Chapter 2 Polymonials

Given:

The polynomials: (i) $x^5 - x^4 + 3$, (ii) $2 - y^2 - y^3 + y^8$, (iii) 2.

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To Find:

The degree of each polynomial.

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Solution:

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

(i) $x^5 - x^4 + 3$

The terms in this polynomial are $x^5$, $-x^4$, and 3.

The power of $x$ in the term $x^5$ is 5.

The power of $x$ in the term $-x^4$ is 4.

The term 3 can be written as $3x^0$, so the power of $x$ is 0.

The highest power of the variable $x$ in the polynomial is 5.

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

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(ii) $2 - y^2 - y^3 + y^8$

The terms in this polynomial are 2, $-y^2$, $-y^3$, and $y^8$.

The power of $y$ in the term 2 ($2y^0$) is 0.

The power of $y$ in the term $-y^2$ is 2.

The power of $y$ in the term $-y^3$ is 3.

The power of $y$ in the term $y^8$ is 8.

The highest power of the variable $y$ in the polynomial is 8.

Therefore, the degree of the polynomial $2 - y^2 - y^3 + y^8$ is 8.

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(iii) 2

This is a constant polynomial. A constant polynomial can be written in the form $ax^0$, where $a$ is a non-zero constant. For example, $2 = 2x^0$.

The power of the variable (which is not explicitly written) is considered to be 0.

Therefore, the degree of the constant polynomial 2 is 0.

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Results:

(i) The degree of $x^5 - x^4 + 3$ is 5.

(ii) The degree of $2 - y^2 - y^3 + y^8$ is 8.

(iii) The degree of 2 is 0.

A polynomial in one variable is an expression that consists of terms, where each term is the product of a constant and a variable raised to a non-negative integer power. The variable is the same in all terms.

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(i) $4x^2 - 3x + 7$

This expression contains only one variable, $x$. The powers of $x$ are 2, 1 (in $3x$), and 0 (in 7, which is $7x^0$). All these powers (2, 1, 0) are non-negative integers.

Therefore, $4x^2 - 3x + 7$ is a polynomial in one variable.

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(ii) $y^2 + \sqrt{2}$

This expression contains only one variable, $y$. The powers of $y$ are 2 (in $y^2$) and 0 (in $\sqrt{2}$, which is $\sqrt{2}y^0$). Both powers (2, 0) are non-negative integers. The coefficient $\sqrt{2}$ is a real number, which is allowed in a polynomial.

Therefore, $y^2 + \sqrt{2}$ is a polynomial in one variable.

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

This expression can be written as $3t^{\frac{1}{2}} + \sqrt{2}t^1$. It contains only one variable, $t$.

However, the power of $t$ in the first term ($3\sqrt{t}$) is $\frac{1}{2}$. The exponent $\frac{1}{2}$ is not a non-negative integer.

Therefore, $3\sqrt{t} + t\sqrt{2}$ is not a polynomial because the power of the variable $t$ is not a non-negative integer.

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(iv) $y + \frac{2}{y}$

This expression can be written as $y^1 + 2y^{-1}$. It contains only one variable, $y$.

However, the power of $y$ in the second term ($\frac{2}{y}$) is -1. The exponent -1 is not a non-negative integer.

Therefore, $y + \frac{2}{y}$ is not a polynomial because the power of the variable $y$ is not a non-negative integer.

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(v) $x^{10} + y^3 + t^{50}$

This expression contains three different variables: $x$, $y$, and $t$. The powers of the variables (10, 3, 50) are all non-negative integers.

However, the expression involves more than one variable.

Therefore, $x^{10} + y^3 + t^{50}$ is not a polynomial in one variable. It is a polynomial in three variables.

The coefficient of $x^2$ in a polynomial is the constant that is multiplied by the $x^2$ term.

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(i) $2 + x^2 + x$

We can rewrite the polynomial in descending powers of $x$: $x^2 + x + 2$.

The term with $x^2$ is $x^2$. This can be written as $1 \cdot x^2$.

The coefficient of $x^2$ is 1.

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

We can rewrite the polynomial in descending powers of $x$: $x^3 - x^2 + 2$.

The term with $x^2$ is $-x^2$. This can be written as $-1 \cdot x^2$.

The coefficient of $x^2$ is -1.

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

The term with $x^2$ is $\frac{\pi}{2} x^2$.

The coefficient of $x^2$ is $\frac{\pi}{2}$.

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(iv) $\sqrt{2} x - 1$

This polynomial can be written as $0 \cdot x^2 + \sqrt{2}x - 1$.

The term with $x^2$ is not present, which means its coefficient is 0.

The coefficient of $x^2$ is 0.

Given:

Request for examples of specific types of polynomials with given degrees.

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To Provide:

An example of a binomial of degree 35.

An example of a monomial of degree 100.

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Solution:

Recall the definitions:

A binomial is a polynomial with exactly two terms.

A monomial is a polynomial with exactly one term.

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

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Example of a binomial of degree 35:

A binomial has two terms. For the degree to be 35, the highest power of the variable in one of the terms must be 35.

Let the variable be $x$. We need a term with $x^{35}$ and another term. The other term can be any non-zero constant or a term with $x$ raised to a power less than 35 (which is a non-negative integer).

A simple example is a term with $x^{35}$ and a constant term:

$x^{35} + 5$

Other examples include $2x^{35} - 1$, $y^{35} + y^{10}$, $-3z^{35} + 7z$, etc. All these have two terms and the highest power is 35.

One example of a binomial of degree 35 is $x^{35} + 5$.

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Example of a monomial of degree 100:

A monomial has one term. For the degree to be 100, the power of the variable in that single term must be 100.

Let the variable be $x$. We need a term with $x^{100}$. This term can have a non-zero coefficient.

A simple example is:

$x^{100}$

Other examples include $7y^{100}$, $-\frac{1}{2}z^{100}$, etc.

One example of a monomial of degree 100 is $x^{100}$.

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Results:

An example of a binomial of degree 35 is $x^{35} + 5$.

An example of a monomial of degree 100 is $x^{100}$.

Given:

The polynomials: (i) $5x^3 + 4x^2 + 7x$, (ii) $4 - y^2$, (iii) $5t - \sqrt{7}$, (iv) 3.

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To Find:

The degree of each polynomial.

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Solution:

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

(i) $5x^3 + 4x^2 + 7x$

The terms are $5x^3$, $4x^2$, and $7x$.

The power of $x$ in $5x^3$ is 3.

The power of $x$ in $4x^2$ is 2.

The power of $x$ in $7x$ (which is $7x^1$) is 1.

The highest power of $x$ is 3.

The degree of $5x^3 + 4x^2 + 7x$ is 3.

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(ii) $4 - y^2$

The terms are 4 and $-y^2$.

The power of $y$ in 4 (which is $4y^0$) is 0.

The power of $y$ in $-y^2$ is 2.

The highest power of $y$ is 2.

The degree of $4 - y^2$ is 2.

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(iii) $5t - \sqrt{7}$

The terms are $5t$ and $-\sqrt{7}$.

The power of $t$ in $5t$ (which is $5t^1$) is 1.

