Chapter 2 Polynomials

We are given the equation $x^2 + kx + 6 = (x + 2)(x + 3)$, which holds true for all values of $x$.

We need to find the value of $k$.

Let's expand the right-hand side of the equation.

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

Using the distributive property (FOIL):

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

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

Combine the like terms ($3x$ and $2x$):

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

$= x^2 + 5x + 6$

Now, substitute this back into the original equation:

Since this equation is true for all values of $x$, the coefficients of corresponding powers of $x$ on both sides of the equation must be equal.

Compare the coefficients of $x^2$:

This is true.

Compare the coefficients of $x$:

Compare the constant terms:

This is true.

From the comparison of the coefficients of $x$, we find that the value of $k$ is 5.

Looking at the given options:

(A) 1

(B) –1

(C) 5

(D) 3

The value of $k$ matches option (C).

The final answer is $\boxed{5}$.

A polynomial in one variable $x$ is an algebraic expression where the powers of $x$ are non-negative integers (0, 1, 2, 3, ...).

Let's examine each option:

(A) $\frac{x^2}{2} - \frac{2}{x^2}$

We can rewrite the second term as $\frac{2}{x^2} = 2x^{-2}$.

So, the expression is $\frac{1}{2}x^2 - 2x^{-2}$.

The power of $x$ in the second term is -2, which is a negative integer. Therefore, this is not a polynomial.

(B) $\sqrt{2x}$ – 1

We can rewrite the first term using exponent notation: $\sqrt{2x} = (2x)^{\frac{1}{2}} = \sqrt{2} \times x^{\frac{1}{2}}$.

So, the expression is $\sqrt{2}x^{\frac{1}{2}} - 1$.

The power of $x$ is $\frac{1}{2}$, which is a fraction and not an integer. Therefore, this is not a polynomial.

(C) x² + $\frac{3x^{\frac{3}{2}}}{\sqrt{x}}$

Let's simplify the second term: $\frac{3x^{\frac{3}{2}}}{\sqrt{x}} = \frac{3x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$.

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

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

So, the expression simplifies to $x^2 + 3x$.

In this expression, the powers of $x$ are 2 and 1, which are non-negative integers. Therefore, this is a polynomial.

(D) $\frac{x \;-\; 1}{x \;+\; 1}$

This is a rational expression, which is a ratio of two polynomials. While the numerator ($x-1$) and the denominator ($x+1$) are polynomials, the entire expression is not a polynomial because the variable appears in the denominator in a way that cannot be simplified to a non-negative integer power. For example, using polynomial long division would result in a remainder term with $x$ in the denominator (e.g., $1 - \frac{2}{x+1}$). Therefore, this is not a polynomial.

Based on the analysis, only option (C) simplifies to a polynomial.

The correct answer is $\boxed{C}$.

A polynomial is an expression of the form $a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x^1 + a_0 x^0$, where $a_i$ are coefficients and the exponents ($n, n-1, ..., 1, 0$) are non-negative integers.

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

The given expression is $\sqrt{2}$. This is a constant number (approximately 1.414). A constant, non-zero number is a special type of polynomial called a constant polynomial.

A constant polynomial can be written in the form $c \times x^0$, where $c$ is the constant and $x^0 = 1$ (for $x \neq 0$).

So, $\sqrt{2}$ can be written as $\sqrt{2}x^0$.

The variable $x$ has an exponent of 0, which is a non-negative integer.

The highest power of the variable $x$ in the expression $\sqrt{2} = \sqrt{2}x^0$ is 0.

Therefore, the degree of the polynomial $\sqrt{2}$ is 0.

Looking at the given options:

(A) 2

(B) 0

(C) 1

(D) $\frac{1}{2}$

The degree matches option (B).

The correct answer is $\boxed{B}$.

The degree of a polynomial is the highest exponent of the variable that has a non-zero coefficient.

The given polynomial is $4x^4 + 0x^3 + 0x^5 + 5x + 7$.

Let's rewrite the polynomial in descending order of the powers of $x$:

$0x^5 + 4x^4 + 0x^3 + 5x^1 + 7x^0$

Identify the terms with non-zero coefficients and their corresponding exponents:

• The term $0x^5$ has a coefficient of 0, so it does not contribute to the degree.
• The term $4x^4$ has a coefficient of 4 (which is non-zero) and an exponent of 4.
• The term $0x^3$ has a coefficient of 0, so it does not contribute to the degree.
• The term $5x$ (or $5x^1$) has a coefficient of 5 (which is non-zero) and an exponent of 1.
• The term $7$ (or $7x^0$) has a coefficient of 7 (which is non-zero) and an exponent of 0.

The powers of $x$ with non-zero coefficients are 4, 1, and 0.

The highest among these exponents is 4.

Therefore, the degree of the polynomial $4x^4 + 0x^3 + 0x^5 + 5x + 7$ is 4.

Looking at the given options:

(A) 4

(B) 5

(C) 3

(D) 7

The degree matches option (A).

The correct answer is $\boxed{A}$.

The zero polynomial is the polynomial where all coefficients are zero. It is represented by the constant 0.

Let's consider the definition of the degree of a polynomial: it is the highest exponent of the variable with a non-zero coefficient.

For the zero polynomial, represented as 0, we can write it in various forms by including any power of $x$ multiplied by a coefficient of 0:

$0 = 0 \times x^0$

$0 = 0 \times x^1$

$0 = 0 \times x^{100}$

$0 = 0 \times x^n$ (for any non-negative integer n)

In the zero polynomial, every coefficient is 0. There is no term with a non-zero coefficient.

Therefore, according to the definition, there is no highest exponent with a non-zero coefficient.

For this reason, the degree of the zero polynomial is generally considered to be undefined.

Looking at the given options:

(A) 0

(B) 1

(C) Any natural number

(D) Not defined

The degree matches option (D).

(Note: Some texts define the degree of the zero polynomial as $-\infty$ or -1 for consistency in certain algebraic properties, but "not defined" is the most common and widely accepted answer in introductory contexts).

The correct answer is $\boxed{D}$.

We are given the polynomial $p(x) = x^2 - 2\sqrt{2}x + 1$.

We need to find the value of $p(x)$ when $x = 2\sqrt{2}$. This means we substitute $2\sqrt{2}$ for $x$ in the polynomial expression.

$p(2\sqrt{2}) = (2\sqrt{2})^2 - 2\sqrt{2}(2\sqrt{2}) + 1$

Evaluate each term:

$(2\sqrt{2})^2 = (2)^2 \times (\sqrt{2})^2 = 4 \times 2 = 8$

$2\sqrt{2}(2\sqrt{2}) = (2 \times 2) \times (\sqrt{2} \times \sqrt{2}) = 4 \times 2 = 8$

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

$p(2\sqrt{2}) = 8 - 8 + 1$

$p(2\sqrt{2}) = 0 + 1$

$p(2\sqrt{2}) = 1$

Looking at the given options:

(A) 0

(B) 1

(C) 4$\sqrt{2}$

(D) 8$\sqrt{2}$ + 1

The value of $p(2\sqrt{2})$ matches option (B).

The correct answer is $\boxed{B}$.

Let the polynomial be $p(x) = 5x - 4x^2 + 3$.

We need to find the value of $p(x)$ when $x = -1$. Substitute $x = -1$ into the polynomial expression.

$p(-1) = 5(-1) - 4(-1)^2 + 3$

Evaluate each term:

$5(-1) = -5$

$(-1)^2 = (-1) \times (-1) = 1$

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

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

$p(-1) = -5 - 4 + 3$

$p(-1) = -9 + 3$

$p(-1) = -6$

Looking at the given options:

(A) – 6

(B) 6

(C) 2

(D) –2

The value of $p(-1)$ matches option (A).

The correct answer is $\boxed{A}$.

We are given the polynomial $p(x) = x + 3$.

We need to find the value of $p(x) + p(-x)$.

We already have $p(x) = x + 3$.

Now, let's find $p(-x)$ by substituting $-x$ for $x$ in the polynomial expression:

$p(-x) = (-x) + 3$

$p(-x) = -x + 3$

Now, add $p(x)$ and $p(-x)$:

$p(x) + p(-x) = (x + 3) + (-x + 3)$

$p(x) + p(-x) = x + 3 - x + 3$

Combine like terms:

$p(x) + p(-x) = (x - x) + (3 + 3)$

$p(x) + p(-x) = 0 + 6$

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

Looking at the given options:

(A) 3

(B) 2x

(C) 0

(D) 6

The value of $p(x) + p(-x)$ matches option (D).

The correct answer is $\boxed{D}$.

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

The zero polynomial is the polynomial $p(x) = 0$ for all values of $x$.

For the zero polynomial $p(x) = 0$, the equation $p(x) = 0$ is true for every real number $x$.

This means that any real number satisfies the condition $p(x) = 0$ for the zero polynomial.

Therefore, every real number is a zero of the zero polynomial.

Looking at the given options:

(A) 0

(B) 1

(C) Any real number

(D) Not defined

The description of the zero matches option (C).

The correct answer is $\boxed{C}$.

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

We are given the polynomial $p(x) = 2x + 5$.

To find the zero of this polynomial, we set $p(x)$ equal to 0 and solve for $x$.

Subtract 5 from both sides of the equation:

Divide both sides by 2:

So, the zero of the polynomial $p(x) = 2x + 5$ is $-\frac{5}{2}$.

Looking at the given options:

(A) $-\frac{2}{5}$

(B) $-\frac{5}{2}$

(C) $\frac{2}{5}$

(D) $\frac{5}{2}$

The zero matches option (B).

The correct answer is $\boxed{B}$.

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

The given polynomial is $p(x) = 2x^2 + 7x - 4$. We need to find a value of $x$ among the options that makes $p(x) = 0$.

Let's test each option:

(A) $x = 2$

$p(2) = 2(2)^2 + 7(2) - 4$

$p(2) = 2(4) + 14 - 4$

$p(2) = 8 + 14 - 4$

$p(2) = 22 - 4 = 18$

$p(2) \neq 0$, so 2 is not a zero.

(B) $x = \frac{1}{2}$

$p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) - 4$

$p\left(\frac{1}{2}\right) = 2\left(\frac{1}{4}\right) + \frac{7}{2} - 4$

$p\left(\frac{1}{2}\right) = \frac{2}{4} + \frac{7}{2} - 4$

$p\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{7}{2} - 4$

$p\left(\frac{1}{2}\right) = \frac{1+7}{2} - 4$

$p\left(\frac{1}{2}\right) = \frac{8}{2} - 4$

$p\left(\frac{1}{2}\right) = 4 - 4 = 0$

$p\left(\frac{1}{2}\right) = 0$, so $\frac{1}{2}$ is a zero.

Since we found a zero, we can stop here. However, for completeness, let's check the other options.

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

$p\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^2 + 7\left(-\frac{1}{2}\right) - 4$

$p\left(-\frac{1}{2}\right) = 2\left(\frac{1}{4}\right) - \frac{7}{2} - 4$

$p\left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{7}{2} - 4$

$p\left(-\frac{1}{2}\right) = \frac{1-7}{2} - 4$

$p\left(-\frac{1}{2}\right) = \frac{-6}{2} - 4$

$p\left(-\frac{1}{2}\right) = -3 - 4 = -7$

$p\left(-\frac{1}{2}\right) \neq 0$, so $-\frac{1}{2}$ is not a zero.

(D) $x = –2$

$p(-2) = 2(-2)^2 + 7(-2) - 4$

$p(-2) = 2(4) - 14 - 4$

$p(-2) = 8 - 14 - 4$

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

$p(-2) \neq 0$, so –2 is not a zero.

The only option that results in $p(x)=0$ is $x = \frac{1}{2}$.

The correct answer is $\boxed{B}$.

We can use the Remainder Theorem to find the remainder when a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$.

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

In this question, the polynomial is $p(x) = x^{51} + 51$.

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

So, the value of $a$ is -1.

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

Substitute $x = -1$ into the polynomial $p(x) = x^{51} + 51$:

Remainder $= p(-1) = (-1)^{51} + 51$

Evaluate $(-1)^{51}$. Since the exponent 51 is an odd number, $(-1)^{51} = -1$.

Remainder $= -1 + 51$

Remainder $= 50$

So, the remainder is 50.

Looking at the given options:

(A) 0

(B) 1

(C) 49

(D) 50

The remainder matches option (D).

The correct answer is $\boxed{D}$.

We can use the Factor Theorem. The Factor Theorem states that $(x - a)$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$.

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

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

So, the value of $a$ is -1.

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

Substitute $x = -1$ into the polynomial $p(x) = 2x^2 + kx$ and set the result equal to 0.

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

$0 = 2(1) - k$

$0 = 2 - k$

Now, solve for $k$. Add $k$ to both sides of the equation:

So, the value of $k$ is 2.

Looking at the given options:

(A) –3

(B) 4

(C) 2

(D) –2

The value of $k$ matches option (C).

The correct answer is $\boxed{C}$.

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

Here, the potential factor is $x + 1$, which can be written as $x - (-1)$. So, we need to check which polynomial $p(x)$ satisfies $p(-1) = 0$.

Let's test each option:

(A) $p(x) = x^3 + x^2 – x + 1$

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

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

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

$p(-1) = 2$

$p(-1) \neq 0$, so $x+1$ is not a factor of this polynomial.

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

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

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

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

$p(-1) = 0$

$p(-1) = 0$, so $x+1$ is a factor of this polynomial.

Since we found the polynomial for which $x+1$ is a factor, we can conclude that option (B) is the correct answer. However, for completeness, let's check the other options.

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

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

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

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

$p(-1) = 2$

$p(-1) \neq 0$, so $x+1$ is not a factor of this polynomial.

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

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

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

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

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

$p(-1) = 5 - 4 = 1$

$p(-1) \neq 0$, so $x+1$ is not a factor of this polynomial.

The only polynomial for which $p(-1) = 0$ is the one in option (B).

The correct answer is $\boxed{B}$.

We are given the expression $(25x^2 – 1) + (1 + 5x)^2$. We need to find one of its factors.

Let's simplify the expression by expanding the terms.

The first term, $25x^2 - 1$, is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$. Here, $a = \sqrt{25x^2} = 5x$ and $b = \sqrt{1} = 1$.

So, $25x^2 - 1 = (5x - 1)(5x + 1)$.

The second term is $(1 + 5x)^2$. We can write this as $(5x + 1)^2$.

So the expression is $(5x - 1)(5x + 1) + (5x + 1)^2$.

We can see that $(5x + 1)$ is a common factor in both terms.

Factor out the common factor $(5x + 1)$:

$(5x + 1)[(5x - 1) + (5x + 1)]$

Simplify the expression inside the square brackets:

$(5x - 1) + (5x + 1) = 5x - 1 + 5x + 1$

$= (5x + 5x) + (-1 + 1)$

$= 10x + 0$

$= 10x$

So, the factored expression is:

$(5x + 1)(10x)$

The factors of the expression are $(5x + 1)$ and $10x$. Also, any constant multiple of these factors is also a factor. Specifically, $10$ and $x$ are also factors.

We are looking for one of the factors among the given options.

(A) 5 + x (equivalent to x + 5)

(B) 5 – x

(C) 5x – 1

(D) 10x

Compare the factors we found ($(5x + 1)$ and $10x$) with the given options.

Option (A) is $x + 5$, which is not $5x+1$ or $10x$.

Option (B) is $5 - x$, which is not a factor.

Option (C) is $5x - 1$, which is not a factor.

Option (D) is $10x$, which is one of the factors we found.

The correct answer is $\boxed{D}$.

We need to find the value of $249^2 - 248^2$.

This expression is in the form of a difference of squares, $a^2 - b^2$, where $a = 249$ and $b = 248$.

