\cancel{(x + 5)}^{1}(x^2 + 25)}{\cancel{(x + 5)}_{1}}$

$\text{Mean} = (x - 5)(x^2 + 25)$

Expanding the result:

$\text{Mean} = x(x^2 + 25) - 5(x^2 + 25)$

$\text{Mean} = x^3 + 25x - 5x^2 - 125$

Hence, the mean of the observations is $x^3 - 5x^2 + 25x - 125$.

Given:

Height of the triangle ($h$) = $x^4 + y^4$

Base of the triangle ($b$) = $14xy$

To Find:

Area of the triangle.

Solution:

The area of a triangle is given by the formula:

Substituting the given values:

Area = $\frac{1}{2} \times (14xy) \times (x^4 + y^4)$

Area = $\frac{\cancel{14}^{7}}{\cancel{2}_{1}}xy \times (x^4 + y^4)$

Area = $7xy(x^4 + y^4)$

Expanding the expression:

Area = $7xy \cdot x^4 + 7xy \cdot y^4$

Area = $7x^5y + 7xy^5$

Hence, the area of the triangle is $7x^5y + 7xy^5$ square units.

Given:

Cost of one chocolate = $\textsf{₹}(x + 4)$

Number of chocolates bought = $(x + 4)$

To Find:

1. Total amount paid in terms of $x$.

2. Amount paid if $x = 10$.

Solution:

The total amount paid is calculated by multiplying the cost of one chocolate by the total number of chocolates bought.

Substituting the given values:

Total Amount = $(x + 4) \times (x + 4)$

Now, we expand the expression using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$:

Here, $a = x$ and $b = 4$.

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

Total Amount = $x^2 + 8x + 16$

Amount paid when $x = 10$:

To find the numerical value of the amount paid, we substitute $x = 10$ into equation (i):

Total Amount = $(10 + 4)^2$

Total Amount = $(14)^2$

Total Amount = $14 \times 14$

Total Amount = $196$

Alternate Calculation for x = 10:

We can also substitute $x = 10$ into the expanded form $x^2 + 8x + 16$:

Total Amount = $(10)^2 + 8(10) + 16$

Total Amount = $100 + 80 + 16$

Total Amount = $196$

Hence, the total amount paid by Rohit in terms of $x$ is $\textsf{₹}(x^2 + 8x + 16)$ and the amount paid when $x = 10$ is $\textsf{₹}196$.

Given:

Base of the parallelogram ($b$) = $(2x + 3)$ units

Height of the parallelogram ($h$) = $(2x - 3)$ units

To Find:

1. Area of the parallelogram in terms of $x$.

2. Area of the parallelogram when $x = 30$ units.

Solution:

The area of a parallelogram is calculated by the product of its base and the corresponding height.

Substituting the given values into the formula:

Area = $(2x + 3) \times (2x - 3)$

Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$:

Area = $4x^2 - 9$

Calculation for x = 30 units:

To find the numerical value of the area, we substitute $x = 30$ into the expression obtained above:

Area = $4(30)^2 - 9$

Area = $4(900) - 9$

Area = $3600 - 9$

Area = $3591$

Hence, the area of the parallelogram in terms of $x$ is $4x^2 - 9$ sq. units and for $x = 30$ units, the area is $3591$ sq. units.

Given:

Radius of the circle ($r$) = $7ab - 7bc - 14ac$

Value of $\pi = \frac{22}{7}$

To Find:

Circumference of the circle.

Solution:

The circumference ($C$) of a circle is given by the formula:

First, we simplify the expression for the radius by taking out the common factor $7$:

$r = 7(ab - bc - 2ac)$

Now, substitute the values of $r$ and $\pi$ into the circumference formula:

$C = 2 \times \frac{22}{7} \times 7(ab - bc - 2ac)$

We can cancel the $7$ in the numerator and the denominator:

$C = 2 \times 22 \times \frac{\cancel{7}^1}{\cancel{7}_1}(ab - bc - 2ac)$

$C = 44(ab - bc - 2ac)$

Expanding the expression:

$C = 44ab - 44bc - 88ac$

Hence, the circumference of the circle is $44(ab - bc - 2ac)$ or $44ab - 44bc - 88ac$ units.