The power of $t$ in $-\sqrt{7}$ (which is $-\sqrt{7}t^0$) is 0.

The highest power of $t$ is 1.

The degree of $5t - \sqrt{7}$ is 1.

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(iv) 3

This is a constant polynomial. It can be written as $3x^0$.

The power of the variable is 0.

The degree of the constant polynomial 3 is 0.

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Results:

(i) The degree of $5x^3 + 4x^2 + 7x$ is 3.

(ii) The degree of $4 - y^2$ is 2.

(iii) The degree of $5t - \sqrt{7}$ is 1.

(iv) The degree of 3 is 0.

We classify polynomials based on their degree:

A polynomial of degree 1 is called a linear polynomial.

A polynomial of degree 2 is called a quadratic polynomial.

A polynomial of degree 3 is called a cubic polynomial.

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(i) $x^2 + x$

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

This is a quadratic polynomial.

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(ii) $x - x^3$

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

This is a cubic polynomial.

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(iii) $y + y^2 + 4$

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

This is a quadratic polynomial.

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(iv) $1 + x$

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

This is a linear polynomial.

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(v) $3t$

The highest power of the variable $t$ is 1 (since $3t = 3t^1$). The degree is 1.

This is a linear polynomial.

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(vi) $r^2$

The highest power of the variable $r$ is 2. The degree is 2.

This is a quadratic polynomial.

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(vii) $7x^3$

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

This is a cubic polynomial.

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Results:

(i) Quadratic

(ii) Cubic

(iii) Quadratic

(iv) Linear

(v) Linear

(vi) Quadratic

(vii) Cubic

Given:

Polynomials and values of variables at which to evaluate them.

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To Find:

The value of each polynomial at the given variable value.

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Solution:

To find the value of a polynomial at a specific value of the variable, we substitute that value into the polynomial expression and calculate the result.

(i) $p(x) = 5x^2 - 3x + 7$ at $x = 1$

Substitute $x=1$ into the polynomial:

$p(1) = 5(1)^2 - 3(1) + 7$

Calculate the powers and products:

$p(1) = 5(1) - 3 + 7$

$p(1) = 5 - 3 + 7$

Perform the addition and subtraction:

$p(1) = 2 + 7$

$p(1) = 9$

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(ii) $q(y) = 3y^3 - 4y + \sqrt{11}$ at $y = 2$

Substitute $y=2$ into the polynomial:

$q(2) = 3(2)^3 - 4(2) + \sqrt{11}$

Calculate the powers and products:

$q(2) = 3(8) - 8 + \sqrt{11}$

$q(2) = 24 - 8 + \sqrt{11}$

Perform the subtraction:

$q(2) = 16 + \sqrt{11}$

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(iii) $p(t) = 4t^4 + 5t^3 - t^2 + 6$ at $t = a$

Substitute $t=a$ into the polynomial:

$p(a) = 4(a)^4 + 5(a)^3 - (a)^2 + 6$

$p(a) = 4a^4 + 5a^3 - a^2 + 6$

The value of the polynomial at $t=a$ is expressed in terms of $a$.

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Results:

(i) The value of $p(x)$ at $x = 1$ is 9.

(ii) The value of $q(y)$ at $y = 2$ is $16 + \sqrt{11}$.

(iii) The value of $p(t)$ at $t = a$ is $4a^4 + 5a^3 - a^2 + 6$.

Given:

The polynomial $p(x) = x + 2$ and the numbers -2 and 2.

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To Check:

Determine if -2 and 2 are zeroes of the polynomial $p(x) = x + 2$.

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Solution:

A number $a$ is a zero of a polynomial $p(x)$ if the value of the polynomial at $x=a$ is 0, i.e., $p(a) = 0$.

Checking for $x = -2$:

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

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

$p(-2) = 0$

Since $p(-2) = 0$, -2 is a zero of the polynomial $x + 2$.

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Checking for $x = 2$:

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

$p(2) = (2) + 2$

$p(2) = 4$

Since $p(2) = 4 \neq 0$, 2 is not a zero of the polynomial $x + 2$.

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Results:

Yes, -2 is a zero of the polynomial $x + 2$.

No, 2 is not a zero of the polynomial $x + 2$.

Given:

The polynomial $p(x) = 2x + 1$.

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To Find:

A zero of the polynomial $p(x) = 2x + 1$.

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Solution:

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

To find a zero of the polynomial $p(x) = 2x + 1$, we set the polynomial equal to zero and solve for $x$.

$p(x) = 0$

Subtract 1 from both sides of the equation:

$2x = -1$

Divide both sides by 2:

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

So, $x = -\frac{1}{2}$ is a value for which $p(x) = 0$.

We can check this by substituting $x = -\frac{1}{2}$ into the polynomial:

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

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

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

Since $p\left(-\frac{1}{2}\right) = 0$, $-\frac{1}{2}$ is indeed a zero of the polynomial $p(x) = 2x + 1$.

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Result:

A zero of the polynomial $p(x) = 2x + 1$ is $-\frac{1}{2}$.

Given:

The polynomial $p(x) = x^2 - 2x$ and the numbers 2 and 0.

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To Verify:

Determine if 2 and 0 are zeroes of the polynomial $p(x) = x^2 - 2x$.

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Solution:

A number $a$ is a zero of a polynomial $p(x)$ if $p(a) = 0$.

Verifying for $x = 2$:

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

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

Calculate the terms:

$p(2) = 4 - 4$

$p(2) = 0$

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

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Verifying for $x = 0$:

Substitute $x = 0$ into the polynomial $p(x) = x^2 - 2x$:

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

Calculate the terms:

$p(0) = 0 - 0$

$p(0) = 0$

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

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Result:

Yes, both 2 and 0 are zeroes of the polynomial $x^2 - 2x$.

Given:

The polynomial $p(x) = 5x - 4x^2 + 3$ and the values of $x$: (i) 0, (ii) -1, (iii) 2.

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To Find:

The value of the polynomial at each given value of $x$.

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Solution:

To find the value of the polynomial at a specific value of $x$, we substitute that value into the polynomial expression and evaluate.

Let the polynomial be $p(x) = 5x - 4x^2 + 3$. We can rewrite it in standard form as $p(x) = -4x^2 + 5x + 3$.

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(i) at $x = 0$

Substitute $x=0$ into the polynomial:

$p(0) = 5(0) - 4(0)^2 + 3$

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

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

$p(0) = 3$

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(ii) at $x = -1$

Substitute $x=-1$ into the polynomial:

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

Calculate the terms:

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

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

Perform the addition and subtraction:

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

$p(-1) = -6$

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(iii) at $x = 2$

Substitute $x=2$ into the polynomial:

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

Calculate the terms:

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

$p(2) = 10 - 16 + 3$

Perform the addition and subtraction:

$p(2) = -6 + 3$

$p(2) = -3$

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Results:

(i) The value of the polynomial at $x = 0$ is 3.

(ii) The value of the polynomial at $x = -1$ is -6.

(iii) The value of the polynomial at $x = 2$ is -3.

Given:

Four polynomials.

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To Find:

The value of each polynomial at the variable values 0, 1, and 2.

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Solution:

We substitute the given values (0, 1, and 2) into each polynomial and evaluate.