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

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

$249^2 - 248^2 = (249 - 248)(249 + 248)$

Calculate the values inside the parentheses:

$249 - 248 = 1$

$249 + 248$

$\begin{array}{cc} & 2 & 4 & 9 \\ + & 2 & 4 & 8 \\ \hline & 4 & 9 & 7 \\ \hline \end{array}$

$249 + 248 = 497$

Now, substitute these values back into the expression:

$(1)(497)$

$= 497$

So, the value of $249^2 - 248^2$ is 497.

Looking at the given options:

(A) 1² = 1

(B) 477

(C) 487

(D) 497

The value matches option (D).

The correct answer is $\boxed{D}$.

We need to factor the quadratic polynomial $4x^2 + 8x + 3$.

We can use the splitting the middle term method or check the given options by expanding them.

Method 1: Splitting the middle term

We need to find two numbers whose product is $(4 \times 3) = 12$ and whose sum is 8 (the coefficient of the middle term).

The pairs of factors of 12 are (1, 12), (2, 6), (3, 4).

We check the sum of these pairs:

$1 + 12 = 13$

$2 + 6 = 8$

$3 + 4 = 7$

The pair (2, 6) gives a sum of 8. So, we split the middle term $8x$ as $2x + 6x$ (or $6x + 2x$).

$4x^2 + 8x + 3 = 4x^2 + 2x + 6x + 3$

Now, group the terms and factor by grouping:

$= (4x^2 + 2x) + (6x + 3)$

Factor out the common monomial factor from each group:

The common factor in $(4x^2 + 2x)$ is $2x$. $4x^2 + 2x = 2x(2x + 1)$.

The common factor in $(6x + 3)$ is 3. $6x + 3 = 3(2x + 1)$.

So, the expression becomes:

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

Now, factor out the common binomial factor $(2x + 1)$:

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

Method 2: Checking the options by expansion

Expand each option and see which one matches the given polynomial.

(A) (x + 1) (x + 3)

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

This does not match $4x^2 + 8x + 3$.

(B) (2x + 1) (2x + 3)

$= 2x(2x) + 2x(3) + 1(2x) + 1(3) = 4x^2 + 6x + 2x + 3 = 4x^2 + 8x + 3$

This matches $4x^2 + 8x + 3$.

Since option (B) is the correct factorization, we don't need to check the remaining options. However, for completeness:

(C) (2x + 2) (2x + 5)

$= 2x(2x) + 2x(5) + 2(2x) + 2(5) = 4x^2 + 10x + 4x + 10 = 4x^2 + 14x + 10$

This does not match $4x^2 + 8x + 3$.

(D) (2x – 1) (2x – 3)

$= 2x(2x) + 2x(-3) - 1(2x) - 1(-3) = 4x^2 - 6x - 2x + 3 = 4x^2 - 8x + 3$

This does not match $4x^2 + 8x + 3$.

Both methods show that the factorization of $4x^2 + 8x + 3$ is $(2x + 1)(2x + 3)$.

The correct answer is $\boxed{B}$.

We need to find a factor of the expression $(x + y)^3 - (x^3 + y^3)$.

First, let's expand the term $(x + y)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Here, $a=x$ and $b=y$.

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

Now, substitute this expansion back into the given expression:

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

Remove the parentheses, distributing the negative sign to the terms in the second parenthesis:

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

Combine like terms:

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

$= 0 + 0 + 3x^2y + 3xy^2$

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

Now, factor the simplified expression. We can see that $3xy$ is a common factor in both terms.

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

$= 3xy(x + y)$

So, the factorization of the expression is $3xy(x + y)$.

The factors of the expression are $3$, $x$, $y$, $(x+y)$, $3x$, $3y$, $xy$, $3xy$, $x(x+y)$, $y(x+y)$, $3x(x+y)$, $3y(x+y)$, and $3xy(x+y)$.

We are looking for one of these factors among the given options:

(A) x² + y² + 2xy (This is $(x+y)^2$)

(B) x² + y² – xy

(C) xy²

(D) 3xy

Compare the factors we found with the given options.

Option (A) is $(x+y)^2$, which is not a factor unless $(x+y)^2$ is equal to $3xy(x+y)$ or a constant multiple of it, which is not generally true.

Option (B) is $x^2 + y^2 - xy$, which is related to the sum of cubes factorization $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$, but it is not a factor of $3xy(x+y)$.

Option (C) is $xy^2$, which is a factor of the expression $3x^2y + 3xy^2$ (since $3x^2y + 3xy^2 = xy^2(3\frac{x}{y} + 3)$, this shows $xy^2$ is a factor if $y \neq 0$), but the question asks for a factor from the options. The factors we explicitly found are $3$, $x$, $y$, $x+y$, and their products.

Option (D) is $3xy$, which is one of the factors we found.

The correct answer is $\boxed{D}$.

We need to find the coefficient of $x$ in the expansion of $(x + 3)^3$.

We can expand $(x + 3)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Here, $a=x$ and $b=3$.

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

Simplify each term:

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

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

The expansion of $(x + 3)^3$ is $x^3 + 9x^2 + 27x + 27$.

We are looking for the coefficient of the term with $x^1$ (which is simply $x$).

The term containing $x$ is $27x$.

The coefficient of $x$ is the numerical factor multiplying $x$. In the term $27x$, the coefficient of $x$ is 27.

Looking at the given options:

(A) 1

(B) 9

(C) 18

(D) 27

The coefficient of $x$ matches option (D).

The correct answer is $\boxed{D}$.

We are given the equation $\frac{x}{y} + \frac{y}{x} = -1$, where $x \neq 0$ and $y \neq 0$.

We need to find the value of $x^3 - y^3$.

Let's simplify the given equation. Find a common denominator for the fractions on the left side, which is $xy$.

$\frac{x}{y} \times \frac{x}{x} + \frac{y}{x} \times \frac{y}{y} = -1$

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

Combine the fractions:

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

Multiply both sides by $xy$:

Add $xy$ to both sides:

Now, consider the expression we need to evaluate: $x^3 - y^3$.

We use the factorization formula for the difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

So, $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$.

From equation (i), we know that $x^2 + xy + y^2 = 0$.

Substitute this value into the factorization of $x^3 - y^3$:

$x^3 - y^3 = (x - y)(0)$

$x^3 - y^3 = 0$

So, the value of $x^3 - y^3$ is 0.

Looking at the given options:

(A) 1

(B) –1

(C) 0

(D) $\frac{1}{2}$

The value matches option (C).

The correct answer is $\boxed{C}$.

We are given the equation $49x^2 - b = \left( 7x + \frac{1}{2} \right)\left( 7x - \frac{1}{2} \right)$.

We need to find the value of $b$.

Let's expand the right-hand side of the equation. The right-hand side is in the form of a difference of squares factorization, $(a+c)(a-c) = a^2 - c^2$. Here, $a = 7x$ and $c = \frac{1}{2}$.

$\left( 7x + \frac{1}{2} \right)\left( 7x - \frac{1}{2} \right) = (7x)^2 - \left(\frac{1}{2}\right)^2$

$= 49x^2 - \frac{1^2}{2^2}$

$= 49x^2 - \frac{1}{4}$

Now, substitute this back into the original equation:

Subtract $49x^2$ from both sides of the equation:

Multiply both sides by -1:

So, the value of $b$ is $\frac{1}{4}$.

Looking at the given options:

(A) 0

(B) $\frac{1}{\sqrt{2}}$

(C) $\frac{1}{4}$

(D) $\frac{1}{2}$

The value of $b$ matches option (C).

The correct answer is $\boxed{C}$.

We are given that $a + b + c = 0$.

We need to find the value of $a^3 + b^3 + c^3$.

We know the identity for the sum of cubes: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Given that $a + b + c = 0$.

Substitute $a + b + c = 0$ into the identity:

$a^3 + b^3 + c^3 - 3abc = (0)(a^2 + b^2 + c^2 - ab - bc - ca)$

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

Add $3abc$ to both sides of the equation:

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

Looking at the given options:

(A) 0

(B) abc

(C) 3abc

(D) 2abc

The value matches option (C).

The correct answer is $\boxed{C}$.

(i) $\frac{1}{\sqrt{5}}x^{\frac{1}{2}} + 1$ is a polynomial.

False.

Justification:

A polynomial in one variable $x$ is an algebraic expression where the powers of $x$ are non-negative integers (0, 1, 2, 3, ...).

In the given expression $\frac{1}{\sqrt{5}}x^{\frac{1}{2}} + 1$, the term $\frac{1}{\sqrt{5}}x^{\frac{1}{2}}$ has the variable $x$ raised to the power of $\frac{1}{2}$.

Since $\frac{1}{2}$ is a fraction and not a non-negative integer, the expression is not a polynomial.

(ii) $\frac{6\sqrt{x} \;+\; x^{\frac{3}{2}}}{\sqrt{x}}$ is a polynomial, x ≠ 0.

True.

Justification:

We can simplify the expression by rewriting the terms using exponent notation and distributing the division by $\sqrt{x}$.

$\sqrt{x} = x^{\frac{1}{2}}$

The expression is $\frac{6x^{\frac{1}{2}} \;+\; x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$.

Divide each term in the numerator by the denominator:

$= \frac{6x^{\frac{1}{2}}}{x^{\frac{1}{2}}} + \frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}}}$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$= 6x^{\frac{1}{2} - \frac{1}{2}} + x^{\frac{3}{2} - \frac{1}{2}}$

$= 6x^0 + x^{\frac{3-1}{2}}$

$= 6x^0 + x^{\frac{2}{2}}$

$= 6x^0 + x^1$

Since $x^0 = 1$ (for $x \neq 0$ as given), the expression simplifies to:

$= 6(1) + x$

$= 6 + x$

The simplified expression is $x + 6$.

In this expression, the powers of $x$ are 1 (for the term $x$) and 0 (for the constant term 6, which can be written as $6x^0$). Both 1 and 0 are non-negative integers.

Therefore, the simplified expression is a polynomial.

An algebraic expression is a polynomial if the exponents of the variables are non-negative integers.

(i) 8

This is a polynomial.

Justification: This is a constant term. A constant can be written as $8x^0$. The exponent of the variable ($0$) is a non-negative integer. It is a constant polynomial.

(ii) $\sqrt{3}$x² – 2x

This is a polynomial.

Justification: The expression is $\sqrt{3}x^2 - 2x^1$. The exponents of the variable $x$ are 2 and 1, which are non-negative integers. The coefficients ($\sqrt{3}$ and -2) can be any real numbers.

(iii) 1 – $\sqrt{5x}$

This is not a polynomial.

Justification: The term $\sqrt{5x} = \sqrt{5} \times \sqrt{x} = \sqrt{5}x^{\frac{1}{2}}$. The expression is $1 - \sqrt{5}x^{\frac{1}{2}}$. The exponent of the variable $x$ is $\frac{1}{2}$, which is a fraction and not a non-negative integer.

(iv) $\frac{1}{5x^{-2}}$ + 5x + 7

This is a polynomial.

Justification: Simplify the first term: $\frac{1}{5x^{-2}} = \frac{1}{5} \times \frac{1}{x^{-2}} = \frac{1}{5} \times x^2$. The expression is $\frac{1}{5}x^2 + 5x + 7$. The exponents of the variable $x$ are 2 and 1 (for $5x = 5x^1$), and 0 (for the constant term $7 = 7x^0$). These exponents (2, 1, 0) are non-negative integers.

(v) $\frac{(x \;-\; 2) (x \;-\; 4)}{x}$

This is not a polynomial.

Justification: Expand the numerator: $(x-2)(x-4) = x^2 - 4x - 2x + 8 = x^2 - 6x + 8$. The expression is $\frac{x^2 - 6x + 8}{x}$. Divide each term in the numerator by $x$:

$\frac{x^2}{x} - \frac{6x}{x} + \frac{8}{x} = x - 6 + \frac{8}{x} = x - 6 + 8x^{-1}$

The term $8x^{-1}$ has the variable $x$ raised to the power of -1, which is a negative integer. Therefore, the expression is not a polynomial.

(vi) $\frac{1}{x \;+\; 1}$

This is not a polynomial.

Justification: The variable appears in the denominator in a way that cannot be simplified to terms with non-negative integer exponents. This is a rational expression, but not a polynomial.

(vii) $\frac{1}{7}$a³ – $\frac{2}{\sqrt{3}}$a² + 4a – 7

This is a polynomial.

Justification: The expression is $\frac{1}{7}a^3 - \frac{2}{\sqrt{3}}a^2 + 4a^1 - 7a^0$. The exponents of the variable $a$ are 3, 2, 1, and 0, which are non-negative integers. The coefficients ($\frac{1}{7}$, $-\frac{2}{\sqrt{3}}$, 4, -7) are real numbers.

(viii) $\frac{1}{2x}$

This is not a polynomial.

Justification: The expression can be written as $\frac{1}{2} \times \frac{1}{x} = \frac{1}{2}x^{-1}$. The exponent of the variable $x$ is -1, which is a negative integer. Therefore, the expression is not a polynomial.

The expressions that are polynomials are (i), (ii), (iv), and (vii).

(i) A binomial can have atmost two terms.

False.

Justification:

A binomial is defined as a polynomial with exactly two non-zero terms. The word "atmost" means "at most", implying a maximum of two terms. However, the definition requires exactly two terms. A polynomial with one term is a monomial, and a polynomial with three terms is a trinomial. A binomial cannot have one term (monomial) or more than two terms (like a trinomial or other polynomials).

Example of a binomial: $x + 1$, $x^2 - 5$, $3y^5 + 2y$. These have exactly two terms.

Example of a polynomial with "at most two terms": This would include monomials (e.g., $x$) and binomials (e.g., $x+1$). But a binomial itself must have two terms.

(ii) Every polynomial is a binomial.

False.

Justification:

A polynomial is a general expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Binomials are a specific type of polynomial that has exactly two non-zero terms.

Examples of polynomials that are not binomials include monomials (e.g., $5x^2$) and trinomials (e.g., $x^2 + x + 1$).

(iii) A binomial may have degree 5.

True.

Justification:

The degree of a polynomial is the highest power of the variable with a non-zero coefficient. A binomial is a polynomial with exactly two terms. The degree is determined by the term with the highest exponent.

Example of a binomial with degree 5: $x^5 + 1$, $2x^5 - 3x$. In both cases, there are exactly two terms, and the highest power of the variable is 5.

(iv) Zero of a polynomial is always 0.

False.

Justification:

A zero of a polynomial $p(x)$ is a value of $x$ for which $p(x) = 0$. This value is not necessarily 0.

Example: Consider the polynomial $p(x) = x - 5$. The zero is found by setting $p(x) = 0$, so $x - 5 = 0$, which gives $x = 5$. Here, the zero is 5, not 0.

Consider the polynomial $p(x) = 2x$. The zero is found by setting $p(x) = 0$, so $2x = 0$, which gives $x = 0$. In this case, the zero is 0. This shows that 0 can be a zero, but it's not always the zero.

(v) A polynomial cannot have more than one zero.

False.

Justification:

A polynomial can have multiple zeros. The number of zeros a polynomial can have is at most equal to its degree.

Example: Consider the polynomial $p(x) = x^2 - 4$. This is a polynomial of degree 2. The zeros are found by setting $p(x) = 0$, so $x^2 - 4 = 0$, which means $x^2 = 4$. The solutions are $x = 2$ and $x = -2$. This polynomial has two distinct zeros (2 and -2).

Example: Consider the polynomial $p(x) = x(x-1)(x+2) = x^3 + x^2 - 2x$. This is a polynomial of degree 3. The zeros are found by setting $p(x) = 0$, so $x(x-1)(x+2) = 0$. The solutions are $x=0$, $x=1$, and $x=-2$. This polynomial has three distinct zeros (0, 1, and -2).

(vi) The degree of the sum of two polynomials each of degree 5 is always 5.

False.

Justification:

When adding two polynomials, the degree of the sum is typically the highest degree of the individual polynomials. However, if the leading terms (the terms with the highest degree) cancel out, the degree of the sum can be less than the degree of the individual polynomials.

