Chapter 2 Polynomials (Additional Questions)

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

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Let's examine each option:

(A) $x^2 + \frac{1}{x}$ can be written as $x^2 + x^{-1}$. The term $x^{-1}$ has an exponent of $-1$, which is a negative integer. Therefore, this is not a polynomial.

(B) $\sqrt{x} + 2$ can be written as $x^{1/2} + 2$. The term $x^{1/2}$ has an exponent of $\frac{1}{2}$, which is a fraction. Therefore, this is not a polynomial.

(D) $|x| + 5$ involves the absolute value function, $|x|$. The absolute value function cannot be expressed as a standard term $ax^n$ where $n$ is a non-negative integer. Therefore, this is not a polynomial.

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(C) $x^3 - 3x + \sqrt{5}$ consists of the terms $x^3$, $-3x$, and $\sqrt{5}$.

The term $x^3$ has a coefficient of $1$ and an exponent of $3$ (a non-negative integer).

The term $-3x$ can be written as $-3x^1$, which has a coefficient of $-3$ and an exponent of $1$ (a non-negative integer).

The term $\sqrt{5}$ can be written as $\sqrt{5}x^0$, which has a coefficient of $\sqrt{5}$ and an exponent of $0$ (a non-negative integer).

All coefficients ($1$, $-3$, $\sqrt{5}$) are real numbers, and all exponents ($3$, $1$, $0$) are non-negative integers. This expression fits the definition of a polynomial.

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Thus, the expression $x^3 - 3x + \sqrt{5}$ is a polynomial.

The correct option is (C).

The degree of a polynomial is the highest power of the variable in the polynomial.

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The given polynomial is $4y^2 - 4y^3 + 5y^7 - 6$.

Let's identify the power of the variable $y$ in each term:

• In the term $4y^2$, the power of $y$ is $2$.
• In the term $-4y^3$, the power of $y$ is $3$.
• In the term $5y^7$, the power of $y$ is $7$.
• In the term $-6$, which can be written as $-6y^0$, the power of $y$ is $0$.

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The powers of the variable $y$ in the given polynomial are $2$, $3$, $7$, and $0$.

The highest power among these is $7$.

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Therefore, the degree of the polynomial $4y^2 - 4y^3 + 5y^7 - 6$ is 7.

The correct option is (C).

An expression is a polynomial if it involves only variables with non-negative integer exponents and coefficients that are real numbers, combined using only addition, subtraction, and multiplication.

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Let's analyze each expression:

(i) $3x^2 + \frac{5}{x} + 1 = 3x^2 + 5x^{-1} + 1$. The term $5x^{-1}$ has an exponent of $-1$, which is a negative integer. Therefore, this expression is not a polynomial.

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(ii) $\sqrt{7}y^2 - 2y + 9$. This expression can be written as $\sqrt{7}y^2 - 2y^1 + 9y^0$. All the exponents ($2$, $1$, and $0$) are non-negative integers, and the coefficients ($\sqrt{7}$, $-2$, and $9$) are real numbers. Therefore, this expression is a polynomial.

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(iii) $z^3 + z^2 + z^{1/2}$. The term $z^{1/2}$ has an exponent of $\frac{1}{2}$, which is a fraction (not a non-negative integer). Therefore, this expression is not a polynomial.

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(iv) $10$. This is a constant term. A constant can be considered as a polynomial of degree 0, written as $10x^0$. The exponent is $0$, which is a non-negative integer, and the coefficient $10$ is a real number. Therefore, this expression is a polynomial.

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Based on the analysis, expressions (ii) and (iv) are polynomials.

The correct option is (C).

To find the value of $p(2)$, we need to substitute $x=2$ into the given polynomial expression $p(x)$.

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The given polynomial is $p(x) = 2x^3 - 3x^2 + 4x - 1$.

Substitute $x = 2$ into the polynomial:

$p(2) = 2(2)^3 - 3(2)^2 + 4(2) - 1$

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Now, evaluate the powers of $2$:

$2^3 = 2 \times 2 \times 2 = 8$

$2^2 = 2 \times 2 = 4$

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Substitute these values back into the expression for $p(2)$:

$p(2) = 2(8) - 3(4) + 4(2) - 1$

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Perform the multiplications:

$p(2) = 16 - 12 + 8 - 1$

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Perform the additions and subtractions from left to right:

$p(2) = (16 - 12) + 8 - 1$

$p(2) = 4 + 8 - 1$

$p(2) = 12 - 1$

$p(2) = 11$

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Thus, the value of $p(2)$ is 11.

The correct option is (A).

The zero of a polynomial $q(x)$ is the value of $x$ for which the polynomial evaluates to zero, i.e., $q(x) = 0$.

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The given polynomial is $q(x) = 3x - 6$.

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To find the zero, we set $q(x) = 0$:

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Now, we solve the linear equation for $x$. Add $6$ to both sides of the equation:

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Divide both sides by $3$:

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Thus, the zero of the polynomial $q(x) = 3x - 6$ is 2.

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Verification: Substitute $x=2$ into $q(x)$:

$q(2) = 3(2) - 6 = 6 - 6 = 0$.

Since $q(2) = 0$, $x=2$ is indeed the zero of the polynomial.

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The correct option is (A).

The Remainder Theorem states that if a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, then the remainder is $p(a)$.

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In this problem, the polynomial is $p(x) = x^3 + x^2 + x + 1$, and the divisor is $(x - 1)$.

Comparing the divisor $(x - 1)$ with the form $(x - a)$, we find that $a = 1$.

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According to the Remainder Theorem, the remainder when $p(x)$ is divided by $(x - 1)$ is $p(1)$.

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We need to substitute $x = 1$ into the polynomial $p(x)$:

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

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

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Now, we evaluate the expression:

$p(1) = 1 + 1 + 1 + 1$

$p(1) = 4$

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Therefore, the remainder when $p(x) = x^3 + x^2 + x + 1$ is divided by $(x - 1)$ is 4.

The correct option is (D).

The Factor Theorem states that if $p(x)$ is a polynomial of degree greater than or equal to $1$, and $a$ is any real number, then $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.

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In this problem, the polynomial is $p(x) = x^3 + 2x^2 + 3x + 2$.

We need to check if $(x+1)$ is a factor of $p(x)$.

We can write $(x+1)$ in the form $(x - a)$ as $(x - (-1))$.

Comparing $(x - (-1))$ with $(x - a)$, we get $a = -1$.

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According to the Factor Theorem, $(x+1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.

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Let's evaluate the polynomial $p(x)$ at $x = -1$ to find $p(-1)$.

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

Substitute $x = -1$:

$p(-1) = (-1)^3 + 2(-1)^2 + 3(-1) + 2$

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Now, we calculate the value:

$(-1)^3 = -1$

$(-1)^2 = 1$

$3(-1) = -3$

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Substitute these values back into the expression for $p(-1)$:

$p(-1) = (-1) + 2(1) + (-3) + 2$

$p(-1) = -1 + 2 - 3 + 2$

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Perform the arithmetic operations:

$p(-1) = (-1 + 2) + (-3 + 2)$

$p(-1) = 1 + (-1)$

$p(-1) = 0$

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Since $p(-1) = 0$, by the converse of the Factor Theorem, $(x - (-1))$, which is $(x+1)$, is a factor of the polynomial $p(x)$.

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The option that correctly identifies this condition is (C).

The correct option is (C).

We can use the algebraic identity for squaring a binomial: $(x + y)^2 = x^2 + 2xy + y^2$.

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In the given expression, $(2a + 3b)^2$, we can consider $x = 2a$ and $y = 3b$.

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Substitute these values into the identity:

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

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Now, evaluate each term:

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

$2(2a)(3b) = 2 \times 2 \times 3 \times a \times b = 12ab$

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

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Combine the results of the terms:

$(2a + 3b)^2 = 4a^2 + 12ab + 9b^2$

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Rearranging the terms, we get $4a^2 + 9b^2 + 12ab$.

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

The correct option is (B).

The given expression is $49x^2 - 25y^2$.

We can rewrite this expression as a difference of two squares.

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The first term, $49x^2$, can be written as $(7x)^2$, since $7^2 = 49$.

The second term, $25y^2$, can be written as $(5y)^2$, since $5^2 = 25$.

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So, the expression becomes $(7x)^2 - (5y)^2$.

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This is in the form of the algebraic identity for the difference of squares: $a^2 - b^2 = (a - b)(a + b)$.

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Comparing $(7x)^2 - (5y)^2$ with $a^2 - b^2$, we have $a = 7x$ and $b = 5y$.

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Using the identity, we substitute $a = 7x$ and $b = 5y$:

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

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Therefore, the factored form of $49x^2 - 25y^2$ is $(7x - 5y)(7x + 5y)$.

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Comparing this with the given options, we find that option (C) matches the factored form.

The correct option is (C).

To complete the identity $(x+a)(x+b)$, we need to multiply the two binomials.

We can use the distributive property (or FOIL method).

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

Distribute $x$ into the second parenthesis and $a$ into the second parenthesis:

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

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Perform the multiplications:

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

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Group the terms with $x$ and factor out $x$:

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

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

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Comparing this result with the given incomplete identity $x^2 + (a+b)x + \dots$, the missing term is $ab$.

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The complete identity is $(x+a)(x+b) = x^2 + (a+b)x + ab$.

The missing term is $ab$, which is option (B).

The correct option is (B).

We need to expand the expression $(p+q)^3$ using an algebraic identity.

The suitable identity for the cube of a sum of two terms is:

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In the given expression $(p+q)^3$, we have $x=p$ and $y=q$.

Substitute $x=p$ and $y=q$ into Identity (i):

$(p+q)^3 = p^3 + q^3 + 3(p)(q)(p+q)$

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Simplify the third term:

$3(p)(q)(p+q) = 3pq(p+q)$

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So, the expanded form is:

$(p+q)^3 = p^3 + q^3 + 3pq(p+q)$

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Alternatively, we can use the expanded form of the identity:

Substitute $x=p$ and $y=q$ into Identity (ii):

$(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$

Notice that the third and fourth terms can be factored:

$3p^2q + 3pq^2 = 3pq(p) + 3pq(q) = 3pq(p+q)$

So, $p^3 + 3p^2q + 3pq^2 + q^3 = p^3 + q^3 + 3pq(p+q)$.

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Both forms of the identity give the same result. The result $p^3 + q^3 + 3pq(p+q)$ matches option (A).

The correct option is (A).

We are given the condition $a+b+c=0$.

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We use the algebraic identity for the sum of cubes: $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$

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Substitute $x=a$, $y=b$, and $z=c$ into the identity:

$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$

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Since we are given that $a+b+c = 0$, we can substitute $0$ for $(a+b+c)$ on the right side of the equation:

Multiplying any expression by zero results in zero:

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So the identity becomes:

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Now, we can solve for $a^3 + b^3 + c^3$ by adding $3abc$ to both sides of the equation:

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Therefore, if $a+b+c=0$, then $a^3+b^3+c^3$ is equal to $3abc$.

This result is a significant special case of the sum of cubes identity.

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Comparing our result with the given options, we find that option (A) matches our result.

The correct option is (A).

To factorise the expression $5xy + 10x$, we need to find the greatest common factor (GCF) of the two terms, $5xy$ and $10x$.

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Consider the coefficients: The coefficients are $5$ and $10$. The GCF of $5$ and $10$ is $5$.

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Consider the variables: The first term has the variables $x$ and $y$ ($xy$). The second term has the variable $x$. The common variable factor is $x$.

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The GCF of the expression $5xy + 10x$ is the product of the GCF of the coefficients and the common variable factor, which is $5 \times x = 5x$.

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Now, we factor out the GCF, $5x$, from each term:

$5xy = 5x \times y$

$10x = 5x \times 2$

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So, $5xy + 10x = (5x \times y) + (5x \times 2)$.

Using the distributive property in reverse, we can write this as:

$5x(y + 2)$

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Therefore, the factored form of $5xy + 10x$ is $5x(y + 2)$.

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Comparing our factored form with the given options, we see that it matches option (C).

The correct option is (C).

To factorise the expression $ab - ac + bd - cd$ by grouping, we group terms that have a common factor.

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Group the first two terms and the last two terms:

$(ab - ac) + (bd - cd)$

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Factor out the common factor from the first group $(ab - ac)$. The common factor is $a$.

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

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Factor out the common factor from the second group $(bd - cd)$. The common factor is $d$.

$bd - cd = d(b - c)$

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Substitute the factored forms back into the expression:

$a(b - c) + d(b - c)$

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Now, we see that the term $(b - c)$ is common to both parts of the expression.

Factor out the common binomial factor $(b - c)$:

$(b - c)(a + d)$

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We can also write the factored form as $(a + d)(b - c)$.

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Therefore, the factored form of $ab - ac + bd - cd$ is $(a + d)(b - c)$.

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Comparing our factored form with the given options, we find that it matches option (C).

The correct option is (C).

To factorise the quadratic polynomial $x^2 + 7x + 12$ by splitting the middle term, we need to find two numbers whose sum is the coefficient of the middle term ($7$) and whose product is the constant term ($12$).

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Let the two numbers be $m$ and $n$. We need to find $m$ and $n$ such that:

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We look for pairs of factors of $12$ and check their sum:

• Factors $1$ and $12$: Sum $= 1 + 12 = 13$ (Not $7$)
• Factors $2$ and $6$: Sum $= 2 + 6 = 8$ (Not $7$)
• Factors $3$ and $4$: Sum $= 3 + 4 = 7$ (This pair works)

The two numbers are $3$ and $4$.

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Now, we split the middle term $7x$ using these two numbers, $3$ and $4$. We can write $7x$ as $3x + 4x$.

Rewrite the polynomial:

$x^2 + 7x + 12 = x^2 + 3x + 4x + 12$

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Now, group the terms and factor by grouping:

Group the first two terms: $(x^2 + 3x)$

Group the last two terms: $(4x + 12)$

$= (x^2 + 3x) + (4x + 12)$

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Factor out the greatest common factor (GCF) from each group:

In the first group, the GCF of $x^2$ and $3x$ is $x$. Factoring $x$ out gives $x(x + 3)$.

In the second group, the GCF of $4x$ and $12$ is $4$. Factoring $4$ out gives $4(x + 3)$.

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So the expression becomes:

$x(x + 3) + 4(x + 3)$

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Now, we see that $(x + 3)$ is a common binomial factor in both terms. Factor out $(x + 3)$.