(i) $p(y) = y^2 - y + 1$

For $y=0$:

$p(0) = (0)^2 - (0) + 1 = 0 - 0 + 1 = 1$

For $y=1$:

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

For $y=2$:

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

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(ii) $p(t) = 2 + t + 2t^2 - t^3$

For $t=0$:

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

For $t=1$:

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

For $t=2$:

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

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(iii) $p(x) = x^3$

For $x=0$:

$p(0) = (0)^3 = 0$

For $x=1$:

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

For $x=2$:

$p(2) = (2)^3 = 2 \times 2 \times 2 = 8$

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

We can first expand the polynomial using the identity $(a-b)(a+b) = a^2 - b^2$ or evaluate directly.

Expanding: $p(x) = x^2 - 1^2 = x^2 - 1$.

Using $p(x) = x^2 - 1$:

For $x=0$:

$p(0) = (0)^2 - 1 = 0 - 1 = -1$

For $x=1$:

$p(1) = (1)^2 - 1 = 1 - 1 = 0$

For $x=2$:

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

Using $p(x) = (x - 1)(x + 1)$ directly:

For $x=0$:

$p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1$

For $x=1$:

$p(1) = (1 - 1)(1 + 1) = (0)(2) = 0$

For $x=2$:

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

Both methods yield the same results.

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Results:

(i) $p(0)=1$, $p(1)=1$, $p(2)=3$

(ii) $p(0)=2$, $p(1)=4$, $p(2)=4$

(iii) $p(0)=0$, $p(1)=1$, $p(2)=8$

(iv) $p(0)=-1$, $p(1)=0$, $p(2)=3$

Given:

Several polynomials and indicated values of the variable.

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To Verify:

Check if the indicated value(s) are zeroes of the corresponding polynomial.

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Solution:

A value $a$ is a zero of a polynomial $p(x)$ if $p(a) = 0$. We substitute the indicated value(s) into the polynomial and check if the result is 0.

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

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

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

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

Since $p\left(-\frac{1}{3}\right) = 0$, $-\frac{1}{3}$ is a zero of the polynomial.

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(ii) $p(x) = 5x - \pi$, $x = \frac{4}{5}$

Substitute $x = \frac{4}{5}$:

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

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

Since $4 - \pi \neq 0$, $\frac{4}{5}$ is not a zero of the polynomial.

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(iii) $p(x) = x^2 - 1$, $x = 1, -1$

For $x=1$:

$p(1) = (1)^2 - 1 = 1 - 1 = 0$

For $x=-1$:

$p(-1) = (-1)^2 - 1 = 1 - 1 = 0$

Since $p(1) = 0$ and $p(-1) = 0$, both 1 and -1 are zeroes of the polynomial.

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(iv) $p(x) = (x + 1)(x - 2)$, $x = -1, 2$

For $x=-1$:

$p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0$

For $x=2$:

$p(2) = (2 + 1)(2 - 2) = (3)(0) = 0$

Since $p(-1) = 0$ and $p(2) = 0$, both -1 and 2 are zeroes of the polynomial.

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(v) $p(x) = x^2$, $x = 0$

Substitute $x=0$:

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

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

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(vi) $p(x) = lx + m$, $x = -\frac{m}{l}$ (given $l \neq 0$ for the expression to be defined)

Substitute $x = -\frac{m}{l}$:

$p\left(-\frac{m}{l}\right) = l\left(-\frac{m}{l}\right) + m$

$p\left(-\frac{m}{l}\right) = -m + m = 0$

Since $p\left(-\frac{m}{l}\right) = 0$, $-\frac{m}{l}$ is a zero of the polynomial.

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(vii) $p(x) = 3x^2 - 1$, $x = -\frac{1}{\sqrt{3}}$, $\frac{2}{\sqrt{3}}$

For $x = -\frac{1}{\sqrt{3}}$:

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

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

$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(\frac{1}{3}\right) - 1$

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

Since $p\left(-\frac{1}{\sqrt{3}}\right) = 0$, $-\frac{1}{\sqrt{3}}$ is a zero of the polynomial.

For $x = \frac{2}{\sqrt{3}}$:

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

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

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

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

Since $p\left(\frac{2}{\sqrt{3}}\right) = 3 \neq 0$, $\frac{2}{\sqrt{3}}$ is not a zero of the polynomial.

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

Substitute $x = \frac{1}{2}$:

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

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

Since $p\left(\frac{1}{2}\right) = 2 \neq 0$, $\frac{1}{2}$ is not a zero of the polynomial.

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Summary of Results:

(i) Yes, $-\frac{1}{3}$ is a zero.

(ii) No, $\frac{4}{5}$ is not a zero.

(iii) Yes, both 1 and -1 are zeroes.

(iv) Yes, both -1 and 2 are zeroes.

(v) Yes, 0 is a zero.

(vi) Yes, $-\frac{m}{l}$ is a zero.

(vii) Yes, $-\frac{1}{\sqrt{3}}$ is a zero. No, $\frac{2}{\sqrt{3}}$ is not a zero.

(viii) No, $\frac{1}{2}$ is not a zero.

Given:

Several polynomials.

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To Find:

The zero of each polynomial.

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Solution:

To find the zero(s) of a polynomial $p(x)$, we set $p(x) = 0$ and solve for $x$.

(i) $p(x) = x + 5$

Set $p(x) = 0$:

$x + 5 = 0$

Subtract 5 from both sides:

$x = -5$

The zero is -5.

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(ii) $p(x) = x - 5$

Set $p(x) = 0$:

$x - 5 = 0$

Add 5 to both sides:

$x = 5$

The zero is 5.

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(iii) $p(x) = 2x + 5$

Set $p(x) = 0$:

$2x + 5 = 0$

Subtract 5 from both sides:

$2x = -5$

Divide both sides by 2:

$x = -\frac{5}{2}$

The zero is $-\frac{5}{2}$.

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

Set $p(x) = 0$:

$3x - 2 = 0$

Add 2 to both sides:

$3x = 2$

Divide both sides by 3:

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

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

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(v) $p(x) = 3x$

Set $p(x) = 0$:

$3x = 0$

Divide both sides by 3:

$x = \frac{0}{3} = 0$

The zero is 0.

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(vi) $p(x) = ax$, $a \neq 0$

Set $p(x) = 0$:

$ax = 0$

Since $a \neq 0$, we can divide both sides by $a$:

$x = \frac{0}{a} = 0$

The zero is 0.

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(vii) $p(x) = cx + d$, $c \neq 0$, c, d are real numbers.

Set $p(x) = 0$:

$cx + d = 0$

Subtract $d$ from both sides:

$cx = -d$

Since $c \neq 0$, we can divide both sides by $c$:

$x = -\frac{d}{c}$

The zero is $-\frac{d}{c}$.

This shows that a linear polynomial $ax+b$ (where $a \neq 0$) has exactly one zero, which is $-\frac{b}{a}$.

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Results:

(i) The zero of $p(x) = x + 5$ is -5.

(ii) The zero of $p(x) = x - 5$ is 5.

(iii) The zero of $p(x) = 2x + 5$ is $-\frac{5}{2}$.