Example 1: Let $p(x) = x^5 + x^2 + 1$ (degree 5) and $q(x) = 2x^5 + x + 3$ (degree 5).

$p(x) + q(x) = (x^5 + x^2 + 1) + (2x^5 + x + 3)$

$= (1+2)x^5 + x^2 + x + (1+3)$

$= 3x^5 + x^2 + x + 4$

The degree of the sum is 5.

Example 2: Let $p(x) = x^5 + x^2 + 1$ (degree 5) and $q(x) = -x^5 + x + 3$ (degree 5).

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

$= (1-1)x^5 + x^2 + x + (1+3)$

$= 0x^5 + x^2 + x + 4$

$= x^2 + x + 4$

The degree of the sum is 2, which is less than 5.

(i) Check whether p(x) is a multiple of g(x) or not, where p(x) = x³ – x + 1, g(x) = 2 – 3x.

For $p(x)$ to be a multiple of $g(x)$, $g(x)$ must be a factor of $p(x)$. According to the Factor Theorem, $g(x) = 2 - 3x$ is a factor of $p(x)$ if and only if $p(a) = 0$, where $a$ is the zero of $g(x)$.

To find the zero of $g(x)$, set $g(x) = 0$:

So, the zero of $g(x)$ is $\frac{2}{3}$.

Now, evaluate $p\left(\frac{2}{3}\right)$:

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

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

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

$p\left(\frac{2}{3}\right) = \frac{8}{27} - \frac{2}{3} + 1$

Find a common denominator, which is 27:

$= \frac{8}{27} - \frac{2 \times 9}{3 \times 9} + \frac{1 \times 27}{1 \times 27}$

$= \frac{8}{27} - \frac{18}{27} + \frac{27}{27}$

$= \frac{8 - 18 + 27}{27}$

$= \frac{-10 + 27}{27}$

$= \frac{17}{27}$

Since $p\left(\frac{2}{3}\right) = \frac{17}{27} \neq 0$, $g(x)$ is not a factor of $p(x)$.

Therefore, $p(x)$ is not a multiple of $g(x)$.

(ii) Check whether g(x) is a factor of p(x) or not, where p(x) = 8x³ – 6x² – 4x + 3, g(x) = $\frac{x}{3}$ - $\frac{1}{4}$.

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

To find the zero of $g(x)$, set $g(x) = 0$:

Multiply both sides by 3:

So, the zero of $g(x)$ is $\frac{3}{4}$.

Now, evaluate $p\left(\frac{3}{4}\right)$:

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

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

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

$p\left(\frac{3}{4}\right) = 8\left(\frac{27}{64}\right) - 6\left(\frac{9}{16}\right) - \cancel{4}\left(\frac{3}{\cancel{4}}\right) + 3$

Simplify the terms:

$8\left(\frac{27}{64}\right) = \frac{\cancel{8}^1 \times 27}{\cancel{64}_8} = \frac{27}{8}$

$6\left(\frac{9}{16}\right) = \frac{\cancel{6}^3 \times 9}{\cancel{16}_8} = \frac{27}{8}$

So, $p\left(\frac{3}{4}\right) = \frac{27}{8} - \frac{27}{8} - 3 + 3$

$p\left(\frac{3}{4}\right) = 0 - 0 = 0$

Since $p\left(\frac{3}{4}\right) = 0$, $g(x)$ is a factor of $p(x)$.

Given:

The polynomial $p(x) = x^3 – ax^2 + 2x + a – 1$.

The linear expression $x - a$ is a factor of $p(x)$.

To Find:

The value of $a$.

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 given as $x - a$. This is already in the form $(x - \text{zero})$, where the zero is $a$.

Substitute $x = a$ into the polynomial $p(x) = x^3 – ax^2 + 2x + a – 1$ and set the result equal to 0.

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

Since $(x - a)$ is a factor, $p(a) = 0$.

Combine like terms:

Now, solve for $a$. Add 1 to both sides of the equation:

Divide both sides by 3:

Thus, the value of $a$ is $\frac{1}{3}$.

(i) Without actually calculating the cubes, find the value of 48³ – 30³ – 18³.

We can rewrite the expression as $48^3 + (-30)^3 + (-18)^3$.

Consider the identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Let $a = 48$, $b = -30$, and $c = -18$.

Check if $a + b + c = 0$:

$a + b + c = 48 + (-30) + (-18)$

$= 48 - 30 - 18$

$= 48 - (30 + 18)$

$= 48 - 48$

$= 0$

Since $a + b + c = 0$, we can apply the identity $a^3 + b^3 + c^3 = 3abc$.

$48^3 + (-30)^3 + (-18)^3 = 3 \times (48) \times (-30) \times (-18)$

$= 3 \times 48 \times (30 \times 18)$

Calculate the products:

$3 \times 48 = 144$

$30 \times 18 = 540$

$= 144 \times 540$

$\begin{array}{cc}& & 1 & 4 & 4 \\ \times & & 5 & 4 & 0 \\ \hline && 0 & 0 & 0 \\ & 5 & 7 & 6 & \times \\ 7 & 2 & 0 & \times & \times \\ \hline 7 & 7 & 7 & 6 & 0 \\ \hline \end{array}$

$144 \times 540 = 77760$

So, $48^3 – 30^3 – 18^3 = 77760$.

(ii) Without finding the cubes, factorise (x – y)³ + (y – z)³ + (z – x)³.

Consider the terms inside the cubes: $(x - y)$, $(y - z)$, and $(z - x)$.

Let $a = x - y$, $b = y - z$, and $c = z - x$.

Check the sum of these terms:

$a + b + c = (x - y) + (y - z) + (z - x)$

$= x - y + y - z + z - x$

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

$= 0 + 0 + 0$

$= 0$

Since $a + b + c = 0$, we can use the identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

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

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

Thus, the factorization is $3(x - y)(y - z)(z - x)$.

To classify a polynomial by the number of variables, we identify the different variables present in the expression.

(i) x² + x + 1

The only variable present is $x$.

This is a polynomial in one variable (x).

(ii) y³ – 5y

The only variable present is $y$.

This is a polynomial in one variable (y).

(iii) xy + yz + zx

The variables present are $x$, $y$, and $z$.

This is a polynomial in three variables (x, y, and z).

(iv) x² – 2xy + y² + 1

The variables present are $x$ and $y$.

This is a polynomial in two variables (x and y).

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

(i) 2x – 1

The polynomial is $2x^1 - 1x^0$.

The powers of the variable $x$ are 1 and 0. The term with the highest power is $2x^1$.

The highest exponent is 1.

The degree of the polynomial is 1.

(ii) –10

The polynomial is $-10$. This is a constant polynomial.

A constant polynomial can be written as $-10x^0$.

The highest exponent of the variable with a non-zero coefficient is 0.

The degree of the polynomial is 0.

(iii) x³ – 9x + 3x⁵

The polynomial is $x^3 - 9x^1 + 3x^5$.

The powers of the variable $x$ are 3, 1, and 5.

Arrange the terms in descending order of exponents: $3x^5 + x^3 - 9x^1$.

The highest exponent is 5.

The degree of the polynomial is 5.

(iv) y³ (1 – y⁴)

First, expand the expression by distributing $y^3$:

$y^3 (1 – y^4) = y^3 \times 1 - y^3 \times y^4$

Using the exponent rule $a^m \times a^n = a^{m+n}$:

$= y^3 - y^{3+4}$

$= y^3 - y^7$

The polynomial is $y^3 - y^7$. Written in descending order of exponents: $-y^7 + y^3$.

The powers of the variable $y$ are 7 and 3.

The highest exponent is 7.

The degree of the polynomial is 7.

The given polynomial is $\frac{x^3 \;+\; 2x \;+\; 1}{5} - \frac{7}{2}x^2 - x^6$.

We can rewrite this polynomial by separating the terms in the first fraction:

$= \frac{x^3}{5} + \frac{2x}{5} + \frac{1}{5} - \frac{7}{2}x^2 - x^6$

Rewrite each term with its coefficient and variable part explicitly:

$= \frac{1}{5}x^3 + \frac{2}{5}x - \frac{7}{2}x^2 + \frac{1}{5} - 1x^6$

Rearrange the terms in descending order of the powers of $x$:

$= -1x^6 + \frac{1}{5}x^3 - \frac{7}{2}x^2 + \frac{2}{5}x + \frac{1}{5}$

Now, let's answer the questions:

(i) the degree of the polynomial

The degree is the highest exponent of $x$ with a non-zero coefficient. In the rearranged polynomial, the powers of $x$ are 6, 3, 2, 1, and 0 (for the constant term).

The highest exponent is 6.

The degree of the polynomial is 6.

(ii) the coefficient of x³

The term with $x^3$ is $\frac{1}{5}x^3$.

The coefficient of $x^3$ is the numerical factor multiplying $x^3$.

The coefficient of $x^3$ is $\frac{1}{5}$.

(iii) the coefficient of x⁶

The term with $x^6$ is $-1x^6$.

The coefficient of $x^6$ is the numerical factor multiplying $x^6$.

The coefficient of $x^6$ is -1.

(iv) the constant term

The constant term is the term that does not have a variable. In the rearranged polynomial, the term with no variable is $\frac{1}{5}$.

The constant term is $\frac{1}{5}$.

(i) The given expression is $\frac{\pi}{6}x + x^2 - 1$.

We can rewrite this in standard polynomial form as $1 \cdot x^2 + \frac{\pi}{6}x - 1$.

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

In this case, the term is $x^2$, which is the same as $1 \cdot x^2$.

Therefore, the coefficient of $x^2$ is $1$.

(ii) The given expression is $3x - 5$.

We can rewrite this expression including an $x^2$ term with a zero coefficient: $0 \cdot x^2 + 3x - 5$.

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

In this case, the coefficient of $x^2$ is $0$.

(iii) The given expression is $(x - 1) (3x - 4)$.

First, we need to expand this expression:

$(x - 1)(3x - 4) = x(3x - 4) - 1(3x - 4)$

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

Combine the like terms:

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

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

The term containing $x^2$ is $3x^2$.

The coefficient of $x^2$ is the number multiplying $x^2$.

Therefore, the coefficient of $x^2$ is $3$.

(iv) The given expression is $(2x - 5) (2x^2 - 3x + 1)$.

First, we need to expand this expression:

$(2x - 5)(2x^2 - 3x + 1) = 2x(2x^2 - 3x + 1) - 5(2x^2 - 3x + 1)$

$= (2x \cdot 2x^2) + (2x \cdot -3x) + (2x \cdot 1) + (-5 \cdot 2x^2) + (-5 \cdot -3x) + (-5 \cdot 1)$

$= 4x^3 - 6x^2 + 2x - 10x^2 + 15x - 5$

Combine the like terms:

$= 4x^3 + (-6x^2 - 10x^2) + (2x + 15x) - 5$

$= 4x^3 - 16x^2 + 17x - 5$

The term containing $x^2$ is $-16x^2$.

The coefficient of $x^2$ is the number multiplying $x^2$.

Therefore, the coefficient of $x^2$ is $-16$.

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

Degree 1: Linear polynomial

Degree 2: Quadratic polynomial

Degree 3: Cubic polynomial

Degree 0 (non-zero constant): Constant polynomial

(i) $2 – x^2 + x^3$

The highest power of the variable $x$ is $3$.

Therefore, it is a cubic polynomial.

(ii) $3x^3$

The highest power of the variable $x$ is $3$.

Therefore, it is a cubic polynomial.

(iii) $5t – \sqrt{7}$

The highest power of the variable $t$ is $1$.

Therefore, it is a linear polynomial.

(iv) $4 – 5y^2$

The highest power of the variable $y$ is $2$.

Therefore, it is a quadratic polynomial.

(v) $3$

This is a constant term, which can be written as $3x^0$.

The highest power of the variable is $0$.

Therefore, it is a constant polynomial.

(vi) $2 + x$

The highest power of the variable $x$ is $1$.

Therefore, it is a linear polynomial.

(vii) $y^3 – y$

The highest power of the variable $y$ is $3$.

Therefore, it is a cubic polynomial.

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

The highest power of the variable $x$ is $2$.

Therefore, it is a quadratic polynomial.

(ix) $t^2$

The highest power of the variable $t$ is $2$.

Therefore, it is a quadratic polynomial.

(x) $\sqrt{2}x – 1$

The highest power of the variable $x$ is $1$.

Therefore, it is a linear polynomial.

(i) A monomial is a polynomial with only one term. The degree of a polynomial is the highest power of the variable.

We need a monomial of degree 1. This means it should have one term and the highest power of the variable should be 1.

An example is a term like $ax$, where $a$ is a non-zero constant.

Example: $5x$ (Here, the term is $5x$ and the degree is 1).

(ii) A binomial is a polynomial with exactly two terms. The degree of a polynomial is the highest power of the variable.

We need a binomial of degree 20. This means it should have two terms and the highest power of the variable in either term should be 20.

Example: One term with a variable raised to the power of 20 and another term with a different power (less than 20) or a constant.

Example: $x^{20} + 9$ (Here, there are two terms $x^{20}$ and $9$, and the highest power is 20).

(iii) A trinomial is a polynomial with exactly three terms. The degree of a polynomial is the highest power of the variable.

We need a trinomial of degree 2. This means it should have three terms and the highest power of the variable in any term should be 2.

Example: One term with a variable raised to the power of 2, another term with a variable raised to the power of 1, and a third term as a constant.

Example: $2x^2 + 3x - 7$ (Here, there are three terms $2x^2$, $3x$, and $-7$, and the highest power is 2).

Let the given polynomial be $P(x) = 3x^3 - 4x^2 + 7x - 5$.

Case 1: When $x = 3$

Substitute $x = 3$ into the polynomial:

$P(3) = 3(3)^3 - 4(3)^2 + 7(3) - 5$

$P(3) = 3(27) - 4(9) + 21 - 5$

$P(3) = 81 - 36 + 21 - 5$

$P(3) = (81 + 21) - (36 + 5)$

$P(3) = 102 - 41$

$P(3) = 61$

So, the value of the polynomial when $x = 3$ is $61$.

Case 2: When $x = -3$

Substitute $x = -3$ into the polynomial:

$P(-3) = 3(-3)^3 - 4(-3)^2 + 7(-3) - 5$

Recall that $(-3)^3 = -27$ and $(-3)^2 = 9$.

$P(-3) = 3(-27) - 4(9) + (-21) - 5$

$P(-3) = -81 - 36 - 21 - 5$

$P(-3) = -(81 + 36 + 21 + 5)$

$P(-3) = -(117 + 21 + 5)$

$P(-3) = -(138 + 5)$

$P(-3) = -143$

So, the value of the polynomial when $x = -3$ is $-143$.

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

We need to evaluate $p(2) - p(-1) + p(\frac{1}{2})$.

First, let's find the value of $p(2)$:

Substitute $x = 2$ into the polynomial:

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

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

$p(2) = -4 + 3$

$p(2) = -1$

Next, let's find the value of $p(-1)$:

Substitute $x = -1$ into the polynomial:

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

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

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

$p(-1) = 5 + 3$

$p(-1) = 8$

Now, let's find the value of $p(\frac{1}{2})$:

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

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

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

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

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

To add $\frac{1}{4}$ and $1$, we find a common denominator:

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

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

$p(\frac{1}{2}) = \frac{5}{4}$

Finally, we evaluate $p(2) - p(-1) + p(\frac{1}{2})$:

$p(2) - p(-1) + p(\frac{1}{2}) = -1 - 8 + \frac{5}{4}$

$= -9 + \frac{5}{4}$

To combine $-9$ and $\frac{5}{4}$, we find a common denominator:

$= -\frac{9 \cdot 4}{4} + \frac{5}{4}$

$= -\frac{36}{4} + \frac{5}{4}$

$= \frac{-36 + 5}{4}$

$= \frac{-31}{4}$

The value of $p(2) - p(-1) + p(\frac{1}{2})$ is $-\frac{31}{4}$.