$(x + 3)(x + 4)$

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Therefore, the factored form of $x^2 + 7x + 12$ is $(x + 3)(x + 4)$.

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Comparing our factored form with the given options, we find that it matches option (A).

The correct option is (A).

We are given that $(x-1)$ is a factor of the polynomial $p(x) = x^3 - 6x^2 + 11x - 6$.

According to the Factor Theorem, if $(x-1)$ is a factor, then $p(1) = 0$.

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To find the other factors, we can divide the polynomial $p(x)$ by the given factor $(x-1)$.

We will use polynomial long division.

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Divide $x^3 - 6x^2 + 11x - 6$ by $x - 1$:

$\begin{array}{r} x^2 - 5x + 6\phantom{)} \\ x-1{\overline{\smash{\big)}\,x^3 - 6x^2 + 11x - 6}} \\ \underline{-~\phantom{(}(x^3 - x^2)}\phantom{-b)} \\ -5x^2 + 11x\phantom{)} \\ \underline{-~\phantom{()}(-5x^2 + 5x)} \\ 6x - 6\phantom{)} \\ \underline{-~\phantom{()}(6x - 6)} \\ 0\phantom{)} \end{array}$

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The quotient is the quadratic polynomial $x^2 - 5x + 6$. The remainder is $0$, which confirms that $(x-1)$ is a factor.

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Now, we need to factorise the quadratic quotient $q(x) = x^2 - 5x + 6$ to find the other factors of the original polynomial.

We factorise the quadratic by splitting the middle term. We need to find two numbers whose sum is the coefficient of the middle term ($-5$) and whose product is the constant term ($6$).

Let the two numbers be $m$ and $n$. We seek $m, n$ such that $m + n = -5$ and $m \times n = 6$.

The pairs of integers whose product is $6$ are $(1, 6), (-1, -6), (2, 3), (-2, -3)$.

Checking their sums:

• $1 + 6 = 7$
• $-1 + (-6) = -7$
• $2 + 3 = 5$
• $-2 + (-3) = -5$

The numbers are $-2$ and $-3$.

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Now, split the middle term $-5x$ as $-2x - 3x$:

$x^2 - 5x + 6 = x^2 - 2x - 3x + 6$

Group the terms:

$= (x^2 - 2x) + (-3x + 6)$

Factor out the common factor from each group:

$= x(x - 2) - 3(x - 2)$

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

$= (x - 2)(x - 3)$

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So, the quadratic quotient $x^2 - 5x + 6$ factors into $(x - 2)(x - 3)$.

The original polynomial is the product of the divisor and the quotient:

$p(x) = (x-1) \times (x^2 - 5x + 6)$

$p(x) = (x-1)(x-2)(x-3)$

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The factors of the polynomial are $(x-1)$, $(x-2)$, and $(x-3)$.

Given that $(x-1)$ is a factor, the other factors are $(x-2)$ and $(x-3)$.

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Comparing this result with the given options, we find that option (A) lists the other factors as $(x-2)$ and $(x-3)$.

The correct option is (A).

A polynomial is an algebraic expression in which the exponents of the variables are restricted to be non-negative integers ($0, 1, 2, 3, \dots$). The coefficients of the terms must be real numbers.

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Let's examine each given expression:

(A) $\frac{2}{3}x^5 - \sqrt{2}x^2 + 7$. The terms are $\frac{2}{3}x^5$, $-\sqrt{2}x^2$, and $7$ (which can be written as $7x^0$). The exponents of the variable $x$ are $5$, $2$, and $0$. These are all non-negative integers. The coefficients $\frac{2}{3}$, $-\sqrt{2}$, and $7$ are real numbers. Therefore, this expression is a polynomial.

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(B) $y^{-3} + y^2 + 1$. The first term is $y^{-3}$. The exponent of the variable $y$ in this term is $-3$. This is a negative integer, not a non-negative integer. Therefore, this expression is NOT a polynomial.

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(C) $t^4 - t^3 + t - 1$. This can be written as $t^4 - t^3 + t^1 - 1t^0$. The exponents of the variable $t$ are $4$, $3$, $1$, and $0$. These are all non-negative integers. The coefficients $1$, $-1$, $1$, and $-1$ are real numbers. Therefore, this expression is a polynomial.

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(D) $5$. This is a constant value, which can be written as $5x^0$. The exponent is $0$, which is a non-negative integer. The coefficient $5$ is a real number. A constant is considered a polynomial of degree 0. Therefore, this expression is a polynomial.

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Based on the analysis, the expression that is NOT a polynomial is $y^{-3} + y^2 + 1$ because it contains a term with a negative exponent.

The correct option is (B).

A zero of a polynomial $p(x)$ is a value of the variable $x$ for which the value of the polynomial is $0$. In other words, if $x=a$ is a zero of $p(x)$, then $p(a) = 0$.

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The given polynomial is $p(x) = x^2 - 4$.

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To find the zeros of the polynomial, we set $p(x) = 0$ and solve for $x$:

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We can factorise the left side of the equation as a difference of squares, since $x^2$ is the square of $x$ and $4$ is the square of $2$ ($4 = 2^2$).

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

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For the product of two factors to be zero, at least one of the factors must be zero.

Set the first factor equal to zero:

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Set the second factor equal to zero:

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The zeros of the polynomial $p(x) = x^2 - 4$ are $x = 2$ and $x = -2$.

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Now, let's check the given options:

(A) Is $2$ a zero? $p(2) = (2)^2 - 4 = 4 - 4 = 0$. Yes, $2$ is a zero.

(B) Is $-2$ a zero? $p(-2) = (-2)^2 - 4 = 4 - 4 = 0$. Yes, $-2$ is a zero.

(C) Is $0$ a zero? $p(0) = (0)^2 - 4 = 0 - 4 = -4$. Since $p(0) = -4 \neq 0$, $0$ is NOT a zero.

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The value among the options that is NOT a zero of the polynomial $p(x) = x^2 - 4$ is $0$.

The correct option is (C).

Let's analyze the Assertion (A) and the Reason (R).

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Assertion (A): When $p(x) = x^2 - 5x + 6$ is divided by $(x-2)$, the remainder is 0.

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by $(x-a)$, the remainder is $p(a)$.

In this case, $p(x) = x^2 - 5x + 6$ and the divisor is $(x-2)$. So, $a = 2$.

The remainder is $p(2)$. Let's calculate $p(2)$:

$p(2) = (2)^2 - 5(2) + 6$

$p(2) = 4 - 10 + 6$

$p(2) = -6 + 6$

$p(2) = 0$

The remainder when $p(x)$ is divided by $(x-2)$ is indeed $0$. Therefore, Assertion (A) is True.

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Reason (R): According to Remainder Theorem, if $p(a)=0$, then $(x-a)$ is a factor of $p(x)$, and the remainder is 0.

The Remainder Theorem states that the remainder when $p(x)$ is divided by $(x-a)$ is $p(a)$.

If $p(a) = 0$, then the remainder is $0$. When the remainder of a division is $0$, it means the divisor is a factor of the dividend. So, if $p(a)=0$, the remainder is 0, and $(x-a)$ is a factor of $p(x)$.

This statement correctly reflects the connection between the Remainder Theorem and the Factor Theorem (which is a special case of the Remainder Theorem). Therefore, Reason (R) is True.

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Now let's check if Reason (R) is the correct explanation of Assertion (A).

Assertion (A) states that the remainder is 0. Our calculation showed that $p(2)=0$. According to Reason (R), if $p(a)=0$, the remainder is 0. In Assertion (A), $a=2$, and we found $p(2)=0$. Reason (R) explains that because $p(2)=0$, the remainder when dividing by $(x-2)$ must be 0. This is exactly why Assertion (A) is true.

Thus, Reason (R) is the correct explanation for Assertion (A).

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Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains why Assertion (A) is true.

The correct option is (A).

Let's analyze the Assertion (A) and the Reason (R).

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Assertion (A): $(x+3)$ is a factor of $p(x) = x^3 + 6x^2 + 11x + 6$.

According to the Factor Theorem, $(x-a)$ is a factor of a polynomial $p(x)$ if and only if $p(a)=0$.

If the factor is $(x+3)$, we can write it as $(x - (-3))$. Comparing this with $(x-a)$, we have $a = -3$.

So, $(x+3)$ is a factor of $p(x)$ if and only if $p(-3) = 0$.

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Let's evaluate $p(x)$ at $x = -3$:

$p(x) = x^3 + 6x^2 + 11x + 6$

$p(-3) = (-3)^3 + 6(-3)^2 + 11(-3) + 6$

$p(-3) = -27 + 6(9) + (-33) + 6$

$p(-3) = -27 + 54 - 33 + 6$

$p(-3) = (54 + 6) + (-27 - 33)$

$p(-3) = 60 - 60$

$p(-3) = 0$

Since $p(-3) = 0$, $(x+3)$ is a factor of $p(x)$. Thus, Assertion (A) is True.

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Reason (R): If $(x+a)$ is a factor of $p(x)$, then $p(a)=0$.

The standard statement of the Factor Theorem is: $(x-a)$ is a factor of $p(x)$ if and only if $p(a) = 0$. Here, $a$ is the value that makes the factor zero.

If the factor is $(x+a)$, the value that makes the factor zero is $x+a=0 \implies x = -a$. So, according to the Factor Theorem, if $(x+a)$ is a factor of $p(x)$, then $p(-a) = 0$.

The statement in Reason (R) says that if $(x+a)$ is a factor, then $p(a)=0$. This is incorrect according to the standard theorem where $a$ in $p(a)$ refers to the zero of the factor. The zero of $(x+a)$ is $-a$, not $a$.

Therefore, Reason (R) is False.

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Assertion (A) is True, and Reason (R) is False.

Comparing this with the given options, option (C) matches our findings.

The correct option is (C).

The classification of polynomials is based on their degree (the highest power of the variable in the polynomial).

• A polynomial of degree 1 is called a Linear Polynomial.
• A polynomial of degree 2 is called a Quadratic Polynomial.
• A polynomial of degree 3 is called a Cubic Polynomial.
• A polynomial of degree 0 (a non-zero constant) is called a Constant Polynomial.

---

Let's determine the degree of each polynomial in Column A:

• (i) $x^3 - 1$: The highest power of the variable $x$ is $3$. The degree is $3$. This is a Cubic Polynomial.
• (ii) $x^2 + 2x + 1$: The highest power of the variable $x$ is $2$. The degree is $2$. This is a Quadratic Polynomial.
• (iii) $5x + 4$: This can be written as $5x^1 + 4$. The highest power of the variable $x$ is $1$. The degree is $1$. This is a Linear Polynomial.
• (iv) $7$: This is a constant, which can be written as $7x^0$. The highest power of the variable is $0$. The degree is $0$. This is a Constant Polynomial.

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Now, we match the polynomials with the descriptions in Column B:

• (i) Cubic Polynomial corresponds to (c). So, (i) - (c).
• (ii) Quadratic Polynomial corresponds to (d). So, (ii) - (d).
• (iii) Linear Polynomial corresponds to (a). So, (iii) - (a).
• (iv) Constant Polynomial corresponds to (b). So, (iv) - (b).

---

The correct matching is (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

Comparing this with the given options, option (B) matches our result.

The correct option is (B).

The length of the rectangular garden is given by $L = (2x+3)$ metres.

The width of the rectangular garden is given by $W = (x+1)$ metres.

The area of the garden is given by $A(x) = (2x+3)(x+1)$.

---

We need to find the area when $x = 5$ metres.

First method: Calculate the length and width when $x=5$ and then multiply them.

Length when $x=5$: $L = 2(5) + 3 = 10 + 3 = 13$ metres.

Width when $x=5$: $W = 5 + 1 = 6$ metres.

Area = Length $\times$ Width

Area $= 13 \times 6 = 78$ square metres.

This matches the calculation shown in option (A).

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Second method: Substitute $x=5$ directly into the area polynomial $A(x) = (2x+3)(x+1)$.

$A(5) = (2(5)+3)(5+1)$

$A(5) = (10+3)(6)$

$A(5) = (13)(6)$

$A(5) = 78$ square metres.

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Alternatively, we can first expand the polynomial $A(x)$:

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

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

$A(x) = 2x^2 + 5x + 3$

Now, substitute $x=5$ into the expanded form:

$A(5) = 2(5)^2 + 5(5) + 3$

$A(5) = 2(25) + 25 + 3$

$A(5) = 50 + 25 + 3$

$A(5) = 75 + 3$

$A(5) = 78$ square metres.

This calculation matches the calculation shown in option (C).

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Both calculation methods (evaluating length and width separately and then multiplying, or evaluating the expanded polynomial directly) give the area as $78$ square metres when $x=5$.

Option (A) shows the first method's calculation. Option (C) shows the second method's calculation using the expanded polynomial.

Option (B) calculates $10+3+5+1 = 19$, which is incorrect as it doesn't represent the area calculation.

Since both A and C correctly show a calculation that results in the area of $78$ square metres, option (D) which states "Both A and C are correct" is the right choice.

---

The correct option is (D).

In a polynomial, the coefficient of a term is the numerical factor (including the sign) that multiplies the variable part of that term.

---

The given polynomial is $5x^3 - 2x^2 + \frac{3}{4}x - 1$.

We are looking for the coefficient of the $x^2$ term.

---

Let's identify the terms in the polynomial:

• The first term is $5x^3$. The variable part is $x^3$, and the coefficient is $5$.
• The second term is $-2x^2$. The variable part is $x^2$, and the coefficient is $-2$.
• The third term is $\frac{3}{4}x$. The variable part is $x$, and the coefficient is $\frac{3}{4}$.
• The fourth term is $-1$. This is a constant term, which can be thought of as $-1x^0$.

---

The term containing $x^2$ is $-2x^2$.

The numerical factor in front of $x^2$ in this term is $-2$.

---

Therefore, the coefficient of $x^2$ in the polynomial $5x^3 - 2x^2 + \frac{3}{4}x - 1$ is -2.

The correct option is (B).

A constant polynomial is a polynomial that consists only of a constant term.

---

The given polynomial is $100$. This is a non-zero constant.

A non-zero constant, say $c$, can be written in terms of a variable $x$ as $c \times x^0$. This is because any non-zero number raised to the power of $0$ is $1$ ($x^0 = 1$ for $x \neq 0$). For the polynomial context, $x^0=1$ is conventionally used.