(iv) The zero of $p(x) = 3x - 2$ is $\frac{2}{3}$.

(v) The zero of $p(x) = 3x$ is 0.

(vi) The zero of $p(x) = ax$ ($a \neq 0$) is 0.

(vii) The zero of $p(x) = cx + d$ ($c \neq 0$) is $-\frac{d}{c}$.

Given:

The divisor polynomial $g(x) = x + 2$ and two dividend polynomials $p(x) = x^3 + 3x^2 + 5x + 6$ and $q(x) = 2x + 4$.

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To Examine:

Check if $x + 2$ is a factor of $p(x)$ and $q(x)$.

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Solution:

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

In this case, the potential factor is $x + 2$. We can write $x + 2$ as $x - (-2)$. So, we need to check if the polynomial evaluates to 0 when $x = -2$.

Checking for $p(x) = x^3 + 3x^2 + 5x + 6$:

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

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

Calculate the terms:

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

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

Perform the addition and subtraction:

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

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

$p(-2) = 0$

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

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Checking for $q(x) = 2x + 4$:

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

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

Calculate the terms:

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

$q(-2) = 0$

Since $q(-2) = 0$, according to the Factor Theorem, $(x - (-2))$, which is $(x + 2)$, is a factor of $2x + 4$.

Alternatively, we can see that $2x + 4 = 2(x + 2)$, which clearly shows that $(x+2)$ is a factor.

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Result:

Yes, $x + 2$ is a factor of $x^3 + 3x^2 + 5x + 6$.

Yes, $x + 2$ is a factor of $2x + 4$.

Given:

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

The fact that $(x - 1)$ is a factor of $p(x)$.

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To Find:

The value of the constant $k$.

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Solution:

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

In this case, the factor is $(x - 1)$, so $a = 1$.

Since $(x - 1)$ is a factor of $p(x)$, we must have $p(1) = 0$.

Substitute $x = 1$ into the polynomial $p(x) = 4x^3 + 3x^2 - 4x + k$:

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

Calculate the terms:

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

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

Simplify the expression:

$p(1) = 7 - 4 + k$

$p(1) = 3 + k$

Since $p(1)$ must be equal to 0 for $(x - 1)$ to be a factor, we set the expression equal to 0:

$3 + k = 0$

Solve for $k$:

$k = -3$

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Result:

The value of $k$ is -3.

Factorisation by splitting the middle term:

We need to factorise the quadratic polynomial $6x^2 + 17x + 5$. This is in the form $ax^2 + bx + c$, where $a=6$, $b=17$, and $c=5$.

In the splitting the middle term method, we need to find two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the middle term ($p+q = b = 17$) and their product is equal to the product of the coefficient of $x^2$ and the constant term ($p \times q = a \times c = 6 \times 5 = 30$).

We need to find $p$ and $q$ such that:

Let's list pairs of factors of 30 and check their sums:

• 1 and 30: $1 + 30 = 31$ (Not 17)
• 2 and 15: $2 + 15 = 17$ (This matches!)

So the two numbers are 2 and 15. We use these numbers to split the middle term $17x$ into $2x + 15x$.

Now, rewrite the polynomial:

$6x^2 + 17x + 5 = 6x^2 + 2x + 15x + 5$

Group the first two terms and the last two terms:

$(6x^2 + 2x) + (15x + 5)$

Factor out the greatest common factor from each group:

From $(6x^2 + 2x)$, the common factor is $2x$. So, $2x(3x + 1)$.

From $(15x + 5)$, the common factor is 5. So, $5(3x + 1)$.

The expression becomes:

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

Now, we can see that $(3x + 1)$ is a common binomial factor in both terms. Factor out $(3x + 1)$:

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

Thus, the factorised form of $6x^2 + 17x + 5$ is $(3x + 1)(2x + 5)$.

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Factorisation by using the Factor Theorem:

Let the polynomial be $P(x) = 6x^2 + 17x + 5$.

According to the Factor Theorem, if $P(a) = 0$ for some value 'a', then $(x-a)$ is a factor of $P(x)$.

For a polynomial of the form $ax^2 + bx + c$, any rational root $\frac{p}{q}$ (in simplest form) must have $p$ as a factor of the constant term ($c=5$) and $q$ as a factor of the leading coefficient ($a=6$).

Factors of the constant term 5 are $\pm 1, \pm 5$.

Factors of the leading coefficient 6 are $\pm 1, \pm 2, \pm 3, \pm 6$.

Possible rational roots $\left(\frac{p}{q}\right)$ are:

$\pm \frac{1}{1}, \pm \frac{5}{1}, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{1}{6}, \pm \frac{5}{6}$.

Let's test some of these possible values for $x$ to see if $P(x) = 0$.

Test $x = -1/3$:

$P\left(-\frac{1}{3}\right) = 6\left(-\frac{1}{3}\right)^2 + 17\left(-\frac{1}{3}\right) + 5$

$P\left(-\frac{1}{3}\right) = 6\left(\frac{1}{9}\right) - \frac{17}{3} + 5$

$P\left(-\frac{1}{3}\right) = \frac{6}{9} - \frac{17}{3} + 5$

Simplify $\frac{6}{9}$ to $\frac{2}{3}$ and write 5 as $\frac{15}{3}$ to get a common denominator:

$P\left(-\frac{1}{3}\right) = \frac{2}{3} - \frac{17}{3} + \frac{15}{3}$

$P\left(-\frac{1}{3}\right) = \frac{2 - 17 + 15}{3} = \frac{-15 + 15}{3} = \frac{0}{3} = 0$

Since $P\left(-\frac{1}{3}\right) = 0$, by the Factor Theorem, $\left(x - \left(-\frac{1}{3}\right)\right) = \left(x + \frac{1}{3}\right)$ is a factor of $P(x)$.

We can write $\left(x + \frac{1}{3}\right) = \frac{3x+1}{3}$. Since $\frac{3x+1}{3}$ is a factor, $(3x+1)$ is also a factor.

Test $x = -5/2$:

$P\left(-\frac{5}{2}\right) = 6\left(-\frac{5}{2}\right)^2 + 17\left(-\frac{5}{2}\right) + 5$

$P\left(-\frac{5}{2}\right) = 6\left(\frac{25}{4}\right) - \frac{85}{2} + 5$

$P\left(-\frac{5}{2}\right) = \frac{150}{4} - \frac{85}{2} + 5$

Simplify $\frac{150}{4}$ to $\frac{75}{2}$ and write 5 as $\frac{10}{2}$ to get a common denominator:

$P\left(-\frac{5}{2}\right) = \frac{75}{2} - \frac{85}{2} + \frac{10}{2}$

$P\left(-\frac{5}{2}\right) = \frac{75 - 85 + 10}{2} = \frac{-10 + 10}{2} = \frac{0}{2} = 0$

Since $P\left(-\frac{5}{2}\right) = 0$, by the Factor Theorem, $\left(x - \left(-\frac{5}{2}\right)\right) = \left(x + \frac{5}{2}\right)$ is a factor of $P(x)$.

We can write $\left(x + \frac{5}{2}\right) = \frac{2x+5}{2}$. Since $\frac{2x+5}{2}$ is a factor, $(2x+5)$ is also a factor.