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 $-12$.

Solution (i): $(ab + bc) (ab – bc) + (bc + ca) (bc – ca) $$ + (ca + ab) (ca – ab) $$ = 0$

To verify this, we will use the identity $(x + y)(x – y) = x^2 – y^2$ for each part of the Left Hand Side (LHS).

LHS = $[(ab)^2 – (bc)^2] + [(bc)^2 – (ca)^2] + [(ca)^2 – (ab)^2]$

LHS = $a^2b^2 – b^2c^2 + b^2c^2 – c^2a^2 + c^2a^2 – a^2b^2$

Grouping the like terms together:

LHS = $(a^2b^2 – a^2b^2) + (b^2c^2 – b^2c^2) + (c^2a^2 – c^2a^2)$

LHS = $0 + 0 + 0 = 0$

Since LHS = RHS, the statement is verified.

Solution (ii): $(a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) = a^3 + b^3 + c^3 – 3abc$

LHS = $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)$

Expanding the brackets:

= $(a^3 + ab^2 + ac^2 – a^2b – abc – a^2c) + (ba^2 + b^3 + bc^2 – ab^2 – b^2c – abc) $$ + (ca^2 + cb^2 + c^3 – abc – bc^2 – ac^2)$

Rearranging and cancelling the terms:

= $a^3 + b^3 + c^3 – abc – abc – abc + (ab^2 – ab^2) + (ac^2 – ac^2) + (a^2b – a^2b) $$ + (a^2c – a^2c) + (b^2c – b^2c) + (bc^2 – bc^2)$

= $a^3 + b^3 + c^3 – 3abc$

Since LHS = RHS, the statement is verified.

Solution (iii): $(p – q) (p^2 + pq + q^2) = p^3 – q^3$

LHS = $p(p^2 + pq + q^2) – q(p^2 + pq + q^2)$

Expanding the brackets:

= $(p^3 + p^2q + pq^2) – (p^2q + pq^2 + q^3)$

= $p^3 + p^2q + pq^2 – p^2q – pq^2 – q^3$

= $p^3 + (p^2q – p^2q) + (pq^2 – pq^2) – q^3$

= $p^3 – q^3$

Since LHS = RHS, the statement is verified.

Solution (iv): $(m + n) (m^2 – mn + n^2) = m^3 + n^3$

LHS = $m(m^2 – mn + n^2) + n(m^2 – mn + n^2)$

Expanding the brackets:

= $(m^3 – m^2n + mn^2) + (m^2n – mn^2 + n^3)$

= $m^3 – m^2n + m^2n + mn^2 – mn^2 + n^3$

= $m^3 + n^3$

Since LHS = RHS, the statement is verified.

Solution (v): $(a + b) (a + b) (a + b) = a^3 + 3a^2b + 3ab^2 + b^3$

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

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

= $a^3 + 2a^2b + ab^2 + ba^2 + 2ab^2 + b^3$

= $a^3 + 3a^2b + 3ab^2 + b^3$

Since LHS = RHS, the statement is verified.

Solution (vi): $(a – b) (a – b) (a – b) = a^3 – 3a^2b + 3ab^2 – b^3$

LHS = $(a – b) (a – b)^2$

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

= $a^3 – 2a^2b + ab^2 – ba^2 + 2ab^2 – b^3$

= $a^3 – 3a^2b + 3ab^2 – b^3$

Since LHS = RHS, the statement is verified.

Solution (vii): $(a^2 – b^2) (a^2 + b^2) + (b^2 – c^2) (b^2 + c^2) + (c^2 – a^2) (c^2 + a^2) = 0$

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

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

LHS = $a^4 – b^4 + b^4 – c^4 + c^4 – a^4$

Grouping the terms:

LHS = $(a^4 – a^4) + (b^4 – b^4) + (c^4 – c^4) = 0$

Since LHS = RHS, the statement is verified.