So, the constant polynomial $100$ can be written as $100x^0$.

---

The degree of a polynomial is the highest power of the variable in the polynomial.

---

In the expression $100x^0$, the only power of the variable $x$ is $0$.

Therefore, the highest power of the variable is $0$.

---

The degree of the constant polynomial $100$ is 0.

Note: The degree of the zero polynomial, $p(x)=0$, is typically considered undefined or sometimes $-\infty$. However, $100$ is a non-zero constant.

---

Comparing our result with the given options, we find that option (B) is 0.

The correct option is (B).

A zero of a polynomial $p(x)$ is the value of the variable $x$ for which $p(x) = 0$.

---

The given polynomial is $p(x) = ax+b$, where $a \neq 0$.

---

To find the zero, we set the polynomial equal to zero:

---

Now, we solve this linear equation for $x$. First, subtract $b$ from both sides of the equation:

---

Since it is given that $a \neq 0$, we can divide both sides of the equation (ii) by $a$:

---

Thus, the zero of the polynomial $p(x) = ax+b$ is $-\frac{b}{a}$.

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Comparing our result with the given options, we find that option (D) matches our result.

The correct option is (D).

Let's check the correctness of each algebraic identity.

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

The correct identity for $(a-b)^2$ is $(a-b)^2 = a^2 - 2ab + b^2$.

The identity $a^2 - b^2$ is for the difference of squares, which factors as $(a-b)(a+b)$.

Comparing $(a-b)^2 = a^2 - 2ab + b^2$ with $a^2 - b^2$, we see they are not equal in general (only if $2ab = -b^2$, which is not always true).

For example, if $a=2$ and $b=1$, $(2-1)^2 = 1^2 = 1$, while $2^2 - 1^2 = 4 - 1 = 3$. $1 \neq 3$.

Therefore, identity (i) is Incorrect.

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

Let's expand the left side: $(a+b+c)^2 = (a+b+c)(a+b+c)$.

Using the distributive property:

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

$= a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2$

$= a^2 + b^2 + c^2 + ab + ba + ac + ca + bc + cb$

Since multiplication is commutative ($ab=ba$, $ac=ca$, $bc=cb$), we combine like terms:

$= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$

Rearranging the terms as given in the identity: $a^2+b^2+c^2+2ab+2bc+2ca$.

This matches the given identity. Therefore, identity (ii) is Correct.

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

Let's expand the right side: $(a+b)(a^2-ab+b^2)$.

Using the distributive property:

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

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

$= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3$

Combine like terms: $-a^2b + a^2b = 0$ and $ab^2 - ab^2 = 0$.

$= a^3 + b^3$

This matches the left side of the identity. Therefore, identity (iii) is Correct.

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(iv) $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$

The identity for the cube of a difference is indeed $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

Let's expand the right side of the given identity: $a^3 - b^3 - 3ab(a-b)$.

Distribute the $-3ab$ into the parenthesis:

$= a^3 - b^3 - (3ab \cdot a) - (3ab \cdot -b)$

$= a^3 - b^3 - 3a^2b + 3ab^2$

Rearranging the terms: $a^3 - 3a^2b + 3ab^2 - b^3$.

This matches the correct expansion of $(a-b)^3$. Therefore, identity (iv) is Correct.

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Based on the evaluation, identities (ii), (iii), and (iv) are correct.

Looking at the options, option (B) lists (ii), (iii), and (iv).

The correct option is (B).

We need to expand the expression $(2x - 1)^3$ using an algebraic identity.

The appropriate identity for the cube of a difference of two terms is $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$ or the expanded form $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.

---

In the expression $(2x - 1)^3$, we have $a = 2x$ and $b = 1$.

---

Using the identity $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$:

Substitute $a=2x$ and $b=1$:

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

$(2x - 1)^3 = 8x^3 - 1 - 6x(2x - 1)$

This matches the expression in option (A).

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Using the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

Substitute $a=2x$ and $b=1$:

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

$(2x - 1)^3 = 8x^3 - 3(4x^2)(1) + 3(2x)(1) - 1$

$(2x - 1)^3 = 8x^3 - 12x^2 + 6x - 1$

This matches the expression in option (C).

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Both option (A) and option (C) represent the correct expanded form of $(2x - 1)^3$. Option (A) shows a partially simplified form derived from one version of the identity, while option (C) shows the fully expanded and simplified polynomial form. Since both expressions are equivalent:

$8x^3 - 1 - 6x(2x-1) = 8x^3 - 1 - (6x \times 2x) - (6x \times -1) = 8x^3 - 1 - 12x^2 + 6x = 8x^3 - 12x^2 + 6x - 1$.

Thus, the expression in (A) is equal to the expression in (C).

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Since both option (A) and option (C) are correct expanded forms, the correct choice that indicates this is option (D).

The correct option is (D).

To match the factored polynomials with their original expressions, we need to expand the factored expressions in Column A and compare them with the expressions in Column B.

---

Let's expand each expression from Column A:

(i) $(x+2)(x-2)$:

Using the difference of squares identity $(a+b)(a-b) = a^2 - b^2$ with $a=x$ and $b=2$:

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

This matches expression (d) in Column B.

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(ii) $(x+3)^2$:

Using the square of a sum identity $(a+b)^2 = a^2 + 2ab + b^2$ with $a=x$ and $b=3$:

$(x+3)^2 = x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9$.

This matches expression (c) in Column B.

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(iii) $(x-1)(x-2)$:

Using the distributive property:

$(x-1)(x-2) = x(x-2) - 1(x-2)$

$= x^2 - 2x - x + 2$

$= x^2 - 3x + 2$.

This matches expression (b) in Column B.

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(iv) $(2x+1)(3x-2)$:

Using the distributive property (FOIL method):

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

$= 6x^2 - 4x + 3x - 2$

$= 6x^2 - x - 2$.

This matches expression (a) in Column B.

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Based on our expansions, the correct matches are:

• (i) - (d)
• (ii) - (c)
• (iii) - (b)
• (iv) - (a)

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Comparing this set of matches with the given options, option (B) provides the correct correspondence.

The correct option is (B).

Let's analyze the Assertion (A) and the Reason (R).

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Assertion (A): The polynomial $x^2 + x + 1$ cannot be factored into linear factors with real coefficients.

A quadratic polynomial $ax^2+bx+c$ with real coefficients can be factored into linear factors with real coefficients if and only if it has real roots.

The nature of the roots of the quadratic equation $ax^2+bx+c=0$ is determined by its discriminant, $\Delta = b^2 - 4ac$.

For the polynomial $x^2 + x + 1$, we have $a=1$, $b=1$, and $c=1$.

The discriminant is $\Delta = (1)^2 - 4(1)(1) = 1 - 4 = -3$.

Since the discriminant $\Delta = -3$ is less than $0$, the quadratic equation $x^2 + x + 1 = 0$ has no real roots.

Therefore, the polynomial $x^2 + x + 1$ cannot be factored into linear factors with real coefficients. Assertion (A) is True.

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Reason (R): The discriminant of the quadratic $ax^2+bx+c$ is $b^2-4ac$. If $b^2-4ac < 0$, the quadratic has no real roots.

This statement correctly defines the discriminant and its relationship to the existence of real roots for a quadratic equation. If the discriminant is negative, there are no real roots; the roots are complex conjugates.

Therefore, Reason (R) is True.

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Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) is true because the polynomial $x^2 + x + 1$ has a discriminant of $-3$, which is less than $0$. According to Reason (R), a negative discriminant implies no real roots. A quadratic polynomial with no real roots cannot be factored into linear factors with real coefficients.

Thus, Reason (R) provides the necessary condition (no real roots due to negative discriminant) that explains why Assertion (A) is true.

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Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A).

The profit $P(x)$ is defined as the revenue $R(x)$ minus the cost $C(x)$.

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We are given the cost polynomial $C(x) = 1000 + 50x + 0.1x^2$ and the revenue polynomial $R(x) = 200x - 0.2x^2$.

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Substitute the expressions for $R(x)$ and $C(x)$ into equation (i):

$P(x) = (200x - 0.2x^2) - (1000 + 50x + 0.1x^2)$

---

Remove the parentheses. Remember to distribute the negative sign to each term inside the second parenthesis:

$P(x) = 200x - 0.2x^2 - 1000 - 50x - 0.1x^2$

---

Group the like terms together:

$P(x) = (-0.2x^2 - 0.1x^2) + (200x - 50x) - 1000$

---

Combine the coefficients of the like terms:

For the $x^2$ terms: $-0.2x^2 - 0.1x^2 = (-0.2 - 0.1)x^2 = -0.3x^2$

For the $x$ terms: $200x - 50x = (200 - 50)x = 150x$

For the constant term: $-1000$

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Combine the results to get the polynomial for $P(x)$:

$P(x) = -0.3x^2 + 150x - 1000$

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This is the polynomial representing the profit from producing and selling $x$ toy cars.

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

The correct option is (A).

The degree of a polynomial is the highest power of the variable in the polynomial.

When two non-zero polynomials are multiplied, the degree of the resulting product polynomial is the sum of the degrees of the individual polynomials.

---

The first polynomial is $(x^3 - 2x + 1)$.

The highest power of $x$ in this polynomial is $3$.

So, the degree of the first polynomial is $3$.

---

The second polynomial is $(x^2 + 5)$.

The highest power of $x$ in this polynomial is $2$ (from the term $x^2$). The constant term $5$ can be written as $5x^0$, which has a degree of $0$).

So, the degree of the second polynomial is $2$.

---

The degree of the product of these two polynomials is the sum of their degrees:

Degree of product $= (\text{Degree of first polynomial}) + (\text{Degree of second polynomial})$

Degree of product $= 3 + 2 = 5$

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To illustrate, consider the multiplication of the terms with the highest powers from each polynomial:

From $(x^3 - 2x + 1)$, the term with the highest power is $x^3$.

From $(x^2 + 5)$, the term with the highest power is $x^2$.

When we multiply the two polynomials, the term with the highest power in the product will be the product of these highest power terms:

$x^3 \times x^2 = x^{3+2} = x^5$

No other combination of terms from the multiplication will result in a power of $x$ higher than $5$.

For example, $(x^3)(5) = 5x^3$, $(-2x)(x^2) = -2x^3$, $(-2x)(5) = -10x$, $(1)(x^2) = x^2$, $(1)(5) = 5$.

The highest power of $x$ in the expanded product $x^5 + 5x^3 - 2x^3 - 10x + x^2 + 5$ (which simplifies to $x^5 + 3x^3 + x^2 - 10x + 5$) is indeed $5$.

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The degree of the product of the polynomials $(x^3 - 2x + 1)$ and $(x^2 + 5)$ is 5.

The correct option is (A).

The cost of $n$ pens is given by the polynomial $p(n) = 2n+5$.

---

To find the cost of 10 pens, we need to substitute $n = 10$ into the polynomial $p(n)$.

---

Substitute $n = 10$ into the expression for $p(n)$:

$p(10) = 2(10) + 5$

---

Perform the multiplication and addition:

$p(10) = 20 + 5$

$p(10) = 25$

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The value of $p(10)$ is $25$. This represents the cost of 10 pens.

The cost is given in $\textsf{₹}$.

Therefore, the cost of 10 pens is $\textsf{₹} 25$.

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Comparing our result with the given options, we find that option (A) is $\textsf{₹} 25$.

The correct option is (A).

According to the Factor Theorem, if $(x-a)$ is a factor of a polynomial $p(x)$, then $p(a) = 0$.

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In this problem, the polynomial is $p(x) = x^2 + kx - 4$, and we are given that $(x-2)$ is a factor.

Comparing the factor $(x-2)$ with the form $(x-a)$, we see that $a = 2$.

---

By the Factor Theorem, if $(x-2)$ is a factor of $p(x)$, then $p(2)$ must be equal to $0$.

---

Now, we evaluate the polynomial $p(x)$ at $x = 2$:

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

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

---

Simplify the expression for $p(2)$:

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

$p(2) = 0 + 2k$

$p(2) = 2k$

---

Now, using the condition from the Factor Theorem (equation (i)), we set $p(2)$ equal to $0$:

---

Solve for $k$ by dividing both sides of the equation by $2$:

---

Thus, the value of $k$ is 0.

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Verification: If $k=0$, the polynomial is $p(x) = x^2 + 0x - 4 = x^2 - 4$. We know that $x^2-4 = (x-2)(x+2)$. Since $(x-2)$ is a factor of $x^2-4$, our value of $k=0$ is correct.

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The correct option is (A).

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, the remainder is $p(a)$.

If the divisor is in the form $(ax + b)$, we first find the zero of the divisor by setting it to zero: $ax+b = 0 \implies ax = -b \implies x = -\frac{b}{a}$. The remainder is then $p(-\frac{b}{a})$.

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In this problem, the polynomial is $p(x) = 4x^3 - 3x^2 + 2x - 1$, and the divisor is $(2x - 1)$.

Set the divisor equal to zero to find its zero:

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According to the Remainder Theorem, the remainder when $p(x)$ is divided by $(2x - 1)$ is $p(\frac{1}{2})$.

---

Substitute $x = \frac{1}{2}$ into the polynomial $p(x)$:

$p(x) = 4x^3 - 3x^2 + 2x - 1$

$p(\frac{1}{2}) = 4(\frac{1}{2})^3 - 3(\frac{1}{2})^2 + 2(\frac{1}{2}) - 1$

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Evaluate the powers and perform the multiplications:

$(\frac{1}{2})^3 = \frac{1^3}{2^3} = \frac{1}{8}$

$(\frac{1}{2})^2 = \frac{1^2}{2^2} = \frac{1}{4}$

$2(\frac{1}{2}) = \cancel{2} \times \frac{1}{\cancel{2}} = 1$

---

Substitute these values back into the expression for $p(\frac{1}{2})$:

$p(\frac{1}{2}) = 4(\frac{1}{8}) - 3(\frac{1}{4}) + 1 - 1$

$p(\frac{1}{2}) = \frac{\cancel{4}}{\cancel{8}_2} - \frac{3}{4} + 0$

$p(\frac{1}{2}) = \frac{1}{2} - \frac{3}{4}$

---

To subtract the fractions, find a common denominator, which is 4:

$p(\frac{1}{2}) = \frac{1 \times 2}{2 \times 2} - \frac{3}{4}$

$p(\frac{1}{2}) = \frac{2}{4} - \frac{3}{4}$

$p(\frac{1}{2}) = \frac{2 - 3}{4}$

$p(\frac{1}{2}) = -\frac{1}{4}$

---

The remainder when $p(x)$ is divided by $(2x - 1)$ is $-\frac{1}{4}$.