Since $P(x)$ is a quadratic polynomial (degree 2), it can have at most two linear factors.

We have found two factors: $(3x+1)$ and $(2x+5)$.

So, the polynomial $P(x)$ must be the product of these factors, possibly multiplied by a constant $k$ (the leading coefficient):

$P(x) = k (3x + 1)(2x + 5)$

Expand the right side:

$k (3x \times 2x + 3x \times 5 + 1 \times 2x + 1 \times 5)$

$k (6x^2 + 15x + 2x + 5)$

$k (6x^2 + 17x + 5)$

So, $6x^2 + 17x + 5 = k (6x^2 + 17x + 5)$.

Comparing the coefficient of $x^2$ on both sides, we have $6 = k \times 6$, which gives $k=1$.

Therefore, $P(x) = 1 \times (3x+1)(2x+5) = (3x+1)(2x+5)$.

Thus, $6x^2 + 17x + 5 = (3x+1)(2x+5)$.

Let the given polynomial be $P(y) = y^2 - 5y + 6$.

According to the Factor Theorem, if $P(a) = 0$, then $(y-a)$ is a factor of $P(y)$.

We need to find the values of $y$ for which $P(y) = 0$. These are the roots of the polynomial.

The constant term of the polynomial is 6. The factors of 6 are $\pm 1, \pm 2, \pm 3, \pm 6$.

Let's test some of these values:

For $y=1$:

$P(1) = (1)^2 - 5(1) + 6 = 1 - 5 + 6 = 2$.

Since $P(1) \neq 0$, $(y-1)$ is not a factor.

For $y=-1$:

$P(-1) = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = 12$.

Since $P(-1) \neq 0$, $(y+1)$ is not a factor.

For $y=2$:

$P(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$.

Since $P(2) = 0$, $(y-2)$ is a factor of $P(y)$.

For $y=3$:

$P(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$.

Since $P(3) = 0$, $(y-3)$ is a factor of $P(y)$.

Since $P(y)$ is a quadratic polynomial, it can have at most two linear factors.

We have found two factors, $(y-2)$ and $(y-3)$.

So, $P(y)$ can be written as $k(y-2)(y-3)$ for some constant $k$.

$y^2 - 5y + 6 = k(y^2 - 3y - 2y + 6)$

$y^2 - 5y + 6 = k(y^2 - 5y + 6)$

Comparing the coefficient of $y^2$ on both sides, we get $1 = k \times 1$, which means $k = 1$.

Therefore, $P(y) = 1 \times (y-2)(y-3) = (y-2)(y-3)$.

Thus, the factorisation of $y^2 - 5y + 6$ is $(y-2)(y-3)$.

Let $P(x) = x^3 – 23x^2 + 142x – 120$.

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

The constant term of $P(x)$ is $-120$. The possible rational roots are the divisors of 120, both positive and negative.

The divisors of 120 are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 40, \pm 60, \pm 120$.

Let's test some simple values:

Test $x=1$:

$P(1) = (1)^3 - 23(1)^2 + 142(1) - 120$

$P(1) = 1 - 23 + 142 - 120$

$P(1) = 143 - 143 = 0$

Since $P(1) = 0$, by the Factor Theorem, $(x-1)$ is a factor of $P(x)$.

Now, we divide the polynomial $P(x)$ by $(x-1)$ using polynomial long division to find the other factor(s):

$\begin{array}{r} x^2 - 22x + 120 \\ x-1{\overline{\smash{\big)}\,x^3 - 23x^2 + 142x - 120}} \\ \underline{-~\phantom{(}(x^3 - x^2)\phantom{-bxxxxxxxx)}} \\ 0 - 22x^2 + 142x - 120\phantom{)} \\ \underline{-~\phantom{()}(-22x^2 + 22x)\phantom{-xxx)}} \\ 0 + 120x - 120\phantom{)} \\ \underline{-~\phantom{()}(120x - 120)} \\ 0 + 0\phantom{)} \end{array}$

The quotient is $x^2 - 22x + 120$. So, $P(x) = (x-1)(x^2 - 22x + 120)$.

Now we need to factorise the quadratic polynomial $Q(x) = x^2 - 22x + 120$.

We can do this by splitting the middle term. We look for two numbers whose sum is $-22$ (the coefficient of $x$) and whose product is $1 \times 120 = 120$ (the product of the coefficient of $x^2$ and the constant term).

The numbers are $-10$ and $-12$, since $(-10) + (-12) = -22$ and $(-10) \times (-12) = 120$.

So, we can rewrite the middle term $-22x$ as $-10x - 12x$:

$x^2 - 22x + 120 = x^2 - 10x - 12x + 120$

Group the terms and factor by grouping:

$= (x^2 - 10x) + (-12x + 120)$

$= x(x - 10) - 12(x - 10)$

Factor out the common binomial factor $(x - 10)$:

$= (x - 10)(x - 12)$

So, the factorisation of $x^2 - 22x + 120$ is $(x-10)(x-12)$.

Combining the factors, we get the complete factorisation of $P(x)$:

$x^3 – 23x^2 + 142x – 120 = (x-1)(x-10)(x-12)$.

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

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

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

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

$P(-1) = -1 + 1 - 1 + 1$

$P(-1) = 0$

Since $P(-1) = 0$, $(x+1)$ is a factor of $x^3 + x^2 + x + 1$.

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

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

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

$P(-1) = 1 + (-1) + 1 + (-1) + 1$

$P(-1) = 1 - 1 + 1 - 1 + 1$

$P(-1) = 1$

Since $P(-1) \neq 0$, $(x+1)$ is not a factor of $x^4 + x^3 + x^2 + x + 1$.

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

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

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

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

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

$P(-1) = 1$

Since $P(-1) \neq 0$, $(x+1)$ is not a factor of $x^4 + 3x^3 + 3x^2 + x + 1$.

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(iv) $P(x) = x^3 − x^2 − (2 + \sqrt{2}) x + \sqrt{2}$

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

$P(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2}) (-1) + \sqrt{2}$

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

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

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

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

$P(-1) = 2\sqrt{2}$

Since $P(-1) \neq 0$, $(x+1)$ is not a factor of $x^3 − x^2 − (2 + \sqrt{2}) x + \sqrt{2}$.

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

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(i) $p(x) = 2x^3 + x^2 − 2x −1$, $g(x) = x + 1$

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

$x + 1 = 0$

$x = -1$

Now, evaluate $p(x)$ at $x = -1$:

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

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

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

$p(-1) = 0$

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

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(ii) $p(x) = x^3 + 3x^2 + 3x + 1$, $g(x) = x + 2$

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

$x + 2 = 0$

$x = -2$

Now, evaluate $p(x)$ at $x = -2$:

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

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

$p(-2) = -8 + 12 - 6 + 1$

$p(-2) = 13 - 14$

$p(-2) = -1$

Since $p(-2) \neq 0$, by the Factor Theorem, $g(x) = x+2$ is not a factor of $p(x) = x^3 + 3x^2 + 3x + 1$.