Solution (viii): $(5x + 8)^2 – 160x = (5x – 8)^2$

LHS = $(5x)^2 + 2(5x)(8) + 8^2 – 160x$

LHS = $25x^2 + 80x + 64 – 160x$

LHS = $25x^2 – 80x + 64$

Now, RHS = $(5x – 8)^2$

RHS = $(5x)^2 – 2(5x)(8) + 8^2 = 25x^2 – 80x + 64$

Since LHS = RHS, the statement is verified.

Solution (ix): $(7p – 13q)^2 + 364pq = (7p + 13q)^2$

LHS = $(7p)^2 – 2(7p)(13q) + (13q)^2 + 364pq$

LHS = $49p^2 – 182pq + 169q^2 + 364pq$

LHS = $49p^2 + 182pq + 169q^2$

Now, RHS = $(7p + 13q)^2$

RHS = $(7p)^2 + 2(7p)(13q) + (13q)^2 = 49p^2 + 182pq + 169q^2$

Since LHS = RHS, the statement is verified.

Solution (x): $\left( \frac{3p}{7} + \frac{7}{6p} \right)^2 - \left( \frac{3p}{7} - \frac{7}{6p} \right)^2 = 2$

Using the identity $(a + b)^2 – (a – b)^2 = 4ab$:

Here, $a = \frac{3p}{7}$ and $b = \frac{7}{6p}$.

LHS = $4 \left( \frac{3p}{7} \right) \left( \frac{7}{6p} \right)$

LHS = $4 \times \left( \frac{3 \times \cancel{p} \times \cancel{7}}{\cancel{7} \times 6 \times \cancel{p}} \right)$

LHS = $4 \times \left( \frac{3}{6} \right) = 4 \times \frac{1}{2} = 2$

Since LHS = RHS, the statement is verified.

Solution (i): $8a = 35^2 – 27^2$

Using the algebraic identity $x^2 – y^2 = (x – y)(x + y)$:

$8a = (8)(62)$

$a = \frac{8 \times 62}{8}$

$a = 62$

Hence, the value of $a$ is 62.

Solution (ii): $9a = 76^2 – 67^2$

Using the algebraic identity $x^2 – y^2 = (x – y)(x + y)$:

$9a = (9)(143)$

$a = \frac{9 \times 143}{9}$

$a = 143$

Hence, the value of $a$ is 143.

Solution (iii): $pqa = (3p + q)^2 – (3p – q)^2$

Using the identity $(x + y)^2 – (x – y)^2 = 4xy$:

$pqa = 12pq$

$a = \frac{12pq}{pq}$

$a = 12$

Hence, the value of $a$ is 12.

Solution (iv): $pq^2a = (4pq + 3q)^2 – (4pq – 3q)^2$

Using the identity $(x + y)^2 – (x – y)^2 = 4xy$:

$pq^2a = 48pq^2$

$a = \frac{48pq^2}{pq^2}$

$a = 48$

Hence, the value of $a$ is 48.

To Find: The expression that must be added.

Solution:

Let the required expression be $X$. According to the question:

$4c(-a + b + c) + X = 3a(a + b + c) - 2b(a - b + c)$

$X = [3a(a + b + c) - 2b(a - b + c)] - [4c(-a + b + c)]$

First, let's simplify the first part (the expression we want to obtain):

$3a(a + b + c) - 2b(a - b + c)$

$= (3a^2 + 3ab + 3ac) - (2ab - 2b^2 + 2bc)$

$= 3a^2 + 3ab + 3ac - 2ab + 2b^2 - 2bc$

Now, let's simplify the given expression (the expression we are adding to):

$4c(-a + b + c)$

To find $X$, we subtract (ii) from (i):

$X = (3a^2 + 2b^2 + ab + 3ac - 2bc) - (-4ac + 4bc + 4c^2)$

$X = 3a^2 + 2b^2 + ab + 3ac - 2bc + 4ac - 4bc - 4c^2$

Grouping and adding the like terms:

$X = 3a^2 + 2b^2 - 4c^2 + ab + (3ac + 4ac) + (-2bc - 4bc)$

$X = 3a^2 + 2b^2 - 4c^2 + ab + 7ac - 6bc$

Hence, the expression to be added is $3a^2 + 2b^2 - 4c^2 + ab + 7ac - 6bc$.