The correct option is (D).

Let's analyze each statement about the zeros of a polynomial.

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Statement (i): A polynomial can have more than one zero.

Consider the polynomial $p(x) = x^2 - 4$. We find its zeros by setting $p(x) = 0$: $x^2 - 4 = 0 \implies (x-2)(x+2) = 0$. The zeros are $x=2$ and $x=-2$. This polynomial has two zeros.

In general, a polynomial of degree $n$ can have up to $n$ distinct zeros. For $n \ge 2$, it is possible to have more than one zero.

Thus, statement (i) is True.

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Statement (ii): Every linear polynomial has exactly one zero.

A linear polynomial is of the form $p(x) = ax+b$, where $a$ and $b$ are real numbers and $a \neq 0$.

To find the zero, we set $p(x) = 0$: $ax + b = 0$.

Solving for $x$: $ax = -b$, which gives $x = -\frac{b}{a}$.

Since $a \neq 0$, $-\frac{b}{a}$ is a unique real number. This is the only value of $x$ for which $p(x) = 0$.

Thus, every linear polynomial has exactly one zero. Statement (ii) is True.

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Statement (iii): The zero of a polynomial is the value of the variable for which the polynomial equals zero.

This is the fundamental definition of a zero of a polynomial. A value $c$ is a zero (or root) of a polynomial $p(x)$ if $p(c) = 0$.

Thus, statement (iii) is True.

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Statement (iv): A constant polynomial has no zero.

A constant polynomial is of the form $p(x) = c$, where $c$ is a real number.

There are two cases for a constant polynomial:

• Case 1: The constant is non-zero ($c \neq 0$), e.g., $p(x) = 5$. To find the zero, we set $p(x) = 0 \implies 5 = 0$. This equation has no solution, so a non-zero constant polynomial has no zeros.
• Case 2: The constant is zero ($c = 0$), which is the zero polynomial, $p(x) = 0$. To find the zero, we set $p(x) = 0 \implies 0 = 0$. This equation is true for every real value of $x$. So, the zero polynomial has infinitely many zeros (every real number is a zero).

The statement "A constant polynomial has no zero" is true only for non-zero constant polynomials. Since the zero polynomial is also a constant polynomial and it *does* have zeros, the general statement "A constant polynomial has no zero" is false.

Thus, statement (iv) is False.

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Based on our analysis, statements (i), (ii), and (iii) are true, while statement (iv) is false.

The option that lists (i), (ii), and (iii) as the only true statements is (C).

The correct option is (C).

We are asked to find which of the given expressions is NOT equal to $(a-b)(a^2+ab+b^2)$.

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Let's expand the given expression $(a-b)(a^2+ab+b^2)$ using the distributive property:

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

Expanding further:

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

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

Removing the parenthesis and changing the signs inside the second parenthesis:

$= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3$

Combine like terms ($a^2b$ and $-a^2b$, $ab^2$ and $-ab^2$):

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

$= a^3 + 0 + 0 - b^3$

$= a^3 - b^3$

So, the expression $(a-b)(a^2+ab+b^2)$ is equal to $a^3 - b^3$. This is a well-known algebraic identity for the difference of cubes.

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Now let's examine each option given:

(A) $a^3 - b^3$

As shown in the expansion above, $(a-b)(a^2+ab+b^2)$ is equal to $a^3 - b^3$. So, option (A) is equal to the given expression.

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

We can factor out $-1$ from $(b-a)$ to get $-(a-b)$.

We can also factor out $-1$ from $(-a^2-ab-b^2)$ to get $-(a^2+ab+b^2)$.

So, $(b-a)(-a^2-ab-b^2) = [-(a-b)][-(a^2+ab+b^2)]$

$= (-1) \cdot (a-b) \cdot (-1) \cdot (a^2+ab+b^2)$

$= (-1) \cdot (-1) \cdot (a-b)(a^2+ab+b^2)$

$= 1 \cdot (a-b)(a^2+ab+b^2)$

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

Since $(a-b)(a^2+ab+b^2) = a^3 - b^3$, option (B) is also equal to $a^3 - b^3$, and thus equal to the given expression.

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(C) $a^3 + b^3$

The expansion of the given expression is $a^3 - b^3$. The expression $a^3 + b^3$ is the sum of cubes, which is equal to $(a+b)(a^2-ab+b^2)$.

Since $a^3 - b^3$ is not equal to $a^3 + b^3$ (unless $b=0$), option (C) is NOT equal to the given expression.

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

This expression is the direct result of applying the distributive property to $(a-b)(a^2+ab+b^2)$ in the first step of our expansion.

This expression is equal to $(a-b)(a^2+ab+b^2)$ by definition of the distributive property.

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Comparing the results, options (A), (B), and (D) are all equal to $a^3 - b^3$, which is the expanded form of the given expression $(a-b)(a^2+ab+b^2)$. Option (C), $a^3 + b^3$, is not equal to $a^3 - b^3$.

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Therefore, the expression which is NOT equal to $(a-b)(a^2+ab+b^2)$ is $a^3 + b^3$.

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The final answer is $\boxed{(C)}$.

The question asks to fill in the blank in the expansion of $(x+y+z)^2 = x^2+y^2+z^2 + 2(\dots)$.

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We need to expand the expression $(x+y+z)^2$. The formula for the square of a trinomial is given by:

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

---

We can derive this by multiplying $(x+y+z)$ by itself:

$(x+y+z)^2 = (x+y+z)(x+y+z)$

Using the distributive property:

$= x(x+y+z) + y(x+y+z) + z(x+y+z)$

$= (x \cdot x + x \cdot y + x \cdot z) + (y \cdot x + y \cdot y + y \cdot z) + (z \cdot x + z \cdot y + z \cdot z)$

$= (x^2 + xy + xz) + (yx + y^2 + yz) + (zx + zy + z^2)$

Combining like terms ($xy = yx$, $xz = zx$, $yz = zy$):

$= x^2 + y^2 + z^2 + xy + yx + xz + zx + yz + zy$

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

We can factor out 2 from the terms $2xy$, $2xz$, and $2yz$:

$= x^2 + y^2 + z^2 + 2(xy + xz + yz)$

Rearranging the terms inside the parenthesis alphabetically:

$= x^2 + y^2 + z^2 + 2(xy + yz + zx)$

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The given incomplete expression is $(x+y+z)^2 = x^2+y^2+z^2 + 2(\dots)$.

Comparing this with the full expansion $x^2 + y^2 + z^2 + 2(xy + yz + zx)$, we see that the expression inside the parenthesis is $xy + yz + zx$.

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Now let's check the given options:

(A) $xy+yz+zx$ - This matches the expression we found.

(B) $x+y+z$ - This is incorrect.

(C) $xyz$ - This is incorrect.

(D) $xy \cdot yz \cdot zx = x^2y^2z^2$ - This is incorrect.

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Therefore, the expression that fills the blank is $xy+yz+zx$.

The correct option is (A).

The question asks us to simplify the expression $(105)^2 - (95)^2$ using a suitable identity.

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

The suitable algebraic identity for the difference of squares is:

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

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In this problem, we have $a = 105$ and $b = 95$.

Applying the identity, we get:

$(105)^2 - (95)^2 = (105 - 95)(105 + 95)$

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Now, we calculate the values inside the parentheses:

$105 - 95 = 10$

$105 + 95 = 200$

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Substitute these values back into the expression:

$(105 - 95)(105 + 95) = 10 \times 200$

Now, perform the multiplication:

$10 \times 200 = 2000$

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So, the simplified value of $(105)^2 - (95)^2$ using the identity is $2000$.

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Let's examine the given options:

(A) $(105-95)^2 = 10^2 = 100$. This option uses the identity $(a-b)^2$, which is not the correct identity for $a^2 - b^2$. This method is incorrect.

(B) $(105+95)(105-95) = 200 \times 10 = 2000$. This option correctly uses the identity $a^2 - b^2 = (a+b)(a-b)$ and performs the calculations correctly. This method and result match our derivation.

(C) $(105-95)(105-95) = 10 \times 10 = 100$. This option incorrectly uses the identity $(a-b)^2$ for $a^2 - b^2$, same as option (A).

(D) $(105)^2 - (95)^2 = 11025 - 9025 = 2000$. This option calculates the squares directly and subtracts, which yields the correct numerical result. However, the question specifically asks to use a suitable identity. While the result is correct, the method shown in this option does not demonstrate the use of an identity for simplification, which is the core requirement of the question.

---

Therefore, option (B) correctly uses the suitable identity $a^2 - b^2 = (a+b)(a-b)$ to simplify the expression and arrives at the correct result.

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The final answer is $\boxed{(B)}$.

The question asks to identify which of the given expressions is a constant polynomial.

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

A constant polynomial is a polynomial whose value is a constant number (a number that does not change). This means it does not contain any variables with non-zero exponents.

A non-zero constant polynomial can be written in the form $c$, where $c$ is a non-zero real number. The degree of a non-zero constant polynomial is 0.

The number $0$ is also considered a constant polynomial, called the zero polynomial. Its degree is often defined as negative infinity or undefined.

---

Let's examine each option:

(A) $x + 1$: This is a polynomial with a variable $x$ raised to the power of 1. Its degree is 1. This is a linear polynomial.

(B) $0$: This expression is simply the number 0. According to the definition, 0 is the zero polynomial, which is a type of constant polynomial.

(C) $2x$: This is a polynomial with a variable $x$ raised to the power of 1. Its degree is 1. This is a linear polynomial.

(D) $y^2 - 1$: This is a polynomial with a variable $y$ raised to the power of 2. Its degree is 2. This is a quadratic polynomial.

---

Comparing the options with the definition of a constant polynomial, only the expression $0$ fits the description of a constant polynomial (specifically, the zero polynomial).

---

Therefore, the constant polynomial among the given options is 0.

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The final answer is $\boxed{(B)}$.

We are given the polynomial function $p(x) = x-1$.

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We need to find the expression for $p(p(x))$. This is a function composition, where we substitute the entire expression for $p(x)$ into the function $p(x)$ wherever $x$ appears.

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Given $p(x) = x-1$.

To find $p(p(x))$, we replace the variable $x$ in the expression for $p(x)$ with the expression for $p(x)$ itself.

So, $p(p(x)) = p(\underbrace{x-1}_{p(x)})$

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Now, substitute $(x-1)$ into the definition of $p(x)$. The definition of $p(x)$ says "take the input and subtract 1". Here, the input is $(x-1)$.

$p(\text{input}) = \text{input} - 1$

$p(x-1) = (x-1) - 1$

---

Simplify the expression:

$(x-1) - 1 = x - 1 - 1 = x - 2$

---

So, $p(p(x)) = x-2$.

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Now let's check the given options:

(A) $x-1$ - This is the expression for $p(x)$, not $p(p(x))$.

(B) $x-2$ - This matches our calculated result.

(C) $x^2-1$ - This is incorrect.

(D) $x^2-2x+1$ - This is equal to $(x-1)^2$, which is $p(x)^2$, not $p(p(x))$.

---

Therefore, $p(p(x))$ is equal to $x-2$.

---

The final answer is $\boxed{(B)}$.

Definition of a Polynomial:

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

In general, a polynomial in one variable $x$ can be written in the form:

$a_n x^n + a_{n-1} x^{n-1} + ... + a_2 x^2 + a_1 x + a_0$

where $a_n, a_{n-1}, ..., a_1, a_0$ are coefficients (real or complex numbers), $x$ is the variable, and $n$ is a non-negative integer representing the highest power of $x$. Note that the exponents of the variable must be whole numbers (non-negative integers).

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Degree of a Polynomial:

The degree of a polynomial is the highest power of the variable in the polynomial with a non-zero coefficient.

For example, in the general form $a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$, if $a_n \neq 0$, the degree of the polynomial is $n$.

The degree of a constant polynomial (a non-zero number) is 0. The degree of the zero polynomial (the number 0) is often considered undefined or sometimes $- \infty$.

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Example of a Polynomial and its Degree:

Consider the expression: $5x^3 - 7x^2 + 2x - 10$

This is a polynomial because it consists of variables ($x$) and coefficients ($5, -7, 2, -10$), involving only addition, subtraction, multiplication, and non-negative integer exponents ($3, 2, 1$, and $0$ for the constant term $-10 = -10x^0$).

To find the degree, we look for the highest power of $x$ with a non-zero coefficient. The terms are $5x^3$, $-7x^2$, $2x^1$, and $-10x^0$.

The powers are $3, 2, 1,$ and $0$. The highest power is $3$, and its coefficient ($5$) is non-zero.

Therefore, the degree of the polynomial $5x^3 - 7x^2 + 2x - 10$ is $3$.

Let's examine each expression based on the definition of a polynomial, which requires that the exponents of the variables must be non-negative integers.

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(a) $x^2 - 5x + \sqrt{2}$

The powers of the variable $x$ in this expression are $2$, $1$ (since $5x = 5x^1$), and $0$ (since $\sqrt{2} = \sqrt{2}x^0$).

These exponents ($2, 1, 0$) are all non-negative integers.

The coefficients ($1, -5, \sqrt{2}$) are real numbers.

Therefore, the expression $x^2 - 5x + \sqrt{2}$ is a polynomial.

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(b) $y + \frac{1}{y}$

The expression can be rewritten as $y + y^{-1}$.

The powers of the variable $y$ are $1$ and $-1$.

One of the exponents, $-1$, is a negative integer.

Since a polynomial requires non-negative integer exponents for the variables, this expression is not a polynomial.

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(c) $3\sqrt{t} + t\sqrt{2}$

The expression can be rewritten using fractional exponents as $3t^{1/2} + \sqrt{2}t^{1}$.

The powers of the variable $t$ are $1/2$ and $1$.

One of the exponents, $1/2$, is not an integer.

Since a polynomial requires only non-negative integer exponents for the variables, this expression is not a polynomial.

The given polynomial is $5x^3 - 7x^2 + \frac{3}{2}x - 11$.

We need to identify the coefficients of $x^2$, $x$, and the constant term.