(i) Given polynomial is $p(x) = 10x - 4x^2 - 3$.

We need to find $p(0)$, $p(1)$, and $p(-2)$.

Find $p(0)$:

Substitute $x = 0$ into the polynomial:

$p(0) = 10(0) - 4(0)^2 - 3$

$p(0) = 0 - 4(0) - 3$

$p(0) = 0 - 0 - 3$

$p(0) = -3$

Find $p(1)$:

Substitute $x = 1$ into the polynomial:

$p(1) = 10(1) - 4(1)^2 - 3$

$p(1) = 10 - 4(1) - 3$

$p(1) = 10 - 4 - 3$

$p(1) = 6 - 3$

$p(1) = 3$

Find $p(-2)$:

Substitute $x = -2$ into the polynomial:

$p(-2) = 10(-2) - 4(-2)^2 - 3$

Recall that $(-2)^2 = 4$.

$p(-2) = -20 - 4(4) - 3$

$p(-2) = -20 - 16 - 3$

$p(-2) = -36 - 3$

$p(-2) = -39$

(ii) Given polynomial is $p(y) = (y + 2)(y - 2)$.

We can expand this using the identity $(a+b)(a-b) = a^2 - b^2$:

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

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

We need to find $p(0)$, $p(1)$, and $p(-2)$.

Find $p(0)$:

Substitute $y = 0$ into the polynomial:

$p(0) = (0)^2 - 4$

$p(0) = 0 - 4$

$p(0) = -4$

Find $p(1)$:

Substitute $y = 1$ into the polynomial:

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

$p(1) = 1 - 4$

$p(1) = -3$

Find $p(-2)$:

Substitute $y = -2$ into the polynomial:

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

$p(-2) = 4 - 4$

$p(-2) = 0$

A number 'a' is a zero of a polynomial $p(x)$ if $p(a) = 0$. We need to evaluate the polynomial at the given value to verify if it is a zero.

(i) Given polynomial is $p(x) = x - 3$. We need to check if $-3$ is a zero.

Substitute $x = -3$:

$p(-3) = (-3) - 3$

$p(-3) = -6$

Since $p(-3) \neq 0$, $-3$ is not a zero of $x - 3$.

Therefore, the statement is False.

(ii) Given polynomial is $p(x) = 3x + 1$. We need to check if $-\frac{1}{3}$ is a zero.

Substitute $x = -\frac{1}{3}$:

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

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

$p(-\frac{1}{3}) = 0$

Since $p(-\frac{1}{3}) = 0$, $-\frac{1}{3}$ is a zero of $3x + 1$.

Therefore, the statement is True.

(iii) Given polynomial is $p(y) = 4 - 5y$. We need to check if $\frac{-4}{5}$ is a zero.

Substitute $y = \frac{-4}{5}$:

$p(\frac{-4}{5}) = 4 - 5(\frac{-4}{5})$

$p(\frac{-4}{5}) = 4 - (\cancel{5} \cdot \frac{-4}{\cancel{5}})$

$p(\frac{-4}{5}) = 4 - (-4)$

$p(\frac{-4}{5}) = 4 + 4$

$p(\frac{-4}{5}) = 8$

Since $p(\frac{-4}{5}) \neq 0$, $\frac{-4}{5}$ is not a zero of $4 - 5y$.

Therefore, the statement is False.

(iv) Given polynomial is $p(t) = t^2 - 2t$. We need to check if $0$ and $2$ are zeroes.

First, check for $t = 0$:

$p(0) = (0)^2 - 2(0)$

$p(0) = 0 - 0$

$p(0) = 0$

Since $p(0) = 0$, $0$ is a zero of $t^2 - 2t$.

Next, check for $t = 2$:

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

$p(2) = 4 - 4$

$p(2) = 0$

Since $p(2) = 0$, $2$ is a zero of $t^2 - 2t$.

Since both $p(0) = 0$ and $p(2) = 0$, both $0$ and $2$ are zeroes of the polynomial.

Therefore, the statement is True.

(v) Given polynomial is $p(y) = y^2 + y - 6$. We need to check if $-3$ is a zero.

Substitute $y = -3$:

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

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

$p(-3) = 6 - 6$

$p(-3) = 0$

Since $p(-3) = 0$, $-3$ is a zero of $y^2 + y - 6$.

Therefore, the statement is True.

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

(i) Given polynomial is $p(x) = x - 4$.

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

Add 4 to both sides:

The zero of the polynomial $p(x) = x - 4$ is $4$.

(ii) Given polynomial is $g(x) = 3 - 6x$.

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

Add $6x$ to both sides:

Divide both sides by 6:

Simplify the fraction:

The zero of the polynomial $g(x) = 3 - 6x$ is $\frac{1}{2}$.

(iii) Given polynomial is $q(x) = 2x - 7$.

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

Add 7 to both sides:

Divide both sides by 2:

The zero of the polynomial $q(x) = 2x - 7$ is $\frac{7}{2}$.

(iv) Given polynomial is $h(y) = 2y$.

To find the zero, set $h(y) = 0$:

Divide both sides by 2:

The zero of the polynomial $h(y) = 2y$ is $0$.

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

To find the zeroes of the polynomial, we set $p(x) = 0$.

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

We can expand the terms or use the difference of squares identity, $a^2 - b^2 = (a - b)(a + b)$.

Let $a = (x - 2)$ and $b = (x + 2)$.

So, $p(x) = [(x - 2) - (x + 2)][(x - 2) + (x + 2)]$

Simplify the terms inside the brackets:

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

Combine like terms in each bracket:

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

$(-4)(2x) = 0$

Multiply the terms:

$-8x = 0$

Now, solve for $x$:

Alternatively, we could expand the squares directly:

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

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

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

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

Combine like terms:

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

$p(x) = 0x^2 - 8x + 0$

$p(x) = -8x$

Setting $p(x) = 0$:

The zero of the polynomial $p(x) = (x - 2)^2 - (x + 2)^2$ is $0$.

The given equation is:

To solve for $x$, we first eliminate the denominators by multiplying the entire equation by the least common multiple (LCM) of the denominators $2, 4, 3,$ and $2$.

The denominators are $2, 4, 3$.

Let's find the LCM of $2, 4, 3$:

$\begin{array}{c|cc} 2 & 2 \;, & 4 \;, & 3 \\ \hline 2 & 1 \; , & 2 \; , & 3 \\ \hline 3 & 1 \; , & 1 \; , & 3 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$

LCM is $2 \times 2 \times 3 = 12$.

Multiply each term of the equation by 12:

$12 \cdot \left(\frac{x}{2}\right) - 12 \cdot \left(\frac{1}{4}\right) = 12 \cdot \left(\frac{x}{3}\right) + 12 \cdot \left(\frac{1}{2}\right)$

Simplify each term:

$\cancel{12}^6 \cdot \frac{x}{\cancel{2}_1} - \cancel{12}^3 \cdot \frac{1}{\cancel{4}_1} = \cancel{12}^4 \cdot \frac{x}{\cancel{3}_1} + \cancel{12}^6 \cdot \frac{1}{\cancel{2}_1}$

$6x - 3 = 4x + 6$

Now, gather the terms with $x$ on one side and the constant terms on the other side. Subtract $4x$ from both sides:

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

$2x - 3 = 6$

Add 3 to both sides:

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

$2x = 9$

Divide both sides by 2:

The solution to the equation is $x = \frac{9}{2}$.

According to the Remainder Theorem, if a polynomial $p(x)$ is divided by a linear polynomial $g(x) = x - a$, then the remainder is $p(a)$. If $g(x) = bx - a$, the zero is $x = \frac{a}{b}$, and the remainder is $p(\frac{a}{b})$.

(i) Given $p(x) = x^3 – 2x^2 – 4x – 1$ and $g(x) = x + 1$.

To find the zero of $g(x)$, set $g(x) = 0$:

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

Substitute $x = -1$ into $p(x)$:

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

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

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

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

$p(-1) = 1 - 1$

$p(-1) = 0$

The remainder is $0$.

(ii) Given $p(x) = x^3 – 3x^2 + 4x + 50$ and $g(x) = x - 3$.

To find the zero of $g(x)$, set $g(x) = 0$:

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $x - 3$ is $p(3)$.

Substitute $x = 3$ into $p(x)$:

$p(3) = (3)^3 - 3(3)^2 + 4(3) + 50$

$p(3) = 27 - 3(9) + 12 + 50$

$p(3) = 27 - 27 + 12 + 50$

$p(3) = 0 + 12 + 50$

$p(3) = 62$

The remainder is $62$.

(iii) Given $p(x) = 4x^3 – 12x^2 + 14x – 3$ and $g(x) = 2x - 1$.

To find the zero of $g(x)$, set $g(x) = 0$:

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 $p(x)$:

$p(\frac{1}{2}) = 4(\frac{1}{2})^3 - 12(\frac{1}{2})^2 + 14(\frac{1}{2}) - 3$

$p(\frac{1}{2}) = 4(\frac{1}{8}) - 12(\frac{1}{4}) + \cancel{14}^7(\frac{1}{\cancel{2}_1}) - 3$

$p(\frac{1}{2}) = \frac{\cancel{4}^1}{\cancel{8}_2} - \frac{\cancel{12}^3}{\cancel{4}_1} + 7 - 3$

$p(\frac{1}{2}) = \frac{1}{2} - 3 + 7 - 3$

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

$p(\frac{1}{2}) = \frac{1}{2} + 1$

$p(\frac{1}{2}) = \frac{1}{2} + \frac{2}{2}$

$p(\frac{1}{2}) = \frac{1+2}{2}$

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

The remainder is $\frac{3}{2}$.

(iv) Given $p(x) = x^3 – 6x^2 + 2x – 4$ and $g(x) = 1 – \frac{3}{2}x$.

To find the zero of $g(x)$, set $g(x) = 0$:

Multiply both sides by $\frac{2}{3}$:

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $1 – \frac{3}{2}x$ is $p(\frac{2}{3})$.

Substitute $x = \frac{2}{3}$ into $p(x)$:

$p(\frac{2}{3}) = (\frac{2}{3})^3 - 6(\frac{2}{3})^2 + 2(\frac{2}{3}) - 4$

$p(\frac{2}{3}) = \frac{2^3}{3^3} - 6(\frac{2^2}{3^2}) + \frac{4}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} - 6(\frac{4}{9}) + \frac{4}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} - \frac{\cancel{6}^2 \cdot 4}{\cancel{9}_3} + \frac{4}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} - \frac{8}{3} + \frac{4}{3} - 4$

Combine the terms with denominator 3:

$p(\frac{2}{3}) = \frac{8}{27} + \frac{4 - 8}{3} - 4$

$p(\frac{2}{3}) = \frac{8}{27} + \frac{-4}{3} - 4$

Find a common denominator for the fractions, which is 27.

$p(\frac{2}{3}) = \frac{8}{27} - \frac{4 \cdot 9}{3 \cdot 9} - \frac{4 \cdot 27}{1 \cdot 27}$

$p(\frac{2}{3}) = \frac{8}{27} - \frac{36}{27} - \frac{108}{27}$

$p(\frac{2}{3}) = \frac{8 - 36 - 108}{27}$

$p(\frac{2}{3}) = \frac{-28 - 108}{27}$

$p(\frac{2}{3}) = \frac{-136}{27}$

The remainder is $-\frac{136}{27}$.

A polynomial $p(x)$ is a multiple of a polynomial $g(x)$ if and only if the remainder when $p(x)$ is divided by $g(x)$ is zero. We can use the Remainder Theorem to find the remainder.

(i) Given $p(x) = x^3 – 5x^2 + 4x – 3$ and $g(x) = x – 2$.

To check if $p(x)$ is a multiple of $g(x)$, we find the zero of $g(x)$ and evaluate $p(x)$ at that value.

Set $g(x) = 0$:

Now, evaluate $p(2)$:

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

$p(2) = 8 - 5(4) + 8 - 3$

$p(2) = 8 - 20 + 8 - 3$

$p(2) = 16 - 20 - 3$

$p(2) = -4 - 3$

$p(2) = -7$

Since the remainder $p(2) = -7 \neq 0$, $p(x)$ is not a multiple of $g(x)$.

Therefore, the statement is False.

(ii) Given $p(x) = 2x^3 – 11x^2 – 4x + 5$ and $g(x) = 2x + 1$.

To check if $p(x)$ is a multiple of $g(x)$, we find the zero of $g(x)$ and evaluate $p(x)$ at that value.

Set $g(x) = 0$:

Now, evaluate $p(-\frac{1}{2})$:

$p(-\frac{1}{2}) = 2(-\frac{1}{2})^3 - 11(-\frac{1}{2})^2 - 4(-\frac{1}{2}) + 5$

$p(-\frac{1}{2}) = 2(-\frac{1}{8}) - 11(\frac{1}{4}) - (\cancel{4}^2)(-\frac{1}{\cancel{2}_1}) + 5$

$p(-\frac{1}{2}) = -\frac{\cancel{2}^1}{\cancel{8}_4} - \frac{11}{4} - (-2) + 5$

$p(-\frac{1}{2}) = -\frac{1}{4} - \frac{11}{4} + 2 + 5$

$p(-\frac{1}{2}) = \frac{-1 - 11}{4} + 7$

$p(-\frac{1}{2}) = \frac{-12}{4} + 7$

$p(-\frac{1}{2}) = -3 + 7$

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

Since the remainder $p(-\frac{1}{2}) = 4 \neq 0$, $p(x)$ is not a multiple of $g(x)$.

Therefore, the statement is False.

According to the Factor Theorem, a polynomial $g(x)$ is a factor of a polynomial $p(x)$ if and only if $p(a) = 0$, where $a$ is the zero of $g(x)$ (i.e., $g(a) = 0$).

(i) Given $p(x) = 69 + 11x – x^2 + x^3$ and $g(x) = x + 3$.

First, find the zero of $g(x)$. Set $g(x) = 0$:

Now, evaluate $p(x)$ at $x = -3$. If $p(-3) = 0$, then $x + 3$ is a factor of $p(x)$.

Substitute $x = -3$ into $p(x)$:

$p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$

$p(-3) = 69 - 33 - (9) + (-27)$

$p(-3) = 69 - 33 - 9 - 27$

$p(-3) = (69 - 33) - 9 - 27$

$p(-3) = 36 - 9 - 27$

$p(-3) = 27 - 27$

$p(-3) = 0$

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

(ii) Given $p(x) = x + 2x^3 – 9x^2 + 12$ and $g(x) = 2x – 3$.

First, find the zero of $g(x)$. Set $g(x) = 0$:

Now, evaluate $p(x)$ at $x = \frac{3}{2}$. If $p(\frac{3}{2}) = 0$, then $2x - 3$ is a factor of $p(x)$.

Substitute $x = \frac{3}{2}$ into $p(x)$:

$p(\frac{3}{2}) = (\frac{3}{2}) + 2(\frac{3}{2})^3 - 9(\frac{3}{2})^2 + 12$

$p(\frac{3}{2}) = \frac{3}{2} + 2(\frac{27}{8}) - 9(\frac{9}{4}) + 12$

$p(\frac{3}{2}) = \frac{3}{2} + \frac{\cancel{2}^1 \cdot 27}{\cancel{8}_4} - \frac{81}{4} + 12$

$p(\frac{3}{2}) = \frac{3}{2} + \frac{27}{4} - \frac{81}{4} + 12$

Find a common denominator for the fractions, which is 4.