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(iii) $p(x) = x^3 − 4x^2 + x + 6$, $g(x) = x − 3$

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

$x - 3 = 0$

$x = 3$

Now, evaluate $p(x)$ at $x = 3$:

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

$p(3) = 27 - 4(9) + 3 + 6$

$p(3) = 27 - 36 + 3 + 6$

$p(3) = 36 - 36$

$p(3) = 0$

Since $p(3) = 0$, by the Factor Theorem, $g(x) = x-3$ is a factor of $p(x) = x^3 − 4x^2 + x + 6$.

According to the Factor Theorem, if $(x-1)$ is a factor of $p(x)$, then $p(1) = 0$. We will use this condition to find the value of $k$ in each case.

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(i) $p(x) = x^2 + x + k$

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

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

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

$p(1) = 2 + k$

Since $(x-1)$ is a factor, $p(1) = 0$.

$2 + k = 0$

Solving for $k$:

$k = -2$

Thus, the value of $k$ is $-2$.

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(ii) $p(x) = 2x^2 + kx + \sqrt{2}$

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

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

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

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

Since $(x-1)$ is a factor, $p(1) = 0$.

$2 + k + \sqrt{2} = 0$

Solving for $k$:

$k = -2 - \sqrt{2}$

Thus, the value of $k$ is $-2 - \sqrt{2}$.

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(iii) $p(x) = kx^2 − \sqrt{2}x + 1$

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

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

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

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

Since $(x-1)$ is a factor, $p(1) = 0$.

$k - \sqrt{2} + 1 = 0$

Solving for $k$:

$k = \sqrt{2} - 1$

Thus, the value of $k$ is $\sqrt{2} - 1$.

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(iv) $p(x) = kx^2 − 3x + k$

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

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

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

$p(1) = 2k - 3$

Since $(x-1)$ is a factor, $p(1) = 0$.

$2k - 3 = 0$

Solving for $k$:

$2k = 3$

$k = \frac{3}{2}$

Thus, the value of $k$ is $\frac{3}{2}$.

We will factorise each quadratic polynomial by splitting the middle term.

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(i) $12x^2 − 7x + 1$

We need to find two numbers whose sum is $-7$ (coefficient of $x$) and whose product is $12 \times 1 = 12$ (product of coefficient of $x^2$ and constant term).

The two numbers are $-3$ and $-4$, since $(-3) + (-4) = -7$ and $(-3) \times (-4) = 12$.

Now, split the middle term $-7x$ as $-3x - 4x$:

$12x^2 − 7x + 1 = 12x^2 - 3x - 4x + 1$

Group the terms:

$= (12x^2 - 3x) + (-4x + 1)$

Factor out common terms from each group:

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

Factor out the common binomial $(4x - 1)$:

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

Thus, $12x^2 − 7x + 1 = (4x - 1)(3x - 1)$.

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(ii) $2x^2 + 7x + 3$

We need to find two numbers whose sum is $7$ (coefficient of $x$) and whose product is $2 \times 3 = 6$ (product of coefficient of $x^2$ and constant term).

The two numbers are $1$ and $6$, since $1 + 6 = 7$ and $1 \times 6 = 6$.

Now, split the middle term $7x$ as $x + 6x$:

$2x^2 + 7x + 3 = 2x^2 + x + 6x + 3$

Group the terms:

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

Factor out common terms from each group:

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

Factor out the common binomial $(2x + 1)$:

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

Thus, $2x^2 + 7x + 3 = (2x + 1)(x + 3)$.

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(iii) $6x^2 + 5x − 6$

We need to find two numbers whose sum is $5$ (coefficient of $x$) and whose product is $6 \times (-6) = -36$ (product of coefficient of $x^2$ and constant term).

The two numbers are $9$ and $-4$, since $9 + (-4) = 5$ and $9 \times (-4) = -36$.

Now, split the middle term $5x$ as $9x - 4x$:

$6x^2 + 5x − 6 = 6x^2 + 9x - 4x - 6$

Group the terms:

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

Factor out common terms from each group:

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

Factor out the common binomial $(2x + 3)$:

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

Thus, $6x^2 + 5x − 6 = (2x + 3)(3x - 2)$.

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(iv) $3x^2 − x − 4$

We need to find two numbers whose sum is $-1$ (coefficient of $x$) and whose product is $3 \times (-4) = -12$ (product of coefficient of $x^2$ and constant term).

The two numbers are $3$ and $-4$, since $3 + (-4) = -1$ and $3 \times (-4) = -12$.

Now, split the middle term $-x$ as $3x - 4x$:

$3x^2 − x − 4 = 3x^2 + 3x - 4x - 4$

Group the terms:

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

Factor out common terms from each group:

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

Factor out the common binomial $(x + 1)$:

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

Thus, $3x^2 − x − 4 = (x + 1)(3x - 4)$.

We will factorise each cubic polynomial using the Factor Theorem and splitting the middle term for the resulting quadratic factor.

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(i) $x^3 − 2x^2 − x + 2$

Let $P(x) = x^3 − 2x^2 − x + 2$.

We can try factoring by grouping directly:

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

Factor out common terms from each group:

$= x^2(x − 2) − 1(x − 2)$

Factor out the common binomial $(x − 2)$:

$= (x^2 − 1)(x − 2)$

The term $(x^2 - 1)$ is a difference of squares, which can be factorised as $(x-1)(x+1)$.

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

Thus, $x^3 − 2x^2 − x + 2 = (x - 1)(x + 1)(x - 2)$.

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(ii) $x^3 − 3x^2 − 9x − 5$

Let $P(x) = x^3 − 3x^2 − 9x − 5$.

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

The constant term is $-5$. The divisors are $\pm 1, \pm 5$.

Let's test $x = -1$:

$P(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5$

$P(-1) = -1 - 3(1) + 9 - 5$

$P(-1) = -1 - 3 + 9 - 5 = 0$

Since $P(-1) = 0$, $(x - (-1)) = (x+1)$ is a factor of $P(x)$.

Now, we divide $P(x)$ by $(x+1)$ using polynomial long division:

$\begin{array}{r} x^2 - 4x - 5 \\ x+1{\overline{\smash{\big)}\,x^3 - 3x^2 - 9x - 5}} \\ \underline{-~\phantom{(}(x^3 + x^2)\phantom{-bxx)}} \\ 0 - 4x^2 - 9x - 5\phantom{)} \\ \underline{-~\phantom{()}(-4x^2 - 4x)\phantom{-})} \\ 0 - 5x - 5\phantom{)} \\ \underline{-~\phantom{()}(-5x - 5)} \\ 0 + 0\phantom{)} \end{array}$

The quotient is $x^2 - 4x - 5$. So, $P(x) = (x+1)(x^2 - 4x - 5)$.

Now we factorise the quadratic $Q(x) = x^2 - 4x - 5$ by splitting the middle term.

We need two numbers whose sum is $-4$ and product is $1 \times (-5) = -5$. The numbers are $-5$ and $1$.

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

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

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

So, $x^2 - 4x - 5 = (x+1)(x-5)$.

Therefore, $P(x) = (x+1)(x+1)(x-5) = (x+1)^2(x-5)$.

Thus, $x^3 − 3x^2 − 9x − 5 = (x+1)(x+1)(x-5)$.