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The coefficient of a term is the numerical factor that multiplies the variable raised to a specific power.

The constant term is the term that does not contain the variable (or contains the variable raised to the power of 0).

---

The term with $x^2$ is $-7x^2$. The coefficient of $x^2$ is the number multiplying $x^2$.

Coefficient of $x^2$ = $-7$

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The term with $x$ (or $x^1$) is $+\frac{3}{2}x$. The coefficient of $x$ is the number multiplying $x$.

Coefficient of $x$ = $\frac{3}{2}$

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The constant term is the term without $x$. In this polynomial, it is $-11$.

Constant term = $-11$

We classify polynomials based on their degree:

A polynomial of degree 1 is called linear.

A polynomial of degree 2 is called quadratic.

A polynomial of degree 3 is called cubic.

A polynomial of degree 4 is called biquadratic.

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(a) $x - 5$

The highest power of the variable $x$ in this polynomial is $1$ (since $x = x^1$).

The degree of the polynomial is $1$.

Therefore, $x - 5$ is a linear polynomial.

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(b) $3y^2 + 2y + 1$

The terms in the polynomial are $3y^2$, $2y^1$, and $1y^0$.

The powers of the variable $y$ are $2$, $1$, and $0$.

The highest power is $2$.

The degree of the polynomial is $2$.

Therefore, $3y^2 + 2y + 1$ is a quadratic polynomial.

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(c) $7z^3 - 4z^2 + 9$

The terms in the polynomial are $7z^3$, $-4z^2$, and $9z^0$.

The powers of the variable $z$ are $3$, $2$, and $0$.

The highest power is $3$.

The degree of the polynomial is $3$.

Therefore, $7z^3 - 4z^2 + 9$ is a cubic polynomial.

We classify polynomials based on the number of terms they contain:

A polynomial with one term is called a monomial.

A polynomial with two terms is called a binomial.

A polynomial with three terms is called a trinomial.

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(a) $4p^3$

This expression consists of a single term, which is $4p^3$.

Therefore, $4p^3$ is a monomial.

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

This expression consists of three terms, which are $a^2$, $-2ab$, and $b^2$. These terms are separated by addition and subtraction signs.

Therefore, $a^2 - 2ab + b^2$ is a trinomial.

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(c) $x + y$

This expression consists of two terms, which are $x$ and $y$. These terms are separated by an addition sign.

Therefore, $x + y$ is a binomial.

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Given:

The given polynomial is $p(x) = 2x^2 - 3x + 5$.

We need to find the value of the polynomial when $x = 2$.

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Solution:

To find the value of the polynomial $p(x)$ when $x = 2$, we substitute $x=2$ into the expression for $p(x)$.

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

$p(2) = 2(4) - 6 + 5$

$p(2) = 8 - 6 + 5$

$p(2) = 2 + 5$

$p(2) = 7$

Thus, the value of the polynomial $p(x) = 2x^2 - 3x + 5$ when $x=2$ is $\mathbf{7}$.

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Given:

The given polynomial is $q(y) = y^3 + 2y^2 - 5y - 6$.

We need to find the value of the polynomial when $y = -1$.

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Solution:

To find the value of the polynomial $q(y)$ when $y = -1$, we substitute $y = -1$ into the expression for $q(y)$.

$q(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6$

$q(-1) = (-1) + 2(1) - (-5) - 6$

$q(-1) = -1 + 2 + 5 - 6$

$q(-1) = 1 + 5 - 6$

$q(-1) = 6 - 6$

$q(-1) = 0$

Thus, the value of the polynomial $q(y) = y^3 + 2y^2 - 5y - 6$ when $y = -1$ is $\mathbf{0}$.

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Definition of a Zero of a Polynomial:

A zero of a polynomial $p(x)$ is a value of the variable (say, $x=a$) such that when this value is substituted into the polynomial, the result is zero. That is, if $p(a) = 0$, then $a$ is called a zero of the polynomial $p(x)$.

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Given:

The given polynomial is $p(x) = x^2 - 5x + 6$.

We need to check if $x = 3$ is a zero of this polynomial.

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Solution:

To check if $x = 3$ is a zero of $p(x)$, we evaluate the polynomial at $x = 3$.

$p(3) = (3)^2 - 5(3) + 6$

$p(3) = 9 - 15 + 6$

$p(3) = -6 + 6$

$p(3) = 0$

Since the value of the polynomial is $0$ when $x = 3$, we conclude that $\mathbf{x = 3}$ is a zero of the polynomial $p(x) = x^2 - 5x + 6$.

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Given:

The given linear polynomial is $p(x) = 5x - 15$.

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To Find:

The zero of the polynomial $p(x)$.

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Solution:

To find the zero of the polynomial $p(x)$, we set the polynomial equal to zero and solve for $x$.

$p(x) = 0$

Substitute the expression for $p(x)$: $5x - 15 = 0$

Add 15 to both sides of the equation:

$5x = 15$

Divide both sides by 5:

$x = \frac{15}{5}$

$x = 3$

To verify, substitute $x=3$ into $p(x)$:

$p(3) = 5(3) - 15 = 15 - 15 = 0$.

Since $p(3) = 0$, $x=3$ is the zero of the polynomial.

The zero of the linear polynomial $p(x) = 5x - 15$ is $\mathbf{3}$.

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Statement of the Remainder Theorem:

Let $p(x)$ be any polynomial of degree greater than or equal to one and let $a$ be any real number. If $p(x)$ is divided by the linear polynomial $(x - a)$, then the remainder is $p(a)$.

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Application of the Remainder Theorem:

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by the linear polynomial $(x - a)$, the remainder is obtained by evaluating the polynomial $p(x)$ at $x = a$.

Therefore, the remainder is $\mathbf{p(a)}$.

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Given:

The given polynomial is $p(x) = x^2 + 3x + 7$.

The divisor is $(x - 1)$.

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To Find:

The remainder when $p(x)$ is divided by $(x - 1)$ using the Remainder Theorem.

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Solution:

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, the remainder is $p(a)$.

In this case, the divisor is $(x - 1)$. Comparing this with $(x - a)$, we find that $a = 1$.

Therefore, the remainder when $p(x) = x^2 + 3x + 7$ is divided by $(x - 1)$ is $p(1)$.

Substitute $x = 1$ into the polynomial $p(x)$:

$p(1) = (1)^2 + 3(1) + 7$

$p(1) = 1 + 3 + 7$

$p(1) = 4 + 7$

$p(1) = 11$

So, the remainder is $\mathbf{11}$.

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Given:

The given polynomial is $p(x) = x^3 - 4x^2 + 5x - 2$.

The divisor is $(x + 2)$.

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To Find:

The remainder when $p(x)$ is divided by $(x + 2)$ using the Remainder Theorem.

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Solution:

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, the remainder is $p(a)$.

In this case, the divisor is $(x + 2)$. We can write $(x + 2)$ as $(x - (-2))$.

Comparing this with $(x - a)$, we find that $a = -2$.

Therefore, the remainder when $p(x) = x^3 - 4x^2 + 5x - 2$ is divided by $(x + 2)$ is $p(-2)$.

Substitute $x = -2$ into the polynomial $p(x)$:

$p(-2) = (-2)^3 - 4(-2)^2 + 5(-2) - 2$

$p(-2) = (-8) - 4(4) + (-10) - 2$

$p(-2) = -8 - 16 - 10 - 2$

$p(-2) = -24 - 10 - 2$

$p(-2) = -34 - 2$

$p(-2) = -36$

So, the remainder is $\mathbf{-36}$.

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Statement of the Factor Theorem:

Let $p(x)$ be a polynomial of degree greater than or equal to one, and let $a$ be a real number.

(i) If $(x - a)$ is a factor of $p(x)$, then $p(a) = 0$.

(ii) If $p(a) = 0$, then $(x - a)$ is a factor of $p(x)$.

These two statements can be combined into a single "if and only if" statement: $(x - a)$ is a factor of the polynomial $p(x)$ if and only if $p(a) = 0$.

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Relationship between the Zero of a Polynomial and its Factors:

The Factor Theorem establishes a direct relationship between the zeros of a polynomial and its linear factors.

Recall that a zero of a polynomial $p(x)$ is a value $a$ such that $p(a) = 0$.

According to the Factor Theorem:

If $a$ is a zero of the polynomial $p(x)$ (i.e., $p(a) = 0$), then $(x - a)$ is a factor of $p(x)$.

Conversely, if $(x - a)$ is a factor of the polynomial $p(x)$, then $a$ is a zero of the polynomial $p(x)$ (i.e., $p(a) = 0$).

In simple terms, finding the zeros of a polynomial is equivalent to finding its linear factors of the form $(x-a)$. If you know a zero $a$, you know a factor $(x-a)$, and if you know a factor $(x-a)$, you know a zero $a$.

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Given:

The given polynomial is $p(x) = x^2 + 2x - 3$.

We need to check if $(x - 1)$ is a factor of $p(x)$.

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To Show:

Using the Factor Theorem, show that $(x - 1)$ is a factor of $p(x)$.

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Solution:

According to the Factor Theorem, a linear polynomial $(x - a)$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$.

In this problem, the potential factor is $(x - 1)$. Comparing this with $(x - a)$, we have $a = 1$.

To show that $(x - 1)$ is a factor of $p(x)$, we need to evaluate $p(x)$ at $x = 1$ and check if the result is 0.

Substitute $x = 1$ into the polynomial $p(x) = x^2 + 2x - 3$:

$p(1) = (1)^2 + 2(1) - 3$

$p(1) = 1 + 2 - 3$

$p(1) = 3 - 3$

$p(1) = 0$

Since $p(1) = 0$, by the Factor Theorem, $(x - 1)$ is a factor of the polynomial $p(x) = x^2 + 2x - 3$.

Hence, it is shown that $\mathbf{(x - 1)}$ is a factor of $\mathbf{p(x) = x^2 + 2x - 3}$.

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Given:

The given polynomial is $p(x) = x^3 + 3x^2 + 5x + 6$.

We need to check if $(x + 2)$ is a factor of $p(x)$.

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To Check:

Using the Factor Theorem, check if $(x + 2)$ is a factor of $p(x)$.

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Solution:

According to the Factor Theorem, a linear polynomial $(x - a)$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$.

In this problem, the potential factor is $(x + 2)$. We can write $(x + 2)$ as $(x - (-2))$.

Comparing this with $(x - a)$, we have $a = -2$.

To check if $(x + 2)$ is a factor of $p(x)$, we need to evaluate $p(x)$ at $x = -2$ and check if the result is 0.

Substitute $x = -2$ into the polynomial $p(x) = x^3 + 3x^2 + 5x + 6$:

$p(-2) = (-2)^3 + 3(-2)^2 + 5(-2) + 6$

$p(-2) = (-8) + 3(4) + (-10) + 6$

$p(-2) = -8 + 12 - 10 + 6$

$p(-2) = 4 - 10 + 6$

$p(-2) = -6 + 6$

$p(-2) = 0$

Since $p(-2) = 0$, by the Factor Theorem, $(x + 2)$ is a factor of the polynomial $p(x) = x^3 + 3x^2 + 5x + 6$.

Therefore, $\mathbf{(x + 2)}$ is a factor of $\mathbf{p(x) = x^3 + 3x^2 + 5x + 6}$.

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Given:

The expressions to expand are:

(a) $(2x + 3)^2$

(b) $(y - 5)^2$

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To Expand:

Expand the given expressions using algebraic identities.

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Solution:

(a) $(2x + 3)^2$

This expression is in the form of $(a + b)^2$. We use the algebraic identity: $(a + b)^2 = a^2 + 2ab + b^2$

Here, $a = 2x$ and $b = 3$. Substituting these values into the identity:

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

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

Thus, the expansion of $(2x + 3)^2$ is $\mathbf{4x^2 + 12x + 9}$.

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(b) $(y - 5)^2$

This expression is in the form of $(a - b)^2$. We use the algebraic identity: $(a - b)^2 = a^2 - 2ab + b^2$

Here, $a = y$ and $b = 5$. Substituting these values into the identity:

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

$(y - 5)^2 = y^2 - 10y + 25$

Thus, the expansion of $(y - 5)^2$ is $\mathbf{y^2 - 10y + 25}$.

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Given:

The expressions to expand are:

(a) $(3p + 4q)(3p - 4q)$

(b) $(x + 2y + 3z)^2$

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To Expand:

Expand the given expressions using algebraic identities.

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Solution:

(a) $(3p + 4q)(3p - 4q)$

This expression is in the form of $(a + b)(a - b)$. We use the algebraic identity:

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

Here, $a = 3p$ and $b = 4q$. Substituting these values into the identity:

$(3p + 4q)(3p - 4q) = (3p)^2 - (4q)^2$

$(3p + 4q)(3p - 4q) = 9p^2 - 16q^2$

Thus, the expansion of $(3p + 4q)(3p - 4q)$ is $\mathbf{9p^2 - 16q^2}$.

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(b) $(x + 2y + 3z)^2$

This expression is in the form of $(a + b + c)^2$. We use the algebraic identity:

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

Here, $a = x$, $b = 2y$, and $c = 3z$. Substituting these values into the identity:

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

$(x + 2y + 3z)^2 = x^2 + 4y^2 + 9z^2 + 4xy + 12yz + 6zx$

Thus, the expansion of $(x + 2y + 3z)^2$ is $\mathbf{x^2 + 4y^2 + 9z^2 + 4xy + 12yz + 6zx}$.

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Given:

The expression to factorise is $6xy^2 - 9x^2y + 12xy$.

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To Factorise:

Factorise the given expression by taking out common factors.

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Solution:

We need to find the greatest common factor (GCF) of the terms $6xy^2$, $-9x^2y$, and $12xy$.

First, find the GCF of the numerical coefficients 6, -9, and 12. The GCF of 6, 9, and 12 is 3.

Next, consider the variables. The lowest power of $x$ present in all terms is $x^1$ (from $6xy^2$, $-9x^2y$, and $12xy$).

The lowest power of $y$ present in all terms is $y^1$ (from $6xy^2$, $-9x^2y$, and $12xy$).

So, the GCF of the terms $6xy^2$, $-9x^2y$, and $12xy$ is $3xy$.

Now, we factor out $3xy$ from each term:

$6xy^2 = 3xy \times (2y)$

$-9x^2y = 3xy \times (-3x)$

$12xy = 3xy \times (4)$

Therefore, the factorised expression is:

$6xy^2 - 9x^2y + 12xy = 3xy(2y - 3x + 4)$

The factorised form of the expression is $\mathbf{3xy(2y - 3x + 4)}$.