$p(\frac{3}{2}) = \frac{3 \cdot 2}{2 \cdot 2} + \frac{27}{4} - \frac{81}{4} + \frac{12 \cdot 4}{1 \cdot 4}$

$p(\frac{3}{2}) = \frac{6}{4} + \frac{27}{4} - \frac{81}{4} + \frac{48}{4}$

$p(\frac{3}{2}) = \frac{6 + 27 - 81 + 48}{4}$

$p(\frac{3}{2}) = \frac{33 - 81 + 48}{4}$

$p(\frac{3}{2}) = \frac{-48 + 48}{4}$

$p(\frac{3}{2}) = \frac{0}{4}$

$p(\frac{3}{2}) = 0$

Since $p(\frac{3}{2}) = 0$, by the Factor Theorem, $2x - 3$ is a factor of $p(x)$.

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 case, we are checking if $(x - 2)$ is a factor. The zero of $(x - 2)$ is found by setting $x - 2 = 0$, which gives $x = 2$.

So, we need to evaluate each given polynomial at $x = 2$. If the value of the polynomial at $x = 2$ is $0$, then $(x - 2)$ is a factor.

(i) Let $p(x) = 3x^2 + 6x - 24$.

Evaluate $p(2)$:

$p(2) = 3(2)^2 + 6(2) - 24$

$p(2) = 3(4) + 12 - 24$

$p(2) = 12 + 12 - 24$

$p(2) = 24 - 24$

$p(2) = 0$

Since $p(2) = 0$, $(x - 2)$ is a factor of the polynomial $3x^2 + 6x - 24$.

(ii) Let $q(x) = 4x^2 + x - 2$.

Evaluate $q(2)$:

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

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

$q(2) = 16 + 2 - 2$

$q(2) = 16 + 0$

$q(2) = 16$

Since $q(2) = 16 \neq 0$, $(x - 2)$ is not a factor of the polynomial $4x^2 + x - 2$.

Therefore, the polynomial that has $(x - 2)$ as a factor is $3x^2 + 6x - 24$.

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 question, the variable is $p$ and the potential factor is $(p - 1)$. So, $a = 1$.

To show that $p - 1$ is a factor of a polynomial, we need to evaluate the polynomial at $p = 1$. If the result is $0$, then $p - 1$ is a factor.

Part 1: Show that $p - 1$ is a factor of $p^{10} - 1$.

Let $P_1(p) = p^{10} - 1$.

The zero of the linear polynomial $p - 1$ is found by setting $p - 1 = 0$, which gives $p = 1$.

Now, evaluate $P_1(p)$ at $p = 1$:

$P_1(1) = (1)^{10} - 1$

$P_1(1) = 1 - 1$

$P_1(1) = 0$

Since $P_1(1) = 0$, by the Factor Theorem, $(p - 1)$ is a factor of $p^{10} - 1$.

Part 2: Show that $p - 1$ is a factor of $p^{11} - 1$.

Let $P_2(p) = p^{11} - 1$.

The zero of the linear polynomial $p - 1$ is $p = 1$.

Now, evaluate $P_2(p)$ at $p = 1$:

$P_2(1) = (1)^{11} - 1$

$P_2(1) = 1 - 1$

$P_2(1) = 0$

Since $P_2(1) = 0$, by the Factor Theorem, $(p - 1)$ is a factor of $p^{11} - 1$.

Therefore, $p - 1$ is a factor of both $p^{10} - 1$ and $p^{11} - 1$.

Let the given polynomial be $p(x) = x^3 - 2mx^2 + 16$.

The polynomial $p(x)$ is divisible by $x + 2$ if and only if $(x + 2)$ is a factor of $p(x)$.

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

In this case, the divisor is $x + 2$, which can be written as $x - (-2)$. So, $a = -2$.

For $p(x)$ to be divisible by $x + 2$, the remainder when $p(x)$ is divided by $x + 2$ must be $0$. By the Remainder Theorem, this remainder is $p(-2)$.

Therefore, we must have $p(-2) = 0$.

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

$p(-2) = (-2)^3 - 2m(-2)^2 + 16$

Calculate the powers:

$(-2)^3 = -8$

$(-2)^2 = 4$

Substitute these values back:

$p(-2) = -8 - 2m(4) + 16$

$p(-2) = -8 - 8m + 16$

Combine the constant terms:

$p(-2) = (-8 + 16) - 8m$

$p(-2) = 8 - 8m$

Since $p(x)$ is divisible by $x + 2$, the remainder must be $0$. Set $p(-2) = 0$:

Solve for $m$. Add $8m$ to both sides:

Divide both sides by 8:

Thus, for the polynomial $x^3 - 2mx^2 + 16$ to be divisible by $x + 2$, the value of $m$ must be $1$.

Let the given polynomial be $p(x) = x^5 – 4a^2x^3 + 2x + 2a + 3$.

We are given that $x + 2a$ is a factor of $p(x)$.

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

In this case, the factor is $x + 2a$, which can be written as $x - (-2a)$. So, $c = -2a$.

For $x + 2a$ to be a factor of $p(x)$, the value of the polynomial at $x = -2a$ must be $0$.

Thus, we must have $p(-2a) = 0$.

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

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

Calculate the terms:

$(-2a)^5 = (-2)^5 \cdot a^5 = -32a^5$

$(-2a)^3 = (-2)^3 \cdot a^3 = -8a^3$

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

$p(-2a) = -32a^5 - 4a^2(-8a^3) + 2(-2a) + 2a + 3$

$p(-2a) = -32a^5 + 32a^5 - 4a + 2a + 3$

Combine the like terms:

$p(-2a) = (-32a^5 + 32a^5) + (-4a + 2a) + 3$

$p(-2a) = 0a^5 - 2a + 3$

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

Since $x + 2a$ is a factor, $p(-2a)$ must be equal to 0.

Solve for $a$. Subtract 3 from both sides:

Divide both sides by -2:

The value of $a$ for which $x + 2a$ is a factor of the given polynomial is $\frac{3}{2}$.

Let the given polynomial be $p(x) = 8x^4 + 4x^3 – 16x^2 + 10x + m$.

We are given that $2x - 1$ is a factor of $p(x)$.

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

First, find the zero of the divisor $g(x) = 2x - 1$. Set $g(x) = 0$:

Since $2x - 1$ is a factor of $p(x)$, by the Factor Theorem, the value of $p(x)$ at $x = \frac{1}{2}$ must be $0$.

Thus, $p(\frac{1}{2}) = 0$.

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

$p(\frac{1}{2}) = 8(\frac{1}{2})^4 + 4(\frac{1}{2})^3 - 16(\frac{1}{2})^2 + 10(\frac{1}{2}) + m$

Calculate the powers of $\frac{1}{2}$:

$(\frac{1}{2})^4 = \frac{1}{16}$

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

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

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

$p(\frac{1}{2}) = 8(\frac{1}{16}) + 4(\frac{1}{8}) - 16(\frac{1}{4}) + 10(\frac{1}{2}) + m$

Simplify the terms:

$p(\frac{1}{2}) = \frac{8}{16} + \frac{4}{8} - \frac{16}{4} + \frac{10}{2} + m$

$p(\frac{1}{2}) = \frac{1}{2} + \frac{1}{2} - 4 + 5 + m$

Combine the constant terms:

$p(\frac{1}{2}) = (\frac{1}{2} + \frac{1}{2}) + (-4 + 5) + m$

$p(\frac{1}{2}) = 1 + 1 + m$

$p(\frac{1}{2}) = 2 + m$

Since $p(\frac{1}{2})$ must be $0$:

Solve for $m$:

The value of $m$ for which $2x - 1$ is a factor of the polynomial is $-2$.

Let the given polynomial be $p(x) = ax^3 + x^2 – 2x + 4a – 9$.

We are given that $x + 1$ is a factor of $p(x)$.

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

In this case, the factor is $x + 1$, which can be written as $x - (-1)$. So, the zero of the factor is $x = -1$.

For $x + 1$ to be a factor of $p(x)$, the value of the polynomial at $x = -1$ must be $0$.

Thus, we must have $p(-1) = 0$.

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

$p(-1) = a(-1)^3 + (-1)^2 - 2(-1) + 4a - 9$

Calculate the terms:

$(-1)^3 = -1$

$(-1)^2 = 1$

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

$p(-1) = a(-1) + 1 - (-2) + 4a - 9$

$p(-1) = -a + 1 + 2 + 4a - 9$

Combine the like terms (terms with $a$ and constant terms):

$p(-1) = (-a + 4a) + (1 + 2 - 9)$

$p(-1) = 3a + (3 - 9)$

$p(-1) = 3a - 6$

Since $x + 1$ is a factor, $p(-1)$ must be equal to 0.

Solve for $a$. Add 6 to both sides:

Divide both sides by 3:

The value of $a$ is $2$.

(i) We need to factorise the quadratic polynomial $x^2 + 9x + 18$.

We look for two numbers whose sum is the coefficient of $x$ (which is 9) and whose product is the constant term (which is 18).

Let the two numbers be $p$ and $q$. We need $p + q = 9$ and $pq = 18$.

The pairs of factors of 18 are $(1, 18), (2, 9), (3, 6), (-1, -18), (-2, -9), (-3, -6)$.

The pair that adds up to 9 is $(3, 6)$.

So we split the middle term $9x$ as $3x + 6x$.

$x^2 + 9x + 18 = x^2 + 3x + 6x + 18$

Now, we factor by grouping the terms:

$= (x^2 + 3x) + (6x + 18)$

Factor out the common factor from each group:

$= x(x + 3) + 6(x + 3)$

Factor out the common binomial factor $(x + 3)$:

$= (x + 3)(x + 6)$

The factorisation is $(x + 3)(x + 6)$.

(ii) We need to factorise the quadratic polynomial $6x^2 + 7x – 3$.

We look for two numbers whose sum is the coefficient of $x$ (which is 7) and whose product is the product of the coefficient of $x^2$ (6) and the constant term (-3), i.e., $6 \times (-3) = -18$.

Let the two numbers be $p$ and $q$. We need $p + q = 7$ and $pq = -18$.

The pairs of factors of -18 are $(1, -18), (-1, 18), (2, -9), (-2, 9), (3, -6), (-3, 6)$.

The pair that adds up to 7 is $(-2, 9)$.

So we split the middle term $7x$ as $-2x + 9x$.

$6x^2 + 7x – 3 = 6x^2 - 2x + 9x - 3$

Now, we factor by grouping the terms:

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

Factor out the common factor from each group:

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

Factor out the common binomial factor $(3x - 1)$:

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

The factorisation is $(3x - 1)(2x + 3)$.

(iii) We need to factorise the quadratic polynomial $2x^2 – 7x – 15$.

We look for two numbers whose sum is the coefficient of $x$ (which is -7) and whose product is the product of the coefficient of $x^2$ (2) and the constant term (-15), i.e., $2 \times (-15) = -30$.

Let the two numbers be $p$ and $q$. We need $p + q = -7$ and $pq = -30$.

The pairs of factors of -30 are $(1, -30), (-1, 30), (2, -15), (-2, 15), (3, -10), (-3, 10), (5, -6), (-5, 6)$.

The pair that adds up to -7 is $(3, -10)$.

So we split the middle term $-7x$ as $3x - 10x$.

$2x^2 – 7x – 15 = 2x^2 + 3x - 10x - 15$

Now, we factor by grouping the terms:

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

Factor out the common factor from each group. Note that in the second group, we factor out -5.

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

Factor out the common binomial factor $(2x + 3)$:

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

The factorisation is $(2x + 3)(x - 5)$.

(iv) We need to factorise the polynomial $84 – 2r – 2r^2$.

First, let's rearrange the terms in descending powers of $r$ and factor out the common numerical factor, which is 2.

$84 – 2r – 2r^2 = -2r^2 - 2r + 84$

Factor out -2:

$= -2(r^2 + r - 42)$

Now we need to factorise the quadratic expression inside the bracket, $r^2 + r - 42$.

We look for two numbers whose sum is the coefficient of $r$ (which is 1) and whose product is the constant term (which is -42).

Let the two numbers be $p$ and $q$. We need $p + q = 1$ and $pq = -42$.

The pairs of factors of -42 are $(1, -42), (-1, 42), (2, -21), (-2, 21), (3, -14), (-3, 14), (6, -7), (-6, 7)$.

The pair that adds up to 1 is $(-6, 7)$.

So we split the middle term $r$ as $-6r + 7r$.

$r^2 + r - 42 = r^2 - 6r + 7r - 42$

Now, we factor by grouping the terms:

$= (r^2 - 6r) + (7r - 42)$

Factor out the common factor from each group:

$= r(r - 6) + 7(r - 6)$

Factor out the common binomial factor $(r - 6)$:

$= (r - 6)(r + 7)$

Substitute this back into the original expression:

$84 – 2r – 2r^2 = -2(r - 6)(r + 7)$

The factorisation is $-2(r - 6)(r + 7)$.

To factorise these polynomials, we can use the Factor Theorem. We test possible rational roots using the Rational Root Theorem (divisors of the constant term divided by divisors of the leading coefficient). If for a value 'a', the polynomial evaluates to zero, then $(x - a)$ is a factor.

(i) Let $p(x) = 2x^3 – 3x^2 – 17x + 30$.

Possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$ and $\pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}$.

Let's test $x = 2$:

$p(2) = 2(2)^3 - 3(2)^2 - 17(2) + 30$

$p(2) = 2(8) - 3(4) - 34 + 30$

$p(2) = 16 - 12 - 34 + 30$

$p(2) = 4 - 34 + 30 = -30 + 30 = 0$.

Since $p(2) = 0$, $(x - 2)$ is a factor of $p(x)$.

Now, we divide $p(x)$ by $(x - 2)$ to find the other factors. Using synthetic division:

$\begin{array}{r|rrrr} 2 & 2 & -3 & -17 & 30 \\ & & 4 & 2 & -30 \\ \hline & 2 & 1 & -15 & 0 \\ \end{array}$

The quotient is $2x^2 + x - 15$.

Now, factorise the quadratic $2x^2 + x - 15$. We look for two numbers that multiply to $2 \times (-15) = -30$ and add to 1. The numbers are 6 and -5.

$2x^2 + x - 15 = 2x^2 + 6x - 5x - 15$

$= (2x^2 + 6x) + (-5x - 15)$

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

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

Therefore, the factorisation of $2x^3 – 3x^2 – 17x + 30$ is $(x - 2)(x + 3)(2x - 5)$.

(ii) Let $p(x) = x^3 – 6x^2 + 11x – 6$.

Possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$.

Let's test $x = 1$:

$p(1) = (1)^3 - 6(1)^2 + 11(1) - 6$

$p(1) = 1 - 6 + 11 - 6 = 12 - 12 = 0$.

Since $p(1) = 0$, $(x - 1)$ is a factor of $p(x)$.

Now, divide $p(x)$ by $(x - 1)$. Using synthetic division:

$\begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array}$

The quotient is $x^2 - 5x + 6$.

Now, factorise the quadratic $x^2 - 5x + 6$. We look for two numbers that multiply to 6 and add to -5. The numbers are -2 and -3.

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

Therefore, the factorisation of $x^3 – 6x^2 + 11x – 6$ is $(x - 1)(x - 2)(x - 3)$.

(iii) Let $p(x) = x^3 + x^2 – 4x – 4$.

We can try factoring by grouping first.

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

Factor out common terms from each group:

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

Factor out the common binomial factor $(x + 1)$:

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

The quadratic factor $x^2 - 4$ is a difference of squares ($a^2 - b^2 = (a - b)(a + b)$).

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

Therefore, the factorisation of $x^3 + x^2 – 4x – 4$ is $(x + 1)(x - 2)(x + 2)$.

(iv) Let $p(x) = 3x^3 – x^2 – 3x + 1$.

We can try factoring by grouping first.

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

Factor out common terms from each group:

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

Factor out the common binomial factor $(3x - 1)$:

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

The quadratic factor $x^2 - 1$ is a difference of squares ($a^2 - b^2 = (a - b)(a + b)$).

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

Therefore, the factorisation of $3x^3 – x^2 – 3x + 1$ is $(3x - 1)(x - 1)(x + 1)$.