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(iii) $x^3 + 13x^2 + 32x + 20$

Let $P(x) = x^3 + 13x^2 + 32x + 20$.

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

The constant term is $20$. The divisors include $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$.

Let's test $x = -1$:

$P(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$

$P(-1) = -1 + 13(1) - 32 + 20$

$P(-1) = -1 + 13 - 32 + 20 = 33 - 33 = 0$

Since $P(-1) = 0$, $(x - (-1)) = (x+1)$ is a factor of $P(x)$.

Now, we divide $P(x)$ by $(x+1)$ using polynomial long division:

$\begin{array}{r} x^2 + 12x + 20 \\ x+1{\overline{\smash{\big)}\,x^3 + 13x^2 + 32x + 20}} \\ \underline{-~\phantom{(}(x^3 + x^2)\phantom{-bxx)}} \\ 0 + 12x^2 + 32x + 20\phantom{)} \\ \underline{-~\phantom{()}(12x^2 + 12x)\phantom{)}} \\ 0 + 20x + 20\phantom{)} \\ \underline{-~\phantom{()}(20x + 20)} \\ 0 + 0\phantom{)} \end{array}$

The quotient is $x^2 + 12x + 20$. So, $P(x) = (x+1)(x^2 + 12x + 20)$.

Now we factorise the quadratic $Q(x) = x^2 + 12x + 20$ by splitting the middle term.

We need two numbers whose sum is $12$ and product is $1 \times 20 = 20$. The numbers are $2$ and $10$.

$x^2 + 12x + 20 = x^2 + 2x + 10x + 20$

$= x(x + 2) + 10(x + 2)$

$= (x + 2)(x + 10)$

So, $x^2 + 12x + 20 = (x+2)(x+10)$.

Therefore, $P(x) = (x+1)(x+2)(x+10)$.

Thus, $x^3 + 13x^2 + 32x + 20 = (x+1)(x+2)(x+10)$.

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(iv) $2y^3 + y^2 − 2y − 1$

Let $P(y) = 2y^3 + y^2 − 2y − 1$.

According to the Factor Theorem, if $P(a) = 0$, then $(y-a)$ is a factor.

The constant term is $-1$ and the leading coefficient is $2$. Possible rational roots are $\pm \frac{1}{1}, \pm \frac{1}{2}$. Let's test these values.

Test $y = 1$:

$P(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$. So $(y-1)$ is a factor.

Test $y = -1$:

$P(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = 2(-1) + 1 + 2 - 1 = -2 + 1 + 2 - 1 = 0$. So $(y+1)$ is a factor.

Test $y = 1/2$:

$P(1/2) = 2(1/2)^3 + (1/2)^2 - 2(1/2) - 1 = 2(1/8) + 1/4 - 1 - 1 = 1/4 + 1/4 - 2 = 1/2 - 2 = -3/2 \neq 0$.

Test $y = -1/2$:

$P(-1/2) = 2(-1/2)^3 + (-1/2)^2 - 2(-1/2) - 1 = 2(-1/8) + 1/4 + 1 - 1 = -1/4 + 1/4 + 0 = 0$. So $(y+1/2)$ or $(2y+1)$ is a factor.

Since we found three roots (-1, 1, -1/2) and the polynomial is cubic, the factors are $(y-1)$, $(y+1)$, and $(y+1/2)$.

So $P(y) = k(y-1)(y+1)(y+1/2)$ for some constant $k$.

$2y^3 + y^2 - 2y - 1 = k(y^2 - 1)(y + 1/2)$

$2y^3 + y^2 - 2y - 1 = k(y^3 + y^2/2 - y - 1/2)$

Comparing the coefficient of $y^3$ on both sides, $2 = k \times 1$, so $k=2$.

$P(y) = 2(y-1)(y+1)(y+1/2) = (y-1)(y+1) \times 2(y+1/2) = (y-1)(y+1)(2y+1)$.

Thus, $2y^3 + y^2 − 2y − 1 = (y-1)(y+1)(2y+1)$.

Alternatively, using polynomial long division after finding one factor, say $(y-1)$:

$\begin{array}{r} 2y^2 + 3y + 1 \\ y-1{\overline{\smash{\big)}\,2y^3 + y^2 - 2y - 1}} \\ \underline{-~\phantom{(}(2y^3 - 2y^2)\phantom{-bx)}} \\ 0 + 3y^2 - 2y - 1\phantom{)} \\ \underline{-~\phantom{()}(3y^2 - 3y)\phantom{-)}} \\ 0 + y - 1\phantom{)} \\ \underline{-~\phantom{()}(y - 1)} \\ 0 + 0\phantom{)} \end{array}$

The quotient is $2y^2 + 3y + 1$. So, $P(y) = (y-1)(2y^2 + 3y + 1)$.

Now we factorise the quadratic $Q(y) = 2y^2 + 3y + 1$ by splitting the middle term.

We need two numbers whose sum is $3$ and product is $2 \times 1 = 2$. The numbers are $1$ and $2$.

$2y^2 + 3y + 1 = 2y^2 + y + 2y + 1$

$= y(2y + 1) + 1(2y + 1)$

$= (2y + 1)(y + 1)$

So, $2y^2 + 3y + 1 = (y+1)(2y+1)$.

Therefore, $P(y) = (y-1)(y+1)(2y+1)$.

Thus, $2y^3 + y^2 − 2y − 1 = (y-1)(y+1)(2y+1)$.

We will use appropriate algebraic identities to find the products.

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(i) $(x + 3) (x + 3)$

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

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

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

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

$= x^2 + 6x + 9$

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

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(ii) $(x – 3) (x + 5)$

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

Here, $a = -3$ and $b = 5$.

$(x – 3) (x + 5) = (x + (-3))(x + 5)$

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

$= x^2 + (2)x + (-15)$

$= x^2 + 2x - 15$

Thus, $(x – 3) (x + 5) = x^2 + 2x - 15$.

We need to evaluate $105 \times 106$ without multiplying directly.

We can write $105$ as $(100 + 5)$ and $106$ as $(100 + 6)$.

So, the product is $(100 + 5)(100 + 6)$.

This expression is in the form $(x+a)(x+b)$, where $x = 100$, $a = 5$, and $b = 6$.

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

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

$(100 + 5)(100 + 6) = (100)^2 + (5 + 6)(100) + (5)(6)$

Calculate each term:

$(100)^2 = 100 \times 100 = 10000$

$(5 + 6)(100) = (11)(100) = 1100$

$(5)(6) = 30$

Now, add the terms together:

$10000 + 1100 + 30$

$= 11100 + 30$

$= 11130$

Thus, $105 \times 106 = 11130$.

We will use appropriate algebraic identities to factorise the given expressions.

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(i) $49a^2 + 70ab + 25b^2$

We can rewrite the terms to see if this expression fits the form of a perfect square trinomial, $A^2 + 2AB + B^2 = (A+B)^2$ or $A^2 - 2AB + B^2 = (A-B)^2$.

The first term $49a^2$ can be written as $(7a)^2$.

The last term $25b^2$ can be written as $(5b)^2$.

Let's check the middle term. If $A = 7a$ and $B = 5b$, then $2AB = 2(7a)(5b) = 70ab$.