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Given:

The expression to factorise is $ab - a - b + 1$.

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To Factorise:

Factorise the given expression by grouping terms.

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Solution:

We group the terms in pairs and factor out common factors from each pair.

Group the first two terms and the last two terms:

$(ab - a) + (-b + 1)$

Factor out the common factor 'a' from the first group:

$a(b - 1)$

Factor out the common factor '-1' from the second group:

$-1(b - 1)$

Now the expression is:

$a(b - 1) - 1(b - 1)$

We can see that $(b - 1)$ is a common factor in both terms. Factor out $(b - 1)$:

$(b - 1)(a - 1)$

Alternatively, we can group the terms differently:

Group the first and third terms, and the second and fourth terms:

$(ab - b) + (-a + 1)$

Factor out the common factor 'b' from the first group:

$b(a - 1)$

Factor out the common factor '-1' from the second group:

$-1(a - 1)$

Now the expression is:

$b(a - 1) - 1(a - 1)$

Factor out the common factor $(a - 1)$:

$(a - 1)(b - 1)$

Both methods yield the same result.

The factorised form of the expression is $\mathbf{(a - 1)(b - 1)}$.

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Given:

The expression to factorise is $49x^2 - 25y^2$.

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To Factorise:

Factorise the given expression using the identity $a^2 - b^2 = (a - b)(a + b)$.

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Solution:

The given expression is $49x^2 - 25y^2$.

We can rewrite the terms to fit the form $a^2 - b^2$.

The first term $49x^2$ can be written as $(7x)^2$, since $(7x)^2 = 7^2 x^2 = 49x^2$.

The second term $25y^2$ can be written as $(5y)^2$, since $(5y)^2 = 5^2 y^2 = 25y^2$.

So, the expression becomes:

$(7x)^2 - (5y)^2$

Now, this is in the form $a^2 - b^2$, where $a = 7x$ and $b = 5y$.

We use the identity:

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

Substitute $a = 7x$ and $b = 5y$ into the identity:

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

Therefore, the factorised form of the expression is $\mathbf{(7x - 5y)(7x + 5y)}$.

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Given:

The expression to factorise is $x^2 + 10x + 25$.

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To Factorise:

Factorise the given expression using algebraic identities.

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Solution:

The given expression is $x^2 + 10x + 25$.

We observe that the first term $x^2$ is a perfect square $(x)^2$, and the last term 25 is a perfect square $(5)^2$.

The middle term is $10x$. Let's check if this matches the form $2ab$ from the identity $(a+b)^2 = a^2 + 2ab + b^2$.

If we consider $a=x$ and $b=5$, then $a^2 = x^2$, $b^2 = 5^2 = 25$, and $2ab = 2(x)(5) = 10x$.

The expression $x^2 + 10x + 25$ matches the form $a^2 + 2ab + b^2$ with $a=x$ and $b=5$.

Using the identity:

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

Substitute $a = x$ and $b = 5$ into the identity:

$x^2 + 10x + 25 = (x + 5)^2$

The factorised form can also be written as $(x + 5)(x + 5)$.

The factorised form of the expression is $\mathbf{(x + 5)^2}$.

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Given:

The quadratic trinomial to factorise is $x^2 + 5x + 6$.

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To Factorise:

Factorise the given trinomial by splitting the middle term.

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Solution:

The given trinomial is in the form $ax^2 + bx + c$, where $a = 1$, $b = 5$, and $c = 6$.

We need to split the middle term $bx$ (which is $5x$) into two terms such that their sum is $5x$ and their product is equal to the product of the first and last terms, i.e., $ac \times x^2 = (1)(6) \times x^2 = 6x^2$.

We need to find two numbers whose sum is 5 and whose product is 6.

Let the two numbers be $p$ and $q$. We need $p + q = 5$ and $p \times q = 6$.

We look for pairs of factors of 6: (1, 6), (2, 3), (-1, -6), (-2, -3).

Check the sums of these pairs:

$1 + 6 = 7$ (Not 5)

$2 + 3 = 5$ (This is 5)

So, the two numbers are 2 and 3.

We split the middle term $5x$ as $2x + 3x$.

Now rewrite the trinomial:

$x^2 + 5x + 6 = x^2 + 2x + 3x + 6$

Now, group the terms in pairs and factor out common factors:

$(x^2 + 2x) + (3x + 6)$

Factor out $x$ from the first group:

$x(x + 2)$

Factor out 3 from the second group:

$3(x + 2)$

Now the expression is:

$x(x + 2) + 3(x + 2)$

We see that $(x + 2)$ is a common factor in both terms. Factor out $(x + 2)$:

$(x + 2)(x + 3)$

The factorised form of the quadratic trinomial $x^2 + 5x + 6$ is $\mathbf{(x + 2)(x + 3)}$.

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Given:

The polynomial is $p(x) = (x-2)(x+3)$.

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To Find:

The zeros of the polynomial $p(x)$.

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Solution:

To find the zeros of a polynomial, we set the polynomial equal to zero and solve for the variable.

Given $p(x) = (x-2)(x+3)$.

Set $p(x) = 0$:

$(x-2)(x+3) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

So, we have two possible cases:

Case 1: $x - 2 = 0$

Add 2 to both sides:

$x = 2$

Case 2: $x + 3 = 0$

Subtract 3 from both sides:

$x = -3$

The values of $x$ for which $p(x) = 0$ are $x = 2$ and $x = -3$.

Therefore, the zeros of the polynomial $p(x) = (x-2)(x+3)$ are $\mathbf{2}$ and $\mathbf{-3}$.

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Given:

The polynomial is $p(x) = x^2 + ax - 2$.

$(x - 1)$ is a factor of $p(x)$.

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To Find:

The value of $a$.

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Solution:

According to the Factor Theorem, if $(x - a)$ is a factor of a polynomial $p(x)$, then $p(a) = 0$.

In this problem, the factor is $(x - 1)$. Comparing this with $(x - a)$, we have $a = 1$.

Since $(x - 1)$ is a factor of $p(x) = x^2 + ax - 2$, by the Factor Theorem, we must have $p(1) = 0$.

Substitute $x = 1$ into the polynomial $p(x)$:

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

$p(1) = 1 + a - 2$

$p(1) = a - 1$

Now, set $p(1) = 0$ and solve for $a$:

$a - 1 = 0$

Add 1 to both sides of the equation:

$a = 1$

Thus, the value of $a$ is $\mathbf{1}$.

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Given:

The expression to expand is $(x+y)^3$.

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To Expand:

Expand the given expression using the identity for $(a+b)^3$.

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Solution:

We use the algebraic identity for the cube of a binomial sum:

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Alternatively, the identity can be written as:

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

In the expression $(x+y)^3$, we have $a = x$ and $b = y$.

Substitute $a = x$ and $b = y$ into the identity:

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

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

Using the alternate form:

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

$(x + y)^3 = x^3 + y^3 + 3xy(x + y)$

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

Both expanded forms are equivalent.

The expanded form of $(x+y)^3$ is $\mathbf{x^3 + 3x^2y + 3xy^2 + y^3}$.

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Given:

The expression to expand is $(a-b)^3$.

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To Expand:

Expand the given expression using the identity for $(a-b)^3$.

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Solution:

We use the algebraic identity for the cube of a binomial difference:

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

Alternatively, the identity can be written as:

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

In the expression $(a-b)^3$, the variables are already $a$ and $b$ as in the identity.

Using the first form of the identity:

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

Using the alternate form of the identity:

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

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

Both expanded forms are equivalent.

The expanded form of $(a-b)^3$ is $\mathbf{a^3 - 3a^2b + 3ab^2 - b^3}$.

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Given:

The expression to evaluate is $103^3$.

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To Evaluate:

Evaluate the given expression using an algebraic identity.

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Solution:

We can write $103$ as the sum of two numbers, for example, $100 + 3$.

So, $103^3 = (100 + 3)^3$.

This expression is in the form of $(a + b)^3$. We use the algebraic identity:

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Here, $a = 100$ and $b = 3$. Substituting these values into the identity:

$(100 + 3)^3 = (100)^3 + 3(100)^2(3) + 3(100)(3)^2 + (3)^3$

Calculate each term:

$(100)^3 = 100 \times 100 \times 100 = 1,000,000$

$3(100)^2(3) = 3(10000)(3) = 9 \times 10000 = 90,000$

$3(100)(3)^2 = 3(100)(9) = 300 \times 9 = 2,700$

$(3)^3 = 3 \times 3 \times 3 = 27$

Now, add the terms:

$103^3 = 1,000,000 + 90,000 + 2,700 + 27$

$103^3 = 1,090,000 + 2,727$

$103^3 = 1,092,727$

The value of $103^3$ is $\mathbf{1,092,727}$.

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Statement of the Remainder Theorem:

Let $p(x)$ be any polynomial of degree greater than or equal to one and let $a$ be any real number. If $p(x)$ is divided by the linear polynomial $(x - a)$, then the remainder is $p(a)$.

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Proof of the Remainder Theorem:

Let $p(x)$ be any polynomial with degree greater than or equal to 1. Suppose that when $p(x)$ is divided by the linear polynomial $(x - a)$, the quotient is $q(x)$ and the remainder is $r(x)$.

By the Polynomial Division Algorithm, we can write:

$p(x) = (x - a)q(x) + r(x)$

where $q(x)$ is the quotient polynomial and $r(x)$ is the remainder polynomial.

The degree of the remainder $r(x)$ must be less than the degree of the divisor $(x - a)$.

The degree of the divisor $(x - a)$ is 1.

Therefore, the degree of $r(x)$ must be less than 1. This means the degree of $r(x)$ is 0.

A polynomial of degree 0 is a constant. So, the remainder $r(x)$ is a constant, let's call it $R$.

Thus, the equation from the division algorithm becomes:

$p(x) = (x - a)q(x) + R$

Now, let's substitute $x = a$ into this equation:

$p(a) = (a - a)q(a) + R$

$p(a) = (0)q(a) + R$

$p(a) = 0 + R$

$p(a) = R$

Since $R$ is the remainder, this shows that the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$.

This proves the Remainder Theorem.

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Finding the Remainder using the Remainder Theorem:

Given:

The polynomial is $p(x) = x^3 + 3x^2 + 3x + 1$.

Case 1: Divided by $(x + 1)$

The divisor is $(x + 1)$. We can write this as $(x - (-1))$.

Comparing with $(x - a)$, we have $a = -1$.

According to the Remainder Theorem, the remainder is $p(a)$, which is $p(-1)$.

Substitute $x = -1$ into $p(x)$:

$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1$

$p(-1) = (-1) + 3(1) + (-3) + 1$

$p(-1) = -1 + 3 - 3 + 1$

$p(-1) = 2 - 3 + 1$

$p(-1) = -1 + 1$

$p(-1) = 0$

The remainder when $p(x) = x^3 + 3x^2 + 3x + 1$ is divided by $(x + 1)$ is $\mathbf{0}$.

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Case 2: Divided by $(x - 1/2)$

The divisor is $(x - 1/2)$.

Comparing with $(x - a)$, we have $a = 1/2$.

According to the Remainder Theorem, the remainder is $p(a)$, which is $p(1/2)$.

Substitute $x = 1/2$ into $p(x)$:

$p(1/2) = (1/2)^3 + 3(1/2)^2 + 3(1/2) + 1$

$p(1/2) = (1/8) + 3(1/4) + (3/2) + 1$

$p(1/2) = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$

To add these fractions, find a common denominator, which is 8.

$p(1/2) = \frac{1}{8} + \frac{3 \times 2}{4 \times 2} + \frac{3 \times 4}{2 \times 4} + \frac{1 \times 8}{1 \times 8}$

$p(1/2) = \frac{1}{8} + \frac{6}{8} + \frac{12}{8} + \frac{8}{8}$

$p(1/2) = \frac{1 + 6 + 12 + 8}{8}$

$p(1/2) = \frac{7 + 12 + 8}{8}$

$p(1/2) = \frac{19 + 8}{8}$

$p(1/2) = \frac{27}{8}$

The remainder when $p(x) = x^3 + 3x^2 + 3x + 1$ is divided by $(x - 1/2)$ is $\mathbf{\frac{27}{8}}$.

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Statement of the Factor Theorem:

Let $p(x)$ be a polynomial of degree greater than or equal to one, and let $a$ be a real number.

(i) If $(x - a)$ is a factor of $p(x)$, then $p(a) = 0$.

(ii) If $p(a) = 0$, then $(x - a)$ is a factor of $p(x)$.

These two statements can be combined into a single "if and only if" statement: $(x - a)$ is a factor of the polynomial $p(x)$ if and only if $p(a) = 0$.

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Proof of the Factor Theorem:

The proof of the Factor Theorem relies on the Remainder Theorem.

Recall the Remainder Theorem: When a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$. This can be expressed by the division algorithm:

$p(x) = (x - a)q(x) + r(x)$

where $q(x)$ is the quotient and $r(x)$ is the remainder. Since the degree of the divisor $(x-a)$ is 1, the degree of the remainder $r(x)$ must be less than 1, meaning $r(x)$ is a constant, say $R$. Thus, $p(x) = (x - a)q(x) + R$. By the Remainder Theorem, $R = p(a)$. So, we have:

$p(x) = (x - a)q(x) + p(a)$

Now, let's prove the two parts of the Factor Theorem:

Part (i): If $(x - a)$ is a factor of $p(x)$, then $p(a) = 0$.

If $(x - a)$ is a factor of $p(x)$, it means that when $p(x)$ is divided by $(x - a)$, the remainder is 0. In other words, $p(x)$ is exactly divisible by $(x - a)$.

From the division algorithm, we have $p(x) = (x - a)q(x) + p(a)$.

If the remainder is 0, then $p(a) = 0$.

Thus, if $(x - a)$ is a factor of $p(x)$, then $p(a) = 0$.

Part (ii): If $p(a) = 0$, then $(x - a)$ is a factor of $p(x)$.

Assume that $p(a) = 0$.

From the division algorithm, we know that $p(x) = (x - a)q(x) + p(a)$.