Solution:

First, we simplify the expression to be subtracted:

$b(b^2 + b - 7) + 5 = b^3 + b^2 - 7b + 5$

Now, we subtract this from $(3b^2 - 8)$:

$(3b^2 - 8) - (b^3 + b^2 - 7b + 5)$

Opening the brackets and changing the signs of the terms being subtracted:

$3b^2 - 8 - b^3 - b^2 + 7b - 5$

Grouping the like terms together:

$-b^3 + (3b^2 - b^2) + 7b + (-8 - 5)$

Now, we find the value of this expression for $b = -3$. Substituting the value in equation (i):

$-(-3)^3 + 2(-3)^2 + 7(-3) - 13$

$-(-27) + 2(9) - 21 - 13$

$27 + 18 - 21 - 13$

$45 - 34 = 11$

Hence, the simplified expression is $-b^3 + 2b^2 + 7b - 13$ and its value for $b = -3$ is 11.

Given:

Solution:

Squaring both sides of equation (i):

$(x - \frac{1}{x})^2 = (7)^2$

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$:

$x^2 - 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 49$

$x^2 - 2 + \frac{1}{x^2} = 49$

Transposing $-2$ to the Right Hand Side:

$x^2 + \frac{1}{x^2} = 49 + 2$

$x^2 + \frac{1}{x^2} = 51$

Hence, the value of $x^2 + \frac{1}{x^2}$ is 51.

Solution:

We can rewrite the expression by grouping the terms:

$(x^2 + \frac{1}{x^2} + 2) - (3x + \frac{3}{x})$

The first part $(x^2 + \frac{1}{x^2} + 2)$ can be expressed as $(x)^2 + (\frac{1}{x})^2 + 2(x)(\frac{1}{x})$, which is the expansion of $(x + \frac{1}{x})^2$.

In the second part, we take 3 as a common factor:

$(x + \frac{1}{x})^2 - 3(x + \frac{1}{x})$

Now, taking $(x + \frac{1}{x})$ as a common factor from the entire expression:

$(x + \frac{1}{x}) [ (x + \frac{1}{x}) - 3 ]$

$(x + \frac{1}{x}) (x + \frac{1}{x} - 3)$

Hence, the factorised form is $(x + \frac{1}{x})(x + \frac{1}{x} - 3)$.

Solution:

To factorise this expression, we use the method of completing the square. We can add and subtract $p^2q^2$:

$p^4 + q^4 + p^2q^2 + p^2q^2 - p^2q^2$

Rearranging the terms:

$(p^4 + q^4 + 2p^2q^2) - p^2q^2$

The terms in the bracket form a perfect square $(p^2 + q^2)^2$:

$(p^2 + q^2)^2 - (pq)^2$

Now, using the identity $a^2 - b^2 = (a + b)(a - b)$, where $a = (p^2 + q^2)$ and $b = pq$:

$(p^2 + q^2 + pq)(p^2 + q^2 - pq)$

Hence, the factorised form is $(p^2 + pq + q^2)(p^2 - pq + q^2)$.

(i) $\frac{6.25 \times 6.25 - 1.75 \times 1.75}{4.5}$

Solution:

The numerator is in the form of $a^2 - b^2$, where $a = 6.25$ and $b = 1.75$.

Using the identity $a^2 - b^2 = (a + b)(a - b)$:

$\frac{(6.25 + 1.75)(6.25 - 1.75)}{4.5}$

$\frac{(8.00)(4.50)}{4.5}$

Cancelling the common factor $4.5$ from the numerator and denominator:

$\frac{8.00 \times \cancel{4.5}^1}{\cancel{4.5}_1} = 8$

Hence, the value is 8.