Solution (i):

We need to evaluate $103^3$ using a suitable identity.

We can write $103$ as $100 + 3$. So, $103^3 = (100+3)^3$.

We use the identity: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Here, $a = 100$ and $b = 3$.

$(100+3)^3 = (100)^3 + 3(100)^2(3) + 3(100)(3)^2 + (3)^3$

$= 1000000 + 3(10000)(3) + 3(100)(9) + 27$

$= 1000000 + 90000 + 2700 + 27$

$= 1092727$

Thus, $103^3 = 1092727$.

Solution (ii):

We need to evaluate $101 \times 102$ using a suitable identity.

We can write $101$ as $100 + 1$ and $102$ as $100 + 2$.

So, $101 \times 102 = (100+1)(100+2)$.

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

Here, $x = 100$, $a = 1$, and $b = 2$.

$(100+1)(100+2) = (100)^2 + (1+2)(100) + (1)(2)$

$= 10000 + (3)(100) + 2$

$= 10000 + 300 + 2$

$= 10302$

Thus, $101 \times 102 = 10302$.

Solution (iii):

We need to evaluate $999^2$ using a suitable identity.

We can write $999$ as $1000 - 1$. So, $999^2 = (1000-1)^2$.

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

Here, $a = 1000$ and $b = 1$.

$(1000-1)^2 = (1000)^2 - 2(1000)(1) + (1)^2$

$= 1000000 - 2000 + 1$

$= 998000 + 1$

$= 998001$

Thus, $999^2 = 998001$.

Solution (i):

We are asked to factorise $4x^2 + 20x + 25$.

We observe that the expression is in the form of a perfect square trinomial, $a^2 + 2ab + b^2$.

We can write $4x^2$ as $(2x)^2$ and $25$ as $(5)^2$.

So, we have $(2x)^2 + 20x + (5)^2$.

Let's check the middle term: $2 \times (2x) \times (5) = 20x$. This matches the given expression.

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

Here, $a = 2x$ and $b = 5$.

Therefore, $4x^2 + 20x + 25 = (2x+5)^2$.

Solution (ii):

We are asked to factorise $9y^2 – 66yz + 121z^2$.

We observe that the expression is in the form of a perfect square trinomial, $a^2 - 2ab + b^2$.

We can write $9y^2$ as $(3y)^2$ and $121z^2$ as $(11z)^2$.

So, we have $(3y)^2 - 66yz + (11z)^2$.

Let's check the middle term: $2 \times (3y) \times (11z) = 66yz$. This matches the middle term with the correct sign.

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

Here, $a = 3y$ and $b = 11z$.

Therefore, $9y^2 – 66yz + 121z^2 = (3y-11z)^2$.

Solution (iii):

We are asked to factorise $\left( 2x + \frac{1}{3} \right)^{2}-\left( x - \frac{1}{2} \right)^{2}$.

This expression is in the form $A^2 - B^2$, where $A = \left( 2x + \frac{1}{3} \right)$ and $B = \left( x - \frac{1}{2} \right)$.

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

First, let's find $A+B$:

$A+B = \left( 2x + \frac{1}{3} \right) + \left( x - \frac{1}{2} \right)$

$A+B = 2x + \frac{1}{3} + x - \frac{1}{2}$

$A+B = (2x + x) + \left( \frac{1}{3} - \frac{1}{2} \right)$

$A+B = 3x + \left( \frac{2}{6} - \frac{3}{6} \right)$

$A+B = 3x - \frac{1}{6}$

Next, let's find $A-B$:

$A-B = \left( 2x + \frac{1}{3} \right) - \left( x - \frac{1}{2} \right)$

$A-B = 2x + \frac{1}{3} - x + \frac{1}{2}$

$A-B = (2x - x) + \left( \frac{1}{3} + \frac{1}{2} \right)$

$A-B = x + \left( \frac{2}{6} + \frac{3}{6} \right)$

$A-B = x + \frac{5}{6}$

Now, substitute $A+B$ and $A-B$ back into the identity $(A+B)(A-B)$:

$\left( 2x + \frac{1}{3} \right)^{2}-\left( x - \frac{1}{2} \right)^{2} = \left( 3x - \frac{1}{6} \right)\left( x + \frac{5}{6} \right)$

Solution (i):

We are asked to factorise $9x^2 – 12x + 3$.

First, we look for a common factor among the terms. The coefficients are 9, -12, and 3. The greatest common divisor is 3.

So, we can take out 3 as a common factor:

$9x^2 – 12x + 3 = 3(3x^2 – 4x + 1)$

Now, we need to factorise the quadratic expression $3x^2 – 4x + 1$. We can use the method of splitting the middle term.

We need to find two numbers whose product is the product of the coefficient of $x^2$ and the constant term, i.e., $3 \times 1 = 3$, and whose sum is the coefficient of the middle term, which is -4.

The two numbers are -3 and -1, because $(-3) \times (-1) = 3$ and $(-3) + (-1) = -4$.

We split the middle term $-4x$ into $-3x - x$:

$3x^2 – 4x + 1 = 3x^2 - 3x - x + 1$

Now, we group the terms and factor by grouping:

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

$3x(x - 1) - 1(x - 1)$

Now, take the common factor $(x-1)$:

$(x - 1)(3x - 1)$

Combining the common factor we initially took out, the factorization of $9x^2 – 12x + 3$ is:

$3(x - 1)(3x - 1)$

Solution (ii):

We are asked to factorise $9x^2 – 12x + 4$.

We observe that this expression might be a perfect square trinomial of the form $a^2 - 2ab + b^2$.

Let's examine the first and the last terms:

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

$4 = (2)^2$

Now, let's check if the middle term $-12x$ matches $-2ab$ with $a=3x$ and $b=2$:

$-2ab = -2(3x)(2) = -12x$

This matches the middle term of the given expression.

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

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

Therefore, $9x^2 – 12x + 4 = (3x - 2)^2$.

Solution (i):

We need to expand $(4a – b + 2c)^2$.

We use the identity: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Here, $x = 4a$, $y = -b$, and $z = 2c$.

Substituting these values into the identity:

$(4a – b + 2c)^2 = (4a)^2 + (-b)^2 + (2c)^2 + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)$

$= 16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac$

Thus, the expansion is $16a^2 + b^2 + 4c^2 - 8ab - 4bc + 16ac$.

Solution (ii):

We need to expand $(3a – 5b – c)^2$.

We use the identity: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

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

Substituting these values into the identity:

$(3a – 5b – c)^2 = (3a)^2 + (-5b)^2 + (-c)^2 + 2(3a)(-5b) + 2(-5b)(-c) + 2(-c)(3a)$

$= 9a^2 + 25b^2 + c^2 - 30ab + 10bc - 6ac$

Thus, the expansion is $9a^2 + 25b^2 + c^2 - 30ab + 10bc - 6ac$.

Solution (iii):

We need to expand $(– x + 2y – 3z)^2$.

We use the identity: $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Here, $x = -x$, $y = 2y$, and $z = -3z$.

Substituting these values into the identity:

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

$= x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6xz$

Thus, the expansion is $x^2 + 4y^2 + 9z^2 - 4xy - 12yz + 6xz$.

Solution (i):

We are asked to factorise $9x^2 + 4y^2 + 16z^2 + 12xy – 16yz – 24xz$.

We observe that this expression is in the form $x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.

We can write the terms as squares:

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

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

$16z^2 = (4z)^2$ or $(-4z)^2$

Let's examine the cross terms to determine the signs:

$12xy = 2(3x)(2y)$. This term is positive, so $3x$ and $2y$ have the same sign.

$-16yz = 2(2y)(-4z)$. This term is negative, so $2y$ and $-4z$ have opposite signs.

$-24xz = 2(3x)(-4z)$. This term is negative, so $3x$ and $-4z$ have opposite signs.

From the above, if we take the positive values for the square roots of the first terms (e.g., $3x$ and $2y$), then to make $12xy$ positive, both must be positive. To make $-16yz$ negative when $2y$ is positive, $-4z$ must be negative, meaning $z$ is negative. To make $-24xz$ negative when $3x$ is positive, $-4z$ must be negative, meaning $z$ is negative.

This suggests we use $x=3x$, $y=2y$, and $z=-4z$.

Using the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Here, $a = 3x$, $b = 2y$, $c = -4z$.

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

$= 9x^2 + 4y^2 + 16z^2 + 12xy - 16yz - 24xz$

This matches the given expression.

Thus, the factorization is $(3x + 2y - 4z)^2$.

Solution (ii):

We are asked to factorise $25x^2 + 16y^2 + 4z^2 – 40xy + 16yz – 20xz$.

We observe that this expression is in the form $x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.

We can write the terms as squares:

$25x^2 = (5x)^2$ or $(-5x)^2$

$16y^2 = (4y)^2$ or $(-4y)^2$

$4z^2 = (2z)^2$ or $(-2z)^2$

Let's examine the cross terms to determine the signs:

$-40xy$: This term is negative, so $5x$ and $4y$ have opposite signs.

$16yz$: This term is positive, so $4y$ and $2z$ have the same sign.

$-20xz$: This term is negative, so $5x$ and $2z$ have opposite signs.

Let's assume $x = 5x$. Then from $-40xy$, $y$ must be negative, so $y = -4y$. From $16yz$, since $y$ is negative, $z$ must also be negative, so $z = -2z$. Let's check with $-20xz$: $2(5x)(-2z) = -20xz$. This combination works.

So we use $x=5x$, $y=-4y$, and $z=-2z$.

Using the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Here, $a = 5x$, $b = -4y$, $c = -2z$.

$(5x - 4y - 2z)^2 = (5x)^2 + (-4y)^2 + (-2z)^2 + 2(5x)(-4y) + 2(-4y)(-2z) + 2(-2z)(5x)$

$= 25x^2 + 16y^2 + 4z^2 - 40xy + 16yz - 20xz$

This matches the given expression.

Thus, the factorization is $(5x - 4y - 2z)^2$.

Solution (iii):

We are asked to factorise $16x^2 + 4y^2 + 9z^2 – 16xy – 12yz + 24xz$.

We observe that this expression is in the form $x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.

We can write the terms as squares:

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

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

$9z^2 = (3z)^2$ or $(-3z)^2$

Let's examine the cross terms to determine the signs:

$-16xy$: This term is negative, so $4x$ and $2y$ have opposite signs.

$-12yz$: This term is negative, so $2y$ and $3z$ have opposite signs.

$24xz$: This term is positive, so $4x$ and $3z$ have the same sign.

Let's assume $x = 4x$. Then from $-16xy$, $y$ must be negative, so $y = -2y$. From $-12yz$, since $y$ is negative, $z$ must be positive, so $z = 3z$. Let's check with $24xz$: $2(4x)(3z) = 24xz$. This combination works.

So we use $x=4x$, $y=-2y$, and $z=3z$.

Using the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

Here, $a = 4x$, $b = -2y$, $c = 3z$.

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

$= 16x^2 + 4y^2 + 9z^2 - 16xy - 12yz + 24xz$

This matches the given expression.

Thus, the factorization is $(4x - 2y + 3z)^2$.

Given:

$a + b + c = 9$

$ab + bc + ca = 26$

To Find:

$a^2 + b^2 + c^2$

Solution:

We use the identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

We are given $a+b+c = 9$ and $ab+bc+ca = 26$.

Substitute these values into the identity:

$(9)^2 = a^2 + b^2 + c^2 + 2(26)$

Calculate the square of 9 and the product of 2 and 26:

$81 = a^2 + b^2 + c^2 + 52$

Now, we need to isolate $a^2 + b^2 + c^2$. Subtract 52 from both sides of the equation:

$81 - 52 = a^2 + b^2 + c^2$

Calculate the difference:

$29 = a^2 + b^2 + c^2$

Therefore, $a^2 + b^2 + c^2 = 29$.

Solution (i):

We need to expand $(3a – 2b)^3$.

We use the identity for the cube of a binomial difference: $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$

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

Substituting these values into the identity:

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

$= 27a^3 - 3(9a^2)(2b) + 3(3a)(4b^2) - 8b^3$

$= 27a^3 - 54a^2b + 36ab^2 - 8b^3$

Thus, the expansion is $27a^3 - 54a^2b + 36ab^2 - 8b^3$.

Solution (ii):

We need to expand $\left( \frac{1}{x} + \frac{y}{3} \right)^{3}$.

We use the identity for the cube of a binomial sum: $(p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3$

Here, $p = \frac{1}{x}$ and $q = \frac{y}{3}$.

Substituting these values into the identity:

$\left( \frac{1}{x} + \frac{y}{3} \right)^{3} = \left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2\left(\frac{y}{3}\right) + 3\left(\frac{1}{x}\right)\left(\frac{y}{3}\right)^2 + \left(\frac{y}{3}\right)^3$

$= \frac{1}{x^3} + 3\left(\frac{1}{x^2}\right)\left(\frac{y}{3}\right) + 3\left(\frac{1}{x}\right)\left(\frac{y^2}{9}\right) + \frac{y^3}{27}$

$= \frac{1}{x^3} + \frac{3y}{3x^2} + \frac{3y^2}{9x} + \frac{y^3}{27}$

$= \frac{1}{x^3} + \frac{y}{x^2} + \frac{y^2}{3x} + \frac{y^3}{27}$

Thus, the expansion is $\frac{1}{x^3} + \frac{y}{x^2} + \frac{y^2}{3x} + \frac{y^3}{27}$.

Solution (iii):

We need to expand $\left( 4 - \frac{1}{3x} \right)^{3}$.

We use the identity for the cube of a binomial difference: $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$

Here, $a = 4$ and $b = \frac{1}{3x}$.

Substituting these values into the identity:

$\left( 4 - \frac{1}{3x} \right)^{3} = (4)^3 - 3(4)^2\left(\frac{1}{3x}\right) + 3(4)\left(\frac{1}{3x}\right)^2 - \left(\frac{1}{3x}\right)^3$

$= 64 - 3(16)\left(\frac{1}{3x}\right) + 12\left(\frac{1}{9x^2}\right) - \frac{1}{27x^3}$

$= 64 - \frac{48}{3x} + \frac{12}{9x^2} - \frac{1}{27x^3}$

$= 64 - \frac{16}{x} + \frac{4}{3x^2} - \frac{1}{27x^3}$

Thus, the expansion is $64 - \frac{16}{x} + \frac{4}{3x^2} - \frac{1}{27x^3}$.

Solution (i):

We are asked to factorise $1 – 64a^3 – 12a + 48a^2$.

Let's rearrange the terms: $1 + 48a^2 - 12a - 64a^3$.

We observe that this expression involves cubic terms and constants, which suggests using the identity for the cube of a binomial difference: $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$.

We can write $1$ as $1^3$ and $64a^3$ as $(4a)^3$. The term $-64a^3$ can be written as $(-4a)^3$.

Let's consider if the expression is of the form $(1 - 4a)^3$.

Using the identity with $x=1$ and $y=4a$:

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

$= 1 - 3(1)(4a) + 3(1)(16a^2) - 64a^3$

$= 1 - 12a + 48a^2 - 64a^3$

Rearranging the terms, we get $1 - 64a^3 - 12a + 48a^2$, which matches the given expression.

Therefore, the factorization is $(1 - 4a)^3$.

Solution (ii):

We are asked to factorise $8p^3 + \frac{12}{5}p^2 + \frac{6}{25}p + \frac{1}{125}$.

We observe that this expression involves cubic terms and constants, which suggests using the identity for the cube of a binomial sum: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

We can write the first term $8p^3$ as $(2p)^3$. So, we can take $a = 2p$.

We can write the last term $\frac{1}{125}$ as $\left(\frac{1}{5}\right)^3$. So, we can take $b = \frac{1}{5}$.

Now, let's check if the middle terms match the expansion of $(2p + \frac{1}{5})^3$:

$3a^2b = 3(2p)^2\left(\frac{1}{5}\right) = 3(4p^2)\left(\frac{1}{5}\right) = \frac{12}{5}p^2$. This matches the second term.