Substitute $p(a) = 0$ into this equation:

$p(x) = (x - a)q(x) + 0$

$p(x) = (x - a)q(x)$

This equation shows that $p(x)$ can be expressed as the product of $(x - a)$ and some polynomial $q(x)$. This means that $(x - a)$ divides $p(x)$ exactly, with a remainder of 0.

Therefore, $(x - a)$ is a factor of $p(x)$.

Both parts of the theorem are proven.

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Using the Factor Theorem to check for factors:

Given:

The polynomial is $p(x) = x^3 - 2x^2 - x + 2$.

We need to check if $(x - 2)$ and $(x + 1)$ are factors of $p(x)$.

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Solution:

According to the Factor Theorem, a linear polynomial $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$.

Checking if $(x - 2)$ is a factor:

The potential factor is $(x - 2)$. Comparing this with $(x - a)$, we have $a = 2$.

By the Factor Theorem, $(x - 2)$ is a factor of $p(x)$ if and only if $p(2) = 0$.

Substitute $x = 2$ into the polynomial $p(x) = x^3 - 2x^2 - x + 2$:

$p(2) = (2)^3 - 2(2)^2 - (2) + 2$

$p(2) = 8 - 2(4) - 2 + 2$

$p(2) = 8 - 8 - 2 + 2$

$p(2) = 0 - 2 + 2$

$p(2) = 0$

Since $p(2) = 0$, by the Factor Theorem, $(x - 2)$ is a factor of the polynomial $p(x) = x^3 - 2x^2 - x + 2$.

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Checking if $(x + 1)$ is a factor:

The potential factor is $(x + 1)$. We can write $(x + 1)$ as $(x - (-1))$.

Comparing this with $(x - a)$, we have $a = -1$.

By the Factor Theorem, $(x + 1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.

Substitute $x = -1$ into the polynomial $p(x) = x^3 - 2x^2 - x + 2$:

$p(-1) = (-1)^3 - 2(-1)^2 - (-1) + 2$

$p(-1) = (-1) - 2(1) - (-1) + 2$

$p(-1) = -1 - 2 + 1 + 2$

$p(-1) = -3 + 1 + 2$

$p(-1) = -2 + 2$

$p(-1) = 0$

Since $p(-1) = 0$, by the Factor Theorem, $(x + 1)$ is a factor of the polynomial $p(x) = x^3 - 2x^2 - x + 2$.

Conclusion: Both $\mathbf{(x - 2)}$ and $\mathbf{(x + 1)}$ are factors of the polynomial $\mathbf{p(x) = x^3 - 2x^2 - x + 2}$.

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Given:

The polynomial is $p(x) = x^3 + ax^2 + bx - 4$.

$(x - 1)$ and $(x + 2)$ are factors of $p(x)$.

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To Find:

The values of $a$ and $b$.

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Solution:

According to the Factor Theorem, if $(x - c)$ is a factor of a polynomial $p(x)$, then $p(c) = 0$.

Since $(x - 1)$ is a factor of $p(x)$, we have $p(1) = 0$.

Substitute $x = 1$ into the polynomial $p(x) = x^3 + ax^2 + bx - 4$:

$p(1) = (1)^3 + a(1)^2 + b(1) - 4$

$p(1) = 1 + a(1) + b - 4$

$p(1) = 1 + a + b - 4$

$p(1) = a + b - 3$

Setting $p(1) = 0$, we get the first equation:

From equation (1), we have $a + b = 3$.

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Since $(x + 2)$ is a factor of $p(x)$, we can write $(x + 2)$ as $(x - (-2))$, so we have $c = -2$. By the Factor Theorem, $p(-2) = 0$.

Substitute $x = -2$ into the polynomial $p(x) = x^3 + ax^2 + bx - 4$:

$p(-2) = (-2)^3 + a(-2)^2 + b(-2) - 4$

$p(-2) = (-8) + a(4) + (-2b) - 4$

$p(-2) = -8 + 4a - 2b - 4$

$p(-2) = 4a - 2b - 12$

Setting $p(-2) = 0$, we get the second equation:

From equation (2), we have $4a - 2b = 12$. We can divide the entire equation by 2 to simplify it:

$2a - b = 6$

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Now we have a system of two linear equations with two variables $a$ and $b$:

We can solve this system using the elimination method. Add equation (1) and equation (2):

$(a + b) + (2a - b) = 3 + 6$

$a + b + 2a - b = 9$

$3a = 9$

Divide by 3:

$a = \frac{9}{3}$

$a = 3$

Now substitute the value of $a=3$ into equation (1) to find the value of $b$:

$a + b = 3$

$3 + b = 3$

Subtract 3 from both sides:

$b = 3 - 3$

$b = 0$

So, the values are $\mathbf{a = 3}$ and $\mathbf{b = 0}$.

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Verification:

Substitute $a=3$ and $b=0$ back into the original polynomial:

$p(x) = x^3 + 3x^2 + 0x - 4 = x^3 + 3x^2 - 4$

Check if $p(1) = 0$:

$p(1) = (1)^3 + 3(1)^2 - 4 = 1 + 3(1) - 4 = 1 + 3 - 4 = 4 - 4 = 0$. (Correct)

Check if $p(-2) = 0$:

$p(-2) = (-2)^3 + 3(-2)^2 - 4 = -8 + 3(4) - 4 = -8 + 12 - 4 = 4 - 4 = 0$. (Correct)

The values $a=3$ and $b=0$ are correct.

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Given:

The quadratic polynomial to factorise is $6x^2 + 17x + 5$.

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To Factorise:

Factorise the given polynomial by splitting the middle term.

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Solution:

The given polynomial is in the standard quadratic form $ax^2 + bx + c$, where $a = 6$, $b = 17$, and $c = 5$.

Step 1: Find the product of the coefficient of $x^2$ and the constant term, i.e., $ac$.

$ac = 6 \times 5 = 30$.

Step 2: Find two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the middle term ($b=17$) and their product is equal to $ac$ (30).

We need $p + q = 17$ and $p \times q = 30$.

Let's consider the pairs of factors of 30 and their sums:

Factors of 30 (p, q)	Sum (p + q)
(1, 30)	31
(2, 15)	17
(3, 10)	13
(5, 6)	11
(-1, -30)	-31
(-2, -15)	-17
(-3, -10)	-13
(-5, -6)	-11

The pair of factors whose sum is 17 is 2 and 15.

So, we can take $p = 2$ and $q = 15$ (or vice versa).

Step 3: Split the middle term $17x$ using the numbers found in Step 2. We write $17x$ as $2x + 15x$.

Rewrite the polynomial:

$6x^2 + 17x + 5 = 6x^2 + 2x + 15x + 5$

Step 4: Group the first two terms and the last two terms.

$(6x^2 + 2x) + (15x + 5)$

Step 5: Factor out the greatest common factor (GCF) from each group.

From the first group $(6x^2 + 2x)$, the GCF is $2x$.

$6x^2 + 2x = 2x(3x + 1)$

From the second group $(15x + 5)$, the GCF is 5.

$15x + 5 = 5(3x + 1)$

Substitute these back into the grouped expression:

$2x(3x + 1) + 5(3x + 1)$

Step 6: Factor out the common binomial factor, which is $(3x + 1)$.

$(3x + 1)(2x + 5)$

The factorised form of the quadratic polynomial $6x^2 + 17x + 5$ is $\mathbf{(3x + 1)(2x + 5)}$.

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Given:

The quadratic polynomial to factorise is $y^2 - 5y - 24$.

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To Factorise:

Factorise the given polynomial by splitting the middle term.

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Solution:

The given polynomial is in the standard quadratic form $ay^2 + by + c$, where $a = 1$, $b = -5$, and $c = -24$.

Step 1: Find the product of the coefficient of $y^2$ and the constant term, i.e., $ac$.

$ac = 1 \times (-24) = -24$.

Step 2: Find two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the middle term ($b=-5$) and their product is equal to $ac$ (-24).

We need $p + q = -5$ and $p \times q = -24$.

Let's consider the pairs of factors of -24 and their sums:

Factors of -24 (p, q)	Sum (p + q)
(1, -24)	-23
(-1, 24)	23
(2, -12)	-10
(-2, 12)	10
(3, -8)	-5
(-3, 8)	5
(4, -6)	-2
(-4, 6)	2

The pair of factors whose sum is -5 is 3 and -8.

So, we can take $p = 3$ and $q = -8$ (or vice versa).

Step 3: Split the middle term $-5y$ using the numbers found in Step 2. We write $-5y$ as $3y - 8y$.

Rewrite the polynomial:

$y^2 - 5y - 24 = y^2 + 3y - 8y - 24$

Step 4: Group the first two terms and the last two terms.

$(y^2 + 3y) + (-8y - 24)$

Step 5: Factor out the greatest common factor (GCF) from each group.

From the first group $(y^2 + 3y)$, the GCF is $y$.

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

From the second group $(-8y - 24)$, the GCF is $-8$.

$-8y - 24 = -8(y + 3)$

Substitute these back into the grouped expression:

$y(y + 3) - 8(y + 3)$

Step 6: Factor out the common binomial factor, which is $(y + 3)$.

$(y + 3)(y - 8)$

The factorised form of the quadratic polynomial $y^2 - 5y - 24$ is $\mathbf{(y + 3)(y - 8)}$.

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Given:

The cubic polynomial to factorise is $p(x) = x^3 - 2x^2 - x + 2$.

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To Factorise:

Factorise the given cubic polynomial.

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Solution:

We will use the Factor Theorem to find one linear factor of the polynomial. The possible rational roots (and thus values of $a$ for which $(x-a)$ might be a factor) are divisors of the constant term (2), which are $\pm 1, \pm 2$.

Let's test $x=1$:

$p(1) = (1)^3 - 2(1)^2 - (1) + 2 = 1 - 2(1) - 1 + 2 = 1 - 2 - 1 + 2 = -1 + 1 = 0$.

Since $p(1) = 0$, by the Factor Theorem, $(x - 1)$ is a factor of $p(x)$.

Now that we have one factor $(x - 1)$, we can find the other factors by dividing $p(x)$ by $(x - 1)$. We can use polynomial long division or synthetic division, or in this case, observe a pattern for grouping.

Method 1: Grouping (Observational)

Rewrite the polynomial and group terms to extract the factor $(x-1)$.

$x^3 - 2x^2 - x + 2$

We can rewrite $-2x^2$ as $-x^2 - x^2$ and $-x$ as $x - 2x$ to facilitate grouping with $(x-1)$ or $(x-2)$ terms. A simpler grouping often works:

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

Factor out $x^2$ from the first group and $-1$ from the second group:

$x^2(x - 2) - 1(x - 2)$

Now, $(x - 2)$ is a common factor:

$(x - 2)(x^2 - 1)$

The factor $x^2 - 1$ is a difference of squares, which can be factorised using the identity $a^2 - b^2 = (a - b)(a + b)$ where $a=x$ and $b=1$.

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

Substitute this back into the expression:

$(x - 2)(x - 1)(x + 1)$

So, the factors are $(x - 1)$, $(x - 2)$, and $(x + 1)$.

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Method 2: Using Polynomial Long Division

Divide $x^3 - 2x^2 - x + 2$ by $(x - 1)$.

$\begin{array}{r} x^2-x-2\phantom{)} \\ x-1{\overline{\smash{\big)}\,x^3-2x^2-x+2\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-x^2)\phantom{-x+2)}} \\ 0-x^2-x+2\phantom{)} \\ \underline{-~\phantom{()}(-x^2+x)\phantom{+2)}} \\ 0-2x+2\phantom{)} \\ \underline{-~\phantom{()}(-2x+2)} \\ 0+0\phantom{)} \end{array}$

The quotient is $x^2 - x - 2$. The remainder is 0, as expected since $(x-1)$ is a factor.

Now, we need to factorise the quadratic quotient $x^2 - x - 2$. We can use the splitting the middle term method.

We need two numbers whose sum is -1 and product is $(1)(-2) = -2$. The numbers are 1 and -2.

$x^2 - x - 2 = x^2 + x - 2x - 2$

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

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

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

So the factors of the cubic polynomial are $(x - 1)$ and the factors of the quotient $(x + 1)(x - 2)$.

The factors are $(x - 1), (x + 1), (x - 2)$.

The factorised form of the cubic polynomial is $\mathbf{(x - 1)(x + 1)(x - 2)}$.

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Given:

The cubic polynomial to factorise is $p(x) = x^3 - 23x^2 + 142x - 120$.

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To Factorise:

Factorise the given cubic polynomial.

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Solution:

Step 1: Find one factor using the Factor Theorem.

According to the Factor Theorem, if $(x - a)$ is a factor of $p(x)$, then $p(a) = 0$. We look for potential rational roots $a$ that are divisors of the constant term (-120).

Let's test some simple integer divisors of 120 ($\pm 1, \pm 2, \pm 3, \pm 4, ...$).

Test $x = 1$:

$p(1) = (1)^3 - 23(1)^2 + 142(1) - 120$

$p(1) = 1 - 23(1) + 142 - 120$

$p(1) = 1 - 23 + 142 - 120$

$p(1) = -22 + 142 - 120$

$p(1) = 120 - 120$

$p(1) = 0$

Since $p(1) = 0$, by the Factor Theorem, $(x - 1)$ is a factor of the polynomial $p(x) = x^3 - 23x^2 + 142x - 120$.

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Step 2: Divide the polynomial $p(x)$ by the factor $(x - 1)$ to find the other factor(s).

We can use polynomial long division:

$\begin{array}{r} x^2-22x+120\phantom{)} \\ x-1{\overline{\smash{\big)}\,x^3-23x^2+142x-120\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-\phantom{0}x^2)\phantom{+142x-120)}} \\ \phantom{()} -22x^2+142x-120\phantom{)} \\ \underline{-~\phantom{()}(-22x^2+22x)\phantom{-120)}} \\ \phantom{()} 120x-120\phantom{)} \\ \underline{-~\phantom{()}(120x-120)} \\ \phantom{()} 0\phantom{)} \end{array}$

The quotient is $x^2 - 22x + 120$. So, $p(x) = (x - 1)(x^2 - 22x + 120)$.

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Step 3: Factorise the quadratic quotient $x^2 - 22x + 120$ by splitting the middle term.

We need to find two numbers whose sum is -22 (the coefficient of the middle term) and whose product is $1 \times 120 = 120$ (the product of the coefficient of $x^2$ and the constant term).

Let the two numbers be $p$ and $q$. We need $p + q = -22$ and $p \times q = 120$.