(ii) $\frac{198 \times 198 - 102 \times 102}{96}$

Solution:

Using the same identity $a^2 - b^2 = (a + b)(a - b)$, where $a = 198$ and $b = 102$:

$\frac{(198 + 102)(198 - 102)}{96}$

$\frac{(300)(96)}{96}$

Cancelling the common factor $96$ from the numerator and denominator:

$\frac{300 \times \cancel{96}^1}{\cancel{96}_1} = 300$

Hence, the value is 300.

Given:

Product of two expressions = $x^5 + x^3 + x$

One expression = $x^2 + x + 1$

To Find:

The other expression.

Solution:

The other expression can be found by dividing the product by the given expression.

Performing long division:

$\begin{array}{r} x^3 - x^2 + x \phantom{x^2+x+1)} \\ x^2+x+1{\overline{\smash{\big)}\,x^5 + 0x^4 + x^3 + 0x^2 + x \phantom{)}}} \\ \underline{-~\phantom{(}(x^5 + x^4 + x^3)\phantom{-b---)}} \\ 0 - x^4 + 0x^3 + 0x^2 \phantom{)} \\ \underline{-~\phantom{()}(-x^4 - x^3 - x^2)} \\ 0 + x^3 + x^2 + x \phantom{)} \\ \underline{-~\phantom{()}(x^3 + x^2 + x)} \\ 0 \phantom{)} \end{array}$

Hence, the other expression is $x^3 - x^2 + x$.

Given:

Area of the square = $625$ square units

From the figure, side of the square = $(4x + 5)$ units

To Find:

1. Length of the side of the square.

2. Value of $x$.

Solution:

We know that the area of a square is the square of its side.

Taking square root on both sides:

$\text{Side} = \sqrt{\text{Area}}$

$\text{Side} = \sqrt{625}$

$\text{Side} = 25$ units

Now, equating the side from the figure to the calculated side:

$4x = 25 - 5$

$4x = 20$

$x = \frac{20}{4}$

$x = 5$

Hence, the length of the side is 25 units and the value of $x$ is 5.

(i) $2x^2 + 6x + 4$

Given: The expression to factorise is $2x^2 + 6x + 4$.

Solution:

To represent this expression using cards, we need:

1. 2 Green cards (G): Area of each is $x \times x = x^2$. Total area = $2x^2$.

2. 6 Red cards (R): Area of each is $x \times 1 = x$. Total area = $6x$.

3. 4 Yellow cards (Y): Area of each is $1 \times 1 = 1$. Total area = $4$.

As shown in the middle figure of the provided diagram, these $12$ cards are arranged to form one large rectangle.

Dimensions of the resulting rectangle:

The total length is the sum of the widths of the cards placed side-by-side:

The total breadth is the sum of the heights of the cards placed one above the other:

Since the Area of a rectangle = $\text{Length} \times \text{Breadth}$, we have:

$2x^2 + 6x + 4 = (2x + 4)(x + 1)$

By taking out the common factor $2$ from the first bracket:

$2x^2 + 6x + 4 = 2(x + 2)(x + 1)$

Hence, the factorised form is $2(x + 2)(x + 1)$.

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

Given: The expression to factorise is $x^2 + 4x + 4$.

Solution:

To represent this expression, we need:

1. 1 Green card (G): Area = $x^2$.

2. 4 Red cards (R): Area = $4x$.

3. 4 Yellow cards (Y): Area = $4$.

To factorise, we arrange these cards to form a square (a special type of rectangle). To make a square of area $x^2 + 4x + 4$, the cards would be arranged such that:

1. The Green card is at a corner.

2. 2 Red cards are placed horizontally next to it, and 2 Red cards are placed vertically below it.

3. The 4 Yellow cards fill the remaining gap to complete the square.

Dimensions of the resulting square:

$\text{Side length} = x + 1 + 1 = x + 2$

$\text{Side height} = x + 1 + 1 = x + 2$

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

Hence, the factorised form is $(x + 2)^2$.