$3ab^2 = 3(2p)\left(\frac{1}{5}\right)^2 = 3(2p)\left(\frac{1}{25}\right) = \frac{6}{25}p$. This matches the third term.

Since all terms match the expansion of $(a+b)^3$ with $a=2p$ and $b=\frac{1}{5}$, the given expression is a perfect cube.

Therefore, the factorization is $\left(2p + \frac{1}{5}\right)^3$.

Solution (i):

We need to find the product $\left( \frac{x}{2} + 2y \right)\left( \frac{x^2}{4} - xy + 4y^2 \right)$.

We observe that this expression is in the form $(a+b)(a^2 - ab + b^2)$, which is the expansion of the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

Let $a = \frac{x}{2}$ and $b = 2y$.

Then, $a^2 = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$, $b^2 = (2y)^2 = 4y^2$, and $ab = \left(\frac{x}{2}\right)(2y) = xy$.

The second factor is $\frac{x^2}{4} - xy + 4y^2$, which is $a^2 - ab + b^2$.

So, the product is $a^3 + b^3$ with $a = \frac{x}{2}$ and $b = 2y$.

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

$= \frac{x^3}{8} + 8y^3$

Thus, the product is $\frac{x^3}{8} + 8y^3$.

Solution (ii):

We need to find the product $(x^2 – 1) (x^4 + x^2 + 1)$.

We observe that this expression is in the form $(a-b)(a^2 + ab + b^2)$, which is the expansion of the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Let $a = x^2$ and $b = 1$.

Then, $a^2 = (x^2)^2 = x^4$, $b^2 = 1^2 = 1$, and $ab = (x^2)(1) = x^2$.

The second factor is $x^4 + x^2 + 1$, which is $a^2 + ab + b^2$.

So, the product is $a^3 - b^3$ with $a = x^2$ and $b = 1$.

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

$= x^{6} - 1$

Thus, the product is $x^6 - 1$.

Solution (i):

We are asked to factorise $1 + 64x^3$.

We observe that this expression is a sum of two cubes. We can write $1$ as $1^3$ and $64x^3$ as $(4x)^3$.

So, the expression is in the form $a^3 + b^3$, where $a = 1$ and $b = 4x$.

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

Substitute $a = 1$ and $b = 4x$ into the identity:

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

$= (1 + 4x)(1 - 4x + 16x^2)$

Thus, the factorization is $(1 + 4x)(1 - 4x + 16x^2)$.

Solution (ii):

We are asked to factorise $a^3 – 2\sqrt{2}b^3$.

We observe that this expression is a difference of two cubes. We can write $a^3$ as $(a)^3$.

For the second term, we notice that $2\sqrt{2} = (\sqrt{2})^2 \times \sqrt{2} = (\sqrt{2})^3$.

So, $2\sqrt{2}b^3 = (\sqrt{2})^3 b^3 = (\sqrt{2}b)^3$.

The expression is in the form $x^3 - y^3$, where $x = a$ and $y = \sqrt{2}b$.

We use the identity: $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

Substitute $x = a$ and $y = \sqrt{2}b$ into the identity:

$a^3 - (\sqrt{2}b)^3 = (a - \sqrt{2}b)(a^2 + a(\sqrt{2}b) + (\sqrt{2}b)^2)$

$= (a - \sqrt{2}b)(a^2 + \sqrt{2}ab + 2b^2)$

Thus, the factorization is $(a - \sqrt{2}b)(a^2 + \sqrt{2}ab + 2b^2)$.

We need to find the product of $(2x – y + 3z)$ and $(4x^2 + y^2 + 9z^2 + 2xy + 3yz – 6xz)$.

We observe that this product is in the form $(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Let $a = 2x$, $b = -y$, and $c = 3z$.

Let's check if the second factor matches $a^2 + b^2 + c^2 - ab - bc - ca$ with these values:

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

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

$c^2 = (3z)^2 = 9z^2$

$-ab = -(2x)(-y) = 2xy$

$-bc = -(-y)(3z) = 3yz$

$-ca = -(3z)(2x) = -6xz$

The second factor is $4x^2 + y^2 + 9z^2 + 2xy + 3yz - 6xz$, which exactly matches $a^2 + b^2 + c^2 - ab - bc - ca$ with $a=2x$, $b=-y$, and $c=3z$.

The first factor is $2x - y + 3z$, which is $a+b+c$.

We use the identity: $(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc$

Substituting $a = 2x$, $b = -y$, and $c = 3z$ into the right side of the identity:

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

$= 8x^3 - y^3 + 27z^3 - 3(-6xyz)$

$= 8x^3 - y^3 + 27z^3 + 18xyz$

Thus, the product is $8x^3 - y^3 + 27z^3 + 18xyz$.

Solution (i):

We are asked to factorise $a^3 – 8b^3 – 64c^3 – 24abc$.

We can rewrite the terms as cubes:

$a^3 = (a)^3$

$-8b^3 = (-2b)^3$

$-64c^3 = (-4c)^3$

The expression can be written as $(a)^3 + (-2b)^3 + (-4c)^3 - 24abc$.

This expression is in the form $x^3 + y^3 + z^3 - 3xyz$, where $x = a$, $y = -2b$, and $z = -4c$.

Let's check the $-3xyz$ term:

$-3xyz = -3(a)(-2b)(-4c) = -3(8abc) = -24abc$. This matches the given term.

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

Substitute $x = a$, $y = -2b$, and $z = -4c$ into the identity:

$(a + (-2b) + (-4c))((a)^2 + (-2b)^2 + (-4c)^2 - (a)(-2b) - (-2b)(-4c) - (-4c)(a))$

$= (a - 2b - 4c)(a^2 + 4b^2 + 16c^2 + 2ab - 8bc + 4ca)$

Thus, the factorization is $(a - 2b - 4c)(a^2 + 4b^2 + 16c^2 + 2ab - 8bc + 4ca)$.

Solution (ii):

We are asked to factorise $2\sqrt{2}a^3 + 8b^3 – 27c^3 + 18\sqrt{2}abc$.

We can rewrite the terms as cubes:

$2\sqrt{2}a^3 = (\sqrt{2})^3 a^3 = (\sqrt{2}a)^3$

$8b^3 = (2b)^3$

$-27c^3 = (-3c)^3$

The expression can be written as $(\sqrt{2}a)^3 + (2b)^3 + (-3c)^3 + 18\sqrt{2}abc$.

This expression is in the form $x^3 + y^3 + z^3 - 3xyz$, where $x = \sqrt{2}a$, $y = 2b$, and $z = -3c$.

Let's check the $-3xyz$ term:

$-3xyz = -3(\sqrt{2}a)(2b)(-3c) = -3(\sqrt{2}a)(-6bc) = 18\sqrt{2}abc$. This matches the given term.

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

Substitute $x = \sqrt{2}a$, $y = 2b$, and $z = -3c$ into the identity:

$(\sqrt{2}a + 2b + (-3c))((\sqrt{2}a)^2 + (2b)^2 + (-3c)^2 - (\sqrt{2}a)(2b) - (2b)(-3c) - (-3c)(\sqrt{2}a))$

$= (\sqrt{2}a + 2b - 3c)(2a^2 + 4b^2 + 9c^2 - 2\sqrt{2}ab + 6bc + 3\sqrt{2}ca)$

Thus, the factorization is $(\sqrt{2}a + 2b - 3c)(2a^2 + 4b^2 + 9c^2 - 2\sqrt{2}ab + 6bc + 3\sqrt{2}ca)$.

Solution (i):

We need to find the value of $\left( \frac{1}{2} \right)^{3}+\left( \frac{1}{3} \right)^{3}-\left( \frac{5}{6} \right)^{3}$ without calculating the cubes.

We can rewrite the expression as $\left( \frac{1}{2} \right)^{3}+\left( \frac{1}{3} \right)^{3}+\left( -\frac{5}{6} \right)^{3}$.

Let $a = \frac{1}{2}$, $b = \frac{1}{3}$, and $c = -\frac{5}{6}$.

Let's check the sum $a+b+c$:

$a+b+c = \frac{1}{2} + \frac{1}{3} + \left(-\frac{5}{6}\right) = \frac{3}{6} + \frac{2}{6} - \frac{5}{6} = \frac{3+2-5}{6} = \frac{0}{6} = 0$

Since $a+b+c = 0$, we can use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity, the value of the expression is $3abc$:

Value $= 3 \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(-\frac{5}{6}\right)$

Value $= \cancel{3} \times \frac{1}{2} \times \frac{1}{\cancel{3}} \times \left(-\frac{5}{6}\right)$

Value $= \frac{1}{2} \times 1 \times \left(-\frac{5}{6}\right) = -\frac{5}{12}$

Thus, $\left( \frac{1}{2} \right)^{3}+\left( \frac{1}{3} \right)^{3}-\left( \frac{5}{6} \right)^{3} = -\frac{5}{12}$.

Solution (ii):

We need to find the value of $(0.2)^3 – (0.3)^3 + (0.1)^3$ without calculating the cubes.

We can rewrite the expression as $(0.2)^3 + (-0.3)^3 + (0.1)^3$.

Let $a = 0.2$, $b = -0.3$, and $c = 0.1$.

Let's check the sum $a+b+c$:

$a+b+c = 0.2 + (-0.3) + 0.1 = 0.2 - 0.3 + 0.1 = 0.3 - 0.3 = 0$

Since $a+b+c = 0$, we can use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity, the value of the expression is $3abc$:

Value $= 3 \times (0.2) \times (-0.3) \times (0.1)$

Value $= 3 \times (0.02 \times -0.3)$

Value $= 3 \times (-0.006)$

Value $= -0.018$

Thus, $(0.2)^3 – (0.3)^3 + (0.1)^3 = -0.018$.

We are asked to factorise $(x – 2y)^3 + (2y – 3z)^3 + (3z – x)^3$ without finding the cubes.

Let $a = x – 2y$, $b = 2y – 3z$, and $c = 3z – x$.

The given expression is in the form $a^3 + b^3 + c^3$.

Let's find the sum of $a$, $b$, and $c$:

$a+b+c = (x – 2y) + (2y – 3z) + (3z – x)$

$a+b+c = x - 2y + 2y - 3z + 3z - x$

$a+b+c = (x-x) + (-2y+2y) + (-3z+3z)$

$a+b+c = 0 + 0 + 0 = 0$

Since $a+b+c = 0$, we can use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity with $a = x – 2y$, $b = 2y – 3z$, and $c = 3z – x$, the factorization of the given expression is $3abc$:

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

Thus, the factorization is $3(x – 2y)(2y – 3z)(3z – x)$.

Solution (i):

We need to find the value of $x^3 + y^3 – 12xy + 64$, when $x + y = – 4$.

The given expression is $x^3 + y^3 – 12xy + 64$.

We can rewrite the expression as $x^3 + y^3 + 4^3 - 12xy$.

This expression is in the form $a^3 + b^3 + c^3 - 3abc$, where $a=x$, $b=y$, and $c=4$.

Let's check the term $-3abc$:

$-3abc = -3(x)(y)(4) = -12xy$. This matches the term in the given expression.

We are given the condition $x + y = -4$.

Let's check the sum $a+b+c$:

$a+b+c = x + y + 4$

Substitute the given condition $x+y = -4$:

$a+b+c = (-4) + 4 = 0$

We use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 - 3abc = 0$.

In our case, $a=x$, $b=y$, $c=4$. Since $a+b+c=0$, the value of the expression $a^3 + b^3 + c^3 - 3abc$ is 0.

Therefore, the value of $x^3 + y^3 + 4^3 - 3(x)(y)(4)$ is 0.

So, $x^3 + y^3 – 12xy + 64 = 0$.

Solution (ii):

We need to find the value of $x^3 – 8y^3 – 36xy – 216$, when $x = 2y + 6$.

The given condition is $x = 2y + 6$. We can rearrange this as $x - 2y - 6 = 0$.

The given expression is $x^3 – 8y^3 – 36xy – 216$.

We can rewrite the expression by grouping terms and considering signs for potential cube roots:

$x^3 + (-8y^3) + (-216) - 36xy$

We can write the first three terms as cubes:

$x^3 = (x)^3$

$-8y^3 = (-2y)^3$

$-216 = (-6)^3$

So the expression is $(x)^3 + (-2y)^3 + (-6)^3 - 36xy$.

This expression is in the form $a^3 + b^3 + c^3 - 3abc$, where $a=x$, $b=-2y$, and $c=-6$.

Let's check the term $-3abc$:

$-3abc = -3(x)(-2y)(-6) = -3(12xy) = -36xy$. This matches the term in the given expression.

Now let's check the sum $a+b+c$:

$a+b+c = x + (-2y) + (-6) = x - 2y - 6$

From the given condition $x = 2y + 6$, we have $x - 2y - 6 = 0$.

So, $a+b+c = 0$.

We use the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 - 3abc = 0$.

In our case, $a=x$, $b=-2y$, $c=-6$. Since $a+b+c=0$, the value of the expression $a^3 + b^3 + c^3 - 3abc$ is 0.

Therefore, the value of $(x)^3 + (-2y)^3 + (-6)^3 - 3(x)(-2y)(-6)$ is 0.

So, $x^3 – 8y^3 – 36xy – 216 = 0$.

The area of the rectangle is given by the expression $4a^2 + 4a – 3$.

The area of a rectangle is the product of its length and breadth. To find the possible expressions for the length and breadth, we need to factorise the given quadratic expression.

We will factorise the expression $4a^2 + 4a – 3$ by splitting the middle term.

We look for two numbers whose product is $(4) \times (-3) = -12$ and whose sum is the coefficient of the middle term, which is 4.

The two numbers are 6 and -2, since $6 \times (-2) = -12$ and $6 + (-2) = 4$.

Now, we split the middle term $4a$ as $6a - 2a$:

$4a^2 + 4a – 3 = 4a^2 + 6a - 2a – 3$

Group the terms:

$= (4a^2 + 6a) + (-2a – 3)$

Factor out common factors from each group:

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

Factor out the common binomial factor $(2a+3)$:

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

The factorization of the area is $(2a + 3)(2a - 1)$.

Since Area = Length $\times$ Breadth, the possible expressions for the length and breadth are the two factors.

Possible expression for Length = $2a + 3$

Possible expression for Breadth = $2a - 1$

(Alternatively, Length = $2a - 1$ and Breadth = $2a + 3$. We assume the dimension with the potentially larger value is the length, but without knowing the value of 'a', either expression can be length or breadth).

Given:

$x + y = 12$

$xy = 27$

To Find:

The value of $x^3 + y^3$.

Solution:

We use the identity for the sum of cubes: $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$

We are given the values of $(x+y)$ and $xy$.

Substitute the given values into the identity:

$x^3 + y^3 = (12)^3 - 3(27)(12)$

First, calculate $(12)^3$:

$(12)^3 = 12 \times 12 \times 12 = 144 \times 12$

$(12)^3 = 1728$

Next, calculate $3(27)(12)$:

$3 \times 27 = 81$

$3(27)(12) = 81 \times 12$

$81 \times 12 = 972$

Now, substitute these values back into the equation for $x^3 + y^3$:

$x^3 + y^3 = 1728 - 972$

$x^3 + y^3 = 756$

Thus, the value of $x^3 + y^3$ is 756.

Given:

Polynomials: $P(z) = az^3 + 4z^2 + 3z – 4$ and $Q(z) = z^3 – 4z + a$.

When divided by $z – 3$, both polynomials leave the same remainder.

To Find:

The value of $a$.

Solution:

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

In this case, the divisor is $z-3$, so $c=3$.

Let $R_1$ be the remainder when $P(z)$ is divided by $z-3$.