Since the product is positive and the sum is negative, both numbers must be negative. We look for negative factors of 120 whose sum is -22. The numbers -10 and -12 satisfy these conditions because $(-10) + (-12) = -22$ and $(-10) \times (-12) = 120$.

Split the middle term $-22x$ into $-10x - 12x$:

$x^2 - 22x + 120 = x^2 - 10x - 12x + 120$

Group the terms and factor:

$= (x^2 - 10x) + (-12x + 120)$

$= x(x - 10) - 12(x - 10)$

$= (x - 10)(x - 12)$

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Step 4: Write the complete factorisation of the cubic polynomial.

The factorised form of $p(x) = x^3 - 23x^2 + 142x - 120$ is the product of the linear factor $(x - 1)$ and the factorised quadratic quotient $(x - 10)(x - 12)$.

$p(x) = (x - 1)(x - 10)(x - 12)$

The factorised form of the polynomial is $\mathbf{(x - 1)(x - 10)(x - 12)}$.

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Given:

The expressions to expand are:

(a) $(2x - y + z)^2$

(b) $(3a + 4b)^3$

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To Expand:

Expand the given expressions using appropriate algebraic identities.

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Solution:

(a) $(2x - y + z)^2$

This expression is in the form of $(p + q + r)^2$. We use the algebraic identity:

$(p + q + r)^2 = p^2 + q^2 + r^2 + 2pq + 2qr + 2rp$

Here, we substitute $p = 2x$, $q = -y$, and $r = z$.

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

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

$= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$

Thus, the expansion of $(2x - y + z)^2$ is $\mathbf{4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx}$.

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(b) $(3a + 4b)^3$

This expression is in the form of $(p + q)^3$. We use the algebraic identity:

$(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$

Here, we substitute $p = 3a$ and $q = 4b$.

$(3a + 4b)^3 = (3a)^3 + 3(3a)^2(4b) + 3(3a)(4b)^2 + (4b)^3$

$= 27a^3 + 3(9a^2)(4b) + 3(3a)(16b^2) + 64b^3$

$= 27a^3 + (3 \times 9 \times 4)a^2b + (3 \times 3 \times 16)ab^2 + 64b^3$

$= 27a^3 + 108a^2b + 144ab^2 + 64b^3$

Thus, the expansion of $(3a + 4b)^3$ is $\mathbf{27a^3 + 108a^2b + 144ab^2 + 64b^3}$.

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Given:

The expressions to factorise are:

(a) $4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

(b) $8a^3 + b^3 + 12a^2b + 6ab^2$

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To Factorise:

Factorise the given expressions using algebraic identities.

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Solution:

(a) $4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

This expression has terms involving the squares of three variables and cross-product terms. It resembles the expansion of $(p + q + r)^2 = p^2 + q^2 + r^2 + 2pq + 2qr + 2rp$.

Let's identify the terms that are perfect squares:

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

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

$16z^2 = (4z)^2$

So, the squares are of $2x$, $3y$, and $4z$. Now let's look at the cross-product terms:

$12xy = 2 \times (2x) \times (3y)$

$-24yz = 2 \times (3y) \times (-4z)$

$-16xz = 2 \times (2x) \times (-4z)$

Notice the negative signs in the $yz$ and $xz$ terms. The terms involving $y$ and $z$, and $x$ and $z$ are negative, while the term involving $x$ and $y$ is positive. This suggests that the $z$ term in the base expression might be negative, or one of $x, y$ is negative and $z$ is positive (but $12xy$ is positive). Let's assume the terms are $2x$, $3y$, and $-4z$.

Let $p = 2x$, $q = 3y$, $r = -4z$.

Check the expansion of $(2x + 3y - 4z)^2$:

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

$= 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

This matches the given expression.

The factorised form is $\mathbf{(2x + 3y - 4z)^2}$.

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(b) $8a^3 + b^3 + 12a^2b + 6ab^2$

This expression has cubic terms and terms with powers of $a$ and $b$. It resembles the expansion of $(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$.

Let's identify the terms that are perfect cubes:

$8a^3 = (2a)^3$

$b^3 = (b)^3$

So, the cubes are of $2a$ and $b$. Let's check the other terms to see if they match $3p^2q$ and $3pq^2$ with $p = 2a$ and $q = b$.

$3p^2q = 3(2a)^2(b) = 3(4a^2)(b) = 12a^2b$

$3pq^2 = 3(2a)(b)^2 = 6ab^2$

The given expression $8a^3 + b^3 + 12a^2b + 6ab^2$ matches the form $p^3 + q^3 + 3p^2q + 3pq^2$ (or $p^3 + 3p^2q + 3pq^2 + q^3$) with $p = 2a$ and $q = b$.

Using the identity:

$p^3 + 3p^2q + 3pq^2 + q^3 = (p + q)^3$

Substitute $p = 2a$ and $q = b$:

$8a^3 + 12a^2b + 6ab^2 + b^3 = (2a + b)^3$

The factorised form is $\mathbf{(2a + b)^3}$.

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Given:

The expression to evaluate is $(-12)^3 + (7)^3 + (5)^3$.

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To Evaluate:

Evaluate the given expression without directly calculating the cubes, using an algebraic identity.

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Solution:

We are asked to evaluate the sum of three cubes: $(-12)^3 + (7)^3 + (5)^3$.

Let $a = -12$, $b = 7$, and $c = 5$. The expression is in the form $a^3 + b^3 + c^3$.

We use the algebraic identity related to the sum of cubes:

$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Consider the condition when $a + b + c = 0$. If $a + b + c = 0$, then the right side of the identity becomes:

$0 \times (a^2 + b^2 + c^2 - ab - bc - ca) = 0$

So, if $a + b + c = 0$, the identity simplifies to:

$a^3 + b^3 + c^3 - 3abc = 0$

which means:

$a^3 + b^3 + c^3 = 3abc$

Let's check if the sum of our given values $a = -12$, $b = 7$, and $c = 5$ is zero:

$a + b + c = -12 + 7 + 5$

$a + b + c = -12 + 12$

$a + b + c = 0$

Since $a + b + c = 0$, we can use the simplified identity $a^3 + b^3 + c^3 = 3abc$ to evaluate the expression.

$(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12) \times (7) \times (5)$

$= 3 \times (-12) \times (35)$

$= (-36) \times (35)$

Now calculate the product:

$\begin{array}{cc}& & 3 & 5 \\ \times & & 3 & 6 \\ \hline && 2 & 1 & 0 \\ & 1 & 0 & 5 & \times \\ \hline 1 & 2 & 6 & 0 \\ \hline \end{array}$

So, $36 \times 35 = 1260$. Since we are multiplying -36 by 35, the result is negative.

$= -1260$

Therefore, $(-12)^3 + (7)^3 + (5)^3 = \mathbf{-1260}$.

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Given:

The identity to verify is $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$.

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To Verify:

Verify the given algebraic identity.

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Verification:

To verify the identity, we can expand the right-hand side (RHS) of the equation and show that it is equal to the left-hand side (LHS).

RHS = $(x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$

Multiply each term in the first parenthesis $(x+y+z)$ by each term in the second parenthesis $(x^2 + y^2 + z^2 - xy - yz - zx)$.

RHS $= x(x^2 + y^2 + z^2 - xy - yz - zx) + y(x^2 + y^2 + z^2 - xy - yz - zx) + z(x^2 + y^2 + z^2 - xy - yz - zx)$

Distribute $x$ to the terms in the second parenthesis:

$x(x^2 + y^2 + z^2 - xy - yz - zx) = x^3 + xy^2 + xz^2 - x^2y - xyz - x^2z$

Distribute $y$ to the terms in the second parenthesis:

$y(x^2 + y^2 + z^2 - xy - yz - zx) = yx^2 + y^3 + yz^2 - xy^2 - y^2z - xyz$

Distribute $z$ to the terms in the second parenthesis:

$z(x^2 + y^2 + z^2 - xy - yz - zx) = zx^2 + zy^2 + z^3 - xyz - yz^2 - z^2x$

Now, add all these expanded terms together:

RHS $= (x^3 + xy^2 + xz^2 - x^2y - xyz - x^2z) + (x^2y + y^3 + yz^2 - xy^2 - y^2z - xyz) + (zx^2 + zy^2 + z^3 - xyz - yz^2 - z^2x)$

Combine like terms and look for terms that cancel each other out:

$x^3$ (appears once)

$y^3$ (appears once)

$z^3$ (appears once)

$xy^2$ and $-xy^2$ (cancel each other)

$xz^2$ and $-z^2x$ (or $-xz^2$) (cancel each other)

$-x^2y$ and $yx^2$ (or $x^2y$) (cancel each other)

$yz^2$ and $-yz^2$ (cancel each other)

$-y^2z$ and $zy^2$ (or $y^2z$) (cancel each other)

$-xyz$ (appears three times)

After cancelling the terms, we are left with:

RHS $= x^3 + y^3 + z^3 - xyz - xyz - xyz$

RHS $= x^3 + y^3 + z^3 - 3xyz$

This is equal to the left-hand side (LHS) of the identity.

Since LHS = RHS, the identity is verified.

$\mathbf{x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)}$

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Given:

The polynomial is $p(x) = x^4 - 5x^3 + 5x^2 - 10x + 2$.

The divisor is $(x - 2)$.

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To Find:

The remainder when $p(x)$ is divided by $(x - 2)$ using long division and verify using the Remainder Theorem.

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Solution (using Polynomial Long Division):

We divide $x^4 - 5x^3 + 5x^2 - 10x + 2$ by $(x - 2)$.

$\begin{array}{r} x^3 - 3x^2 - x - 12 \\ x-2{\overline{\smash{\big)}\,x^4 - 5x^3 + 5x^2 - 10x + 2}} \\ \underline{-~\phantom{(}(x^4 - 2x^3)\phantom{+ 5x^2 - 10x + 2)}} \\ \phantom{()} -3x^3 + 5x^2 - 10x + 2 \\ \underline{-~\phantom{()}(-3x^3 + 6x^2)\phantom{- 10x + 2)}} \\ \phantom{()} -x^2 - 10x + 2 \\ \underline{-~\phantom{()}(-x^2 + 2x)\phantom{+ 2)}} \\ \phantom{()} -12x + 2 \\ \underline{-~\phantom{()}(-12x + 24)} \\ \phantom{()} -22 \end{array}$

The quotient is $q(x) = x^3 - 3x^2 - x - 12$ and the remainder is $R = -22$.

So, the remainder from the long division is $\mathbf{-22}$.

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Solution (using Remainder Theorem):

According to the Remainder Theorem, when a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, the remainder is $p(a)$.

In this case, the divisor is $(x - 2)$. Comparing this with $(x - a)$, we find that $a = 2$.

Therefore, the remainder when $p(x) = x^4 - 5x^3 + 5x^2 - 10x + 2$ is divided by $(x - 2)$ is $p(2)$.

Substitute $x = 2$ into the polynomial $p(x)$:

$p(2) = (2)^4 - 5(2)^3 + 5(2)^2 - 10(2) + 2$

$p(2) = 16 - 5(8) + 5(4) - 20 + 2$

$p(2) = 16 - 40 + 20 - 20 + 2$

$p(2) = -24 + 20 - 20 + 2$

$p(2) = -4 - 20 + 2$

$p(2) = -24 + 2$

$p(2) = -22$

So, the remainder found using the Remainder Theorem is $\mathbf{-22}$.

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Verification:

The remainder found using polynomial long division is -22.

The remainder found using the Remainder Theorem is $p(2) = -22$.

Since both methods yield the same remainder, the answer is verified.

The remainder when $p(x) = x^4 - 5x^3 + 5x^2 - 10x + 2$ is divided by $(x - 2)$ is $\mathbf{-22}$.

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Given:

The polynomial to factorise is $2y^3 + y^2 - 2y - 1$.

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To Factorise:

Factorise the given polynomial.

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Solution (using Grouping):

We can factorise the polynomial by grouping the terms.

Group the first two terms and the last two terms:

$(2y^3 + y^2) + (-2y - 1)$

Factor out the common factor from the first group $(2y^3 + y^2)$. The common factor is $y^2$.

$y^2(2y + 1)$

Factor out the common factor from the second group $(-2y - 1)$. The common factor is $-1$.

$-1(2y + 1)$

Now the expression is written as:

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

We can see that $(2y + 1)$ is a common binomial factor in both terms.

Factor out the common factor $(2y + 1)$:

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

The second factor $y^2 - 1$ is in the form of a difference of squares, $a^2 - b^2$. We use the identity $a^2 - b^2 = (a - b)(a + b)$ with $a = y$ and $b = 1$.

So, $y^2 - 1 = (y - 1)(y + 1)$.

Substitute this factorisation back into the expression:

$(2y + 1)(y - 1)(y + 1)$

The factorised form of the polynomial is $\mathbf{(2y + 1)(y - 1)(y + 1)}$.

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Alternate Solution (using Factor Theorem):

Let $p(y) = 2y^3 + y^2 - 2y - 1$. By the Factor Theorem, if $p(a) = 0$, then $(y-a)$ is a factor.

Let's test some possible rational roots (divisors of the constant term -1 divided by divisors of the leading coefficient 2). Possible values are $\pm 1, \pm 1/2$.

Test $y=1$:

$p(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$.

Since $p(1) = 0$, $(y - 1)$ is a factor.

Test $y=-1$:

$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0$.

Since $p(-1) = 0$, $(y - (-1)) = (y + 1)$ is a factor.

Test $y=-1/2$:

$p(-1/2) = 2(-1/2)^3 + (-1/2)^2 - 2(-1/2) - 1 = 2(-1/8) + (1/4) + 1 - 1 = -1/4 + 1/4 + 0 = 0$.

Since $p(-1/2) = 0$, $(y - (-1/2)) = (y + 1/2)$ is a factor. We can write $(y + 1/2) = \frac{1}{2}(2y + 1)$. Thus, $(2y + 1)$ is also a factor.

The factors found are $(y - 1)$, $(y + 1)$, and $(2y + 1)$. Since the polynomial is cubic, these are all the linear factors.

The factorised form is the product of these factors:

$(y - 1)(y + 1)(2y + 1)$

The factorised form of the polynomial is $\mathbf{(y - 1)(y + 1)(2y + 1)}$ or $\mathbf{(2y + 1)(y^2 - 1)}$.

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