Calculation of Area for the Top Figure (Large Rectangle)

Given:

The top figure is a rectangle composed of three types of tiles:

1. Green tiles (G): Two squares with side $x$, representing $x^2$ each.

2. Red tiles (R): Six rectangles with dimensions $x$ and $1$, representing $x$ each.

3. Yellow tiles (Y): Four unit squares with side $1$, representing $1$ each.

To Find:

Total area of the top rectangular arrangement.

Solution:

The area can be calculated using two methods: summing the individual parts or multiplying the total dimensions.

Method 1: Sum of individual tiles

$\text{Area} = 2x^2 + 6x + 4$

Method 2: Product of Dimensions (Length $\times$ Breadth)

From the diagram, the total dimensions are:

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

$\text{Area} = (2x + 4)(x + 1)$

$\text{Area} = 2x(x + 1) + 4(x + 1)$

$\text{Area} = 2x^2 + 2x + 4x + 4$

$\text{Area} = 2x^2 + 6x + 4$

Hence, the total area of the top figure is $2x^2 + 6x + 4$ square units.

Calculation of Area for the Bottom Figure (Composite Shape)

Given:

The bottom figure is a composite shape consisting of:

1. Green tiles (G): Two squares of side $x$.

2. Blue tiles (B): Four small unit squares of side $1$.

To Find:

Total area of the bottom arrangement.

Solution:

The total area is the sum of the areas of all the constituent tiles shown in the figure.

Area of one Green tile (G) = $x \times x = x^2$

Area of two Green tiles = $2x^2$

Area of one Blue tile (B) = $1 \times 1 = 1$

Area of four Blue tiles = $4 \times 1 = 4$

The total area is the sum of these two components:

We can also express this in factorised form:

$\text{Total Area} = 2(x^2 + 2)$

Hence, the total area of the bottom figure is $2x^2 + 4$ square units.

Given:

From the figure, the dimensions are:

Length of the wall = $(5x + 2)$ units

Height of the wall = $5x$ units

Width of the door = $x$ units

Height of the door = $3x$ units

Width of the window = $2x$ units

Height of the window = $x$ units

To Find:

The area of the wall to be painted (excluding the area of the door and the window).

Solution:

The area to be painted is equal to the total area of the wall minus the areas of the door and the window.

Step 1: Calculate the Area of the Wall

Area of Wall = $\text{Length} \times \text{Height}$

Area of Wall = $(5x + 2) \times 5x$

Step 2: Calculate the Area of the Door

Area of Door = $\text{Width} \times \text{Height}$

Step 3: Calculate the Area of the Window

Area of Window = $\text{Width} \times \text{Height}$

Step 4: Calculate the Area to be painted

Substitute values from (ii), (iii), and (iv) into equation (i):

Area to be painted = $(25x^2 + 10x) - (3x^2 + 2x^2)$

Area to be painted = $25x^2 + 10x - 5x^2$

Area to be painted = $20x^2 + 10x$

Hence, the algebraic expression for the area of the wall to be painted is $20x^2 + 10x$ square units.

Solution:

We will expand each expression in Column I using algebraic identities to find the corresponding match in Column II.

For (1): $(21x + 13y)^2$

Using identity $(a + b)^2 = a^2 + 2ab + b^2$:

$a = 21x, b = 13y$

$(21x)^2 + 2(21x)(13y) + (13y)^2$

$441x^2 + 546xy + 169y^2$

This matches with (b) in Column II.

For (2): $(21x - 13y)^2$

Using identity $(a - b)^2 = a^2 - 2ab + b^2$:

$a = 21x, b = 13y$

$(21x)^2 - 2(21x)(13y) + (13y)^2$

$441x^2 - 546xy + 169y^2$

This matches with (c) in Column II.

For (3): $(21x - 13y)(21x + 13y)$

Using identity $(a - b)(a + b) = a^2 - b^2$:

$a = 21x, b = 13y$

$(21x)^2 - (13y)^2$

$441x^2 - 169y^2$

This matches with (a) in Column II.

Final Match Table:

Hence, the correct matching is: (1) $\rightarrow$ (b), (2) $\rightarrow$ (c), (3) $\rightarrow$ (a).