$R_1 = P(3)$

$P(3) = a(3)^3 + 4(3)^2 + 3(3) - 4$

$P(3) = a(27) + 4(9) + 9 - 4$

$P(3) = 27a + 36 + 9 - 4$

$R_1 = 27a + 41$

Let $R_2$ be the remainder when $Q(z)$ is divided by $z-3$.

$R_2 = Q(3)$

$Q(3) = (3)^3 - 4(3) + a$

$Q(3) = 27 - 12 + a$

$R_2 = 15 + a$

We are given that the remainders are the same, so $R_1 = R_2$.

$27a + 41 = 15 + a$

Now, we solve this linear equation for $a$.

Subtract $a$ from both sides:

$27a - a + 41 = 15$

$26a + 41 = 15$

Subtract 41 from both sides:

$26a = 15 - 41$

$26a = -26$

Divide both sides by 26:

$a = \frac{-26}{26}$

$a = -1$

Thus, the value of $a$ is -1.

Given:

Polynomial: $p(x) = x^4 – 2x^3 + 3x^2 – ax + 3a – 7$.

When $p(x)$ is divided by $x + 1$, the remainder is 19.

To Find:

The value of $a$.

The remainder when $p(x)$ is divided by $x + 2$.

Solution:

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

For the first part, the divisor is $x+1$. This can be written as $x - (-1)$. So, $c=-1$. The remainder is given as 19.

Using the Remainder Theorem, the remainder is $p(-1)$.

Now, substitute $x = -1$ into the polynomial $p(x) = x^4 – 2x^3 + 3x^2 – ax + 3a – 7$:

$p(-1) = (-1)^4 – 2(-1)^3 + 3(-1)^2 – a(-1) + 3a – 7$

Calculate the terms:

$(-1)^4 = 1$

$(-1)^3 = -1$

$(-1)^2 = 1$

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

$p(-1) = 1 – 2(-1) + 3(1) – a(-1) + 3a – 7$

$p(-1) = 1 + 2 + 3 + a + 3a – 7$

Combine the constant terms and the terms with $a$:

$p(-1) = (1 + 2 + 3 – 7) + (a + 3a)$

$p(-1) = (6 – 7) + 4a$

$p(-1) = -1 + 4a$

From equation (i), we have $p(-1) = 19$. So,

$4a - 1 = 19$

Now, solve for $a$:

$4a = 19 + 1$

$4a = 20$

$a = \frac{20}{4}$

$a = 5$

The value of $a$ is 5.

Now we need to find the remainder when $p(x)$ is divided by $x+2$.

First, substitute the value $a=5$ into the polynomial $p(x)$:

$p(x) = x^4 – 2x^3 + 3x^2 – (5)x + 3(5) – 7$

$p(x) = x^4 – 2x^3 + 3x^2 – 5x + 15 – 7$

$p(x) = x^4 – 2x^3 + 3x^2 – 5x + 8$

For the second part, the divisor is $x+2$, which means $c=-2$. The remainder is $p(-2)$.

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

$p(-2) = (-2)^4 – 2(-2)^3 + 3(-2)^2 – 5(-2) + 8$

Calculate the terms:

$(-2)^4 = 16$

$(-2)^3 = -8$

$(-2)^2 = 4$

Substitute these values back into $p(-2)$:

$p(-2) = 16 – 2(-8) + 3(4) – 5(-2) + 8$

$p(-2) = 16 + 16 + 12 + 10 + 8$

Sum the terms:

$p(-2) = 32 + 12 + 10 + 8$

$p(-2) = 44 + 10 + 8$

$p(-2) = 54 + 8$

$p(-2) = 62$

The remainder when $p(x)$ is divided by $x+2$ is 62.

Given:

The polynomial is $P(x) = px^2 + 5x + r$.

$(x – 2)$ and $(x – \frac{1}{2})$ are factors of the polynomial $P(x)$.

To Show:

$p = r$

Solution:

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

Since $(x – 2)$ is a factor of $P(x)$, we have $P(2) = 0$.

Substitute $x = 2$ into the polynomial $P(x) = px^2 + 5x + r$:

$P(2) = p(2)^2 + 5(2) + r$

$P(2) = 4p + 10 + r$

Since $P(2) = 0$, we have:

This can be rearranged as:

Since $(x – \frac{1}{2})$ is a factor of $P(x)$, we have $P(\frac{1}{2}) = 0$.

Substitute $x = \frac{1}{2}$ into the polynomial $P(x) = px^2 + 5x + r$:

$P\left(\frac{1}{2}\right) = p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r$

$P\left(\frac{1}{2}\right) = p\left(\frac{1}{4}\right) + \frac{5}{2} + r$

$P\left(\frac{1}{2}\right) = \frac{p}{4} + \frac{5}{2} + r$

Since $P(\frac{1}{2}) = 0$, we have:

To eliminate the fractions, multiply the entire equation by 4:

$4 \times \left(\frac{p}{4}\right) + 4 \times \left(\frac{5}{2}\right) + 4 \times r = 4 \times 0$

$p + 10 + 4r = 0$

This can be rearranged as:

Now we have a system of two linear equations:

Equation (1): $4p + r = -10$

Equation (2): $p + 4r = -10$

From Equation (1), we can express $r$ in terms of $p$:

Substitute the expression for $r$ from Equation (3) into Equation (2):

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

$p - 40 - 16p = -10$

Combine like terms:

$-15p - 40 = -10$

Add 40 to both sides:

$-15p = -10 + 40$

$-15p = 30$

Divide both sides by -15:

$p = \frac{30}{-15}$

$p = -2$

Now substitute the value of $p = -2$ into Equation (3) to find the value of $r$:

$r = -10 - 4(-2)$

$r = -10 + 8$

$r = -2$

We found that $p = -2$ and $r = -2$.

Therefore, $p = r$.

Hence, it is shown that $p = r$.

Given:

Polynomial: $P(x) = 2x^4 – 5x^3 + 2x^2 – x + 2$.

Divisor: $D(x) = x^2 – 3x + 2$.

To Prove:

$P(x)$ is divisible by $D(x)$ without actual division.

Proof:

For $P(x)$ to be divisible by $D(x) = x^2 – 3x + 2$, $D(x)$ must be a factor of $P(x)$.

First, we factorise the divisor $D(x) = x^2 – 3x + 2$.

We find two numbers whose product is $(1) \times (2) = 2$ and whose sum is $-3$. These numbers are $-1$ and $-2$.

$x^2 – 3x + 2 = x^2 - x - 2x + 2$

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

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

So, the divisor $x^2 – 3x + 2$ is the product of two linear factors: $(x-1)$ and $(x-2)$.

For $P(x)$ to be divisible by $(x-1)(x-2)$, $P(x)$ must be divisible by each of these factors individually.

According to the Factor Theorem, a polynomial $P(x)$ is divisible by $(x-c)$ if and only if $P(c) = 0$.

We need to check if $P(1) = 0$ and $P(2) = 0$.

Evaluate $P(1)$ by substituting $x=1$ into $P(x) = 2x^4 – 5x^3 + 2x^2 – x + 2$:

$P(1) = 2(1)^4 – 5(1)^3 + 2(1)^2 – (1) + 2$

$P(1) = 2(1) – 5(1) + 2(1) – 1 + 2$

$P(1) = 2 – 5 + 2 – 1 + 2$

$P(1) = (2+2+2) - (5+1) = 6 - 6 = 0$

Since $P(1) = 0$, by the Factor Theorem, $(x-1)$ is a factor of $P(x)$.

Evaluate $P(2)$ by substituting $x=2$ into $P(x) = 2x^4 – 5x^3 + 2x^2 – x + 2$:

$P(2) = 2(2)^4 – 5(2)^3 + 2(2)^2 – (2) + 2$

$P(2) = 2(16) – 5(8) + 2(4) – 2 + 2$

$P(2) = 32 – 40 + 8 – 2 + 2$

$P(2) = (32+8+2) - (40+2) = 42 - 42 = 0$

Since $P(2) = 0$, by the Factor Theorem, $(x-2)$ is a factor of $P(x)$.

Since $P(x)$ is divisible by both $(x-1)$ and $(x-2)$, and these are distinct linear factors, $P(x)$ is divisible by their product $(x-1)(x-2) = x^2 – 3x + 2$.

Therefore, $2x^4 – 5x^3 + 2x^2 – x + 2$ is divisible by $x^2 – 3x + 2$.

We need to simplify $(2x – 5y)^3 – (2x + 5y)^3$.

Let $a = 2x$ and $b = 5y$. The expression can be written as $(a-b)^3 - (a+b)^3$.

We use the identities for the cube of a binomial:

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

Now, subtract the second expansion from the first:

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

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

Combine like terms:

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

$= 0 - 6a^2b + 0 - 2b^3$

$= -6a^2b - 2b^3$

Now, substitute back $a = 2x$ and $b = 5y$ into the simplified expression:

$-6a^2b - 2b^3 = -6(2x)^2(5y) - 2(5y)^3$

$= -6(4x^2)(5y) - 2(125y^3)$

$= -6(20x^2y) - 250y^3$

$= -120x^2y - 250y^3$

Thus, the simplified expression is $-120x^2y - 250y^3$.

We need to multiply the expression $(x^2 + 4y^2 + z^2 + 2xy + xz – 2yz)$ by $(– z + x – 2y)$.

Let's rearrange the terms in the second factor: $(x - 2y - z)$.

The first factor is $x^2 + 4y^2 + z^2 + 2xy + xz – 2yz$.

Let's compare the given expressions with the identity for the sum of cubes minus 3 times the product: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

Let's try to identify $a$, $b$, and $c$ from the second factor $(x - 2y - z)$.

Let $a = x$, $b = -2y$, and $c = -z$.

Then $a+b+c = x + (-2y) + (-z) = x - 2y - z$, which matches the second given factor.

Now, let's calculate the terms in the second part of the identity $(a^2 + b^2 + c^2 - ab - bc - ca)$ using $a=x$, $b=-2y$, and $c=-z$:

$a^2 = (x)^2 = x^2$

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

$c^2 = (-z)^2 = z^2$

$-ab = -(x)(-2y) = 2xy$

$-bc = -(-2y)(-z) = -(2yz) = -2yz$

$-ca = -(-z)(x) = xz$

So, $a^2 + b^2 + c^2 - ab - bc - ca = x^2 + 4y^2 + z^2 + 2xy - 2yz + xz$.

This matches the first given factor, $x^2 + 4y^2 + z^2 + 2xy + xz – 2yz$.

Therefore, the product of the two given expressions is equal to $a^3 + b^3 + c^3 - 3abc$, with $a=x$, $b=-2y$, and $c=-z$.

Let's calculate $a^3$, $b^3$, $c^3$, and $3abc$:

$a^3 = (x)^3 = x^3$

$b^3 = (-2y)^3 = -8y^3$

$c^3 = (-z)^3 = -z^3$

$3abc = 3(x)(-2y)(-z) = 6xyz$

The product is $a^3 + b^3 + c^3 - 3abc$:

Product $= x^3 + (-8y^3) + (-z^3) - 6xyz$

Product $= x^3 - 8y^3 - z^3 - 6xyz$

Thus, the product is $x^3 - 8y^3 - z^3 - 6xyz$.

Given:

$a, b, c$ are all non–zero.

$a + b + c = 0$

To Prove:

$\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3$

Proof:

Consider the left-hand side (LHS) of the equation we need to prove:

LHS $= \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$

To add these fractions, we find a common denominator, which is the least common multiple of $bc, ca, ab$. The LCM is $abc$.

We rewrite each term with the common denominator $abc$:

$\frac{a^2}{bc} = \frac{a^2 \times a}{bc \times a} = \frac{a^3}{abc}$

$\frac{b^2}{ca} = \frac{b^2 \times b}{ca \times b} = \frac{b^3}{abc}$

$\frac{c^2}{ab} = \frac{c^2 \times c}{ab \times c} = \frac{c^3}{abc}$

Now, substitute these back into the expression for the LHS:

LHS $= \frac{a^3}{abc} + \frac{b^3}{abc} + \frac{c^3}{abc}$

LHS $= \frac{a^3 + b^3 + c^3}{abc}$

We are given that $a + b + c = 0$.

We know the identity: If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Using this identity, we can substitute $a^3 + b^3 + c^3$ in the numerator of the LHS expression:

LHS $= \frac{3abc}{abc}$

Since $a, b, c$ are all non-zero, their product $abc$ is also non-zero. Thus, we can cancel $abc$ from the numerator and the denominator.

LHS $= 3$

This is equal to the right-hand side (RHS) of the equation.

LHS $= 3 =$ RHS

Therefore, it is proven that $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3$ when $a+b+c=0$ and $a, b, c$ are non-zero.

Given:

$a + b + c = 5$

$ab + bc + ca = 10$

To Prove:

$a^3 + b^3 + c^3 – 3abc = – 25$

Proof:

We use the identity: $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$

We are given $a+b+c = 5$ and $ab+bc+ca = 10$. However, we need the value of $a^2 + b^2 + c^2$.

We use another identity: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

Substitute the given values into this identity:

$(5)^2 = a^2 + b^2 + c^2 + 2(10)$

$25 = a^2 + b^2 + c^2 + 20$

Now, solve for $a^2 + b^2 + c^2$:

$a^2 + b^2 + c^2 = 25 - 20$

$a^2 + b^2 + c^2 = 5$

Now we have all the required components for the main identity: $a+b+c = 5$, $a^2+b^2+c^2 = 5$, and $ab+bc+ca = 10$.

Substitute these values into the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$:

$a^3 + b^3 + c^3 - 3abc = (5)(5 - 10)$

$a^3 + b^3 + c^3 - 3abc = (5)(-5)$

$a^3 + b^3 + c^3 - 3abc = -25$

Thus, it is proven that $a^3 + b^3 + c^3 – 3abc = – 25$.

To Prove:

$(a + b + c)^3 – a^3 – b^3 – c^3 = 3(a + b ) (b + c) (c + a)$

Proof:

Consider the Left Hand Side (LHS) of the identity:

LHS $= (a + b + c)^3 – a^3 – b^3 – c^3$

We can group the terms and use the identity for the difference of cubes, $X^3 - Y^3 = (X-Y)(X^2+XY+Y^2)$.

Let $X = a+b+c$ and $Y = c$. Then the first part of the LHS is $X^3 - Y^3$.

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

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

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

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

Now substitute this back into the LHS expression:

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

We know the identity for the sum of cubes, $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

Substitute this into the LHS expression:

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

Factor out the common term $(a+b)$:

LHS $= (a+b)[ (a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2) - (a^2 - ab + b^2) ]$

Remove the parenthesis inside the square bracket, changing the signs of the terms from the second group:

LHS $= (a+b)[ a^2 + 2ab + b^2 + 3ac + 3bc + 3c^2 - a^2 + ab - b^2 ]$

Combine like terms inside the square bracket ($a^2$ with $-a^2$, $b^2$ with $-b^2$, $2ab$ with $ab$):

LHS $= (a+b)[ (a^2 - a^2) + (b^2 - b^2) + (2ab + ab) + 3ac + 3bc + 3c^2 ]$

LHS $= (a+b)[ 0 + 0 + 3ab + 3ac + 3bc + 3c^2 ]$

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

Factor out 3 from the terms inside the second parenthesis:

LHS $= (a+b)[ 3(ab + ac + bc + c^2) ]$

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

Now, factor the expression inside the second parenthesis by grouping the terms:

$ab + ac + bc + c^2 = (ab + ac) + (bc + c^2)$

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

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

Substitute this back into the expression for the LHS:

LHS $= 3(a+b)(a+c)(b+c)$

Rearrange the factors in the order given in the RHS:

LHS $= 3(a+b)(b+c)(c+a)$

This is the Right Hand Side (RHS) of the identity.

LHS $=$ RHS

Hence, the identity $(a + b + c)^3 – a^3 – b^3 – c^3 = 3(a + b ) (b + c) (c + a)$ is proven.