The given expression matches the form $A^2 + 2AB + B^2$ with $A = 7a$ and $B = 5b$.

So, we can use the identity $A^2 + 2AB + B^2 = (A+B)^2$.

$49a^2 + 70ab + 25b^2 = (7a)^2 + 2(7a)(5b) + (5b)^2$

$= (7a + 5b)^2$

$= (7a + 5b)(7a + 5b)$

Thus, the factorisation is $(7a + 5b)^2$ or $(7a + 5b)(7a + 5b)$.

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(ii) $\frac{25}{4} x^2 - \frac{y^2}{9}$

We can rewrite the terms to see if this expression fits the form of a difference of squares, $A^2 - B^2 = (A-B)(A+B)$.

The first term $\frac{25}{4} x^2$ can be written as $(\frac{5}{2})^2 x^2 = (\frac{5}{2} x)^2$.

The second term $\frac{y^2}{9}$ can be written as $(\frac{y}{3})^2$.

The given expression matches the form $A^2 - B^2$ with $A = \frac{5}{2} x$ and $B = \frac{y}{3}$.

So, we can use the identity $A^2 - B^2 = (A-B)(A+B)$.

$\frac{25}{4} x^2 - \frac{y^2}{9} = (\frac{5}{2} x)^2 - (\frac{y}{3})^2$

$= (\frac{5}{2} x - \frac{y}{3})(\frac{5}{2} x + \frac{y}{3})$

Thus, the factorisation is $(\frac{5}{2} x - \frac{y}{3})(\frac{5}{2} x + \frac{y}{3})$.

We need to write $(3a + 4b + 5c)^2$ in expanded form.

We can use the algebraic identity for the square of a trinomial:

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

In the given expression, we have $x = 3a$, $y = 4b$, and $z = 5c$.

Substitute these values into the identity:

$(3a + 4b + 5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)$

Calculate each term:

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

$(4b)^2 = 4^2 b^2 = 16b^2$

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

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

$2(4b)(5c) = 2 \times 4 \times 5 \times b \times c = 40bc$

$2(5c)(3a) = 2 \times 5 \times 3 \times c \times a = 30ca$

Combine all the terms:

$(3a + 4b + 5c)^2 = 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca$

Thus, the expanded form of $(3a + 4b + 5c)^2$ is $9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca$.

We need to expand $(4a – 2b – 3c)^2$.

We can rewrite this expression as $(4a + (-2b) + (-3c))^2$.

This is in the form of $(x + y + z)^2$, where $x = 4a$, $y = -2b$, and $z = -3c$.

We use the algebraic identity for the square of a trinomial:

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

Substitute the values of $x$, $y$, and $z$ into the identity:

$(4a – 2b – 3c)^2 = (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) + 2(-2b)(-3c) + 2(-3c)(4a)$

Calculate each term:

$(4a)^2 = 16a^2$

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

$(-3c)^2 = 9c^2$

$2(4a)(-2b) = 2 \times 4 \times (-2) \times a \times b = -16ab$

$2(-2b)(-3c) = 2 \times (-2) \times (-3) \times b \times c = 12bc$

$2(-3c)(4a) = 2 \times (-3) \times 4 \times c \times a = -24ca$

Combine all the terms:

$(4a – 2b – 3c)^2 = 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ca$

Thus, the expanded form of $(4a – 2b – 3c)^2$ is $16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ca$.

We need to factorise the expression $4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz$.

We observe that this expression resembles the expansion of a trinomial squared:

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

Let's try to match the terms of the given expression with the terms in the identity.

The squared terms are $4x^2$, $y^2$, and $z^2$. These can be written as $(2x)^2$, $(y)^2$, and $(z)^2$.

The cross terms are $-4xy$, $-2yz$, and $4xz$. Let's consider the possibilities for the base terms $a, b, c$ based on the squared terms.

Let's assume the base terms are related to $2x$, $y$, and $z$, allowing for negative signs.

Consider the form $(A+B+C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA$.

We have $A^2 = 4x^2 \implies A = \pm 2x$

$B^2 = y^2 \implies B = \pm y$

$C^2 = z^2 \implies C = \pm z$

Now match the cross terms:

• $2AB = -4xy$
• $2BC = -2yz$
• $2CA = 4xz$

Let's try $A=2x$. Then $2(2x)B = -4xy$, which means $4xB = -4xy$, so $B = -y$.

Now use this $B$ in the second cross term: $2(-y)C = -2yz$, which means $-2yC = -2yz$, so $C = z$.

Let's check the third cross term with $A=2x$ and $C=z$: $2CA = 2(z)(2x) = 4xz$. This matches the given term.

So, the terms are $A=2x$, $B=-y$, and $C=z$. Let's verify the squared terms:

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

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

$C^2 = (z)^2 = z^2$

All terms match the given expression when we use $A=2x$, $B=-y$, and $C=z$.

Therefore, the expression can be factorised as $(A+B+C)^2 = (2x + (-y) + z)^2 = (2x - y + z)^2$.

Alternatively, we could have chosen $A=-2x$. Then $2(-2x)B = -4xy$, which means $-4xB = -4xy$, so $B = y$.

Now use this $B$ in the second cross term: $2(y)C = -2yz$, which means $2yC = -2yz$, so $C = -z$.

Let's check the third cross term with $A=-2x$ and $C=-z$: $2CA = 2(-z)(-2x) = 4xz$. This matches the given term.

So, the terms are $A=-2x$, $B=y$, and $C=-z$. The factorisation is $(-2x + y - z)^2$.

Note that $(2x - y + z)^2 = (-( -2x + y - z))^2 = (-1)^2 (-2x + y - z)^2 = (-2x + y - z)^2$. Both factorisations are equivalent.

Using the first set of terms ($2x, -y, z$):

$4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz = (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$

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

Thus, the factorisation is $(2x - y + z)^2$.

We will write the given cubes in the expanded form using appropriate algebraic identities.

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(i) $(3a + 4b)^3$

This expression is in the form $(x+y)^3$. We use the identity:

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

Here, $x = 3a$ and $y = 4b$. Substitute these values into the identity:

$(3a + 4b)^3 = (3a)^3 + 3(3a)^2(4b) + 3(3a)(4b)^2 + (4b)^3$

Calculate each term:

$(3a)^3 = 3^3 a^3 = 27a^3$

$3(3a)^2(4b) = 3(9a^2)(4b) = 3 \times 9 \times 4 \times a^2 b = 108a^2 b$

$3(3a)(4b)^2 = 3(3a)(16b^2) = 3 \times 3 \times 16 \times a b^2 = 144ab^2$

$(4b)^3 = 4^3 b^3 = 64b^3$

Combine the terms:

$(3a + 4b)^3 = 27a^3 + 108a^2b + 144ab^2 + 64b^3$

Thus, the expanded form of $(3a + 4b)^3$ is $27a^3 + 108a^2b + 144ab^2 + 64b^3$.

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(ii) $(5p – 3q)^3$

This expression is in the form $(x-y)^3$. We use the identity:

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

Here, $x = 5p$ and $y = 3q$. Substitute these values into the identity:

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

Calculate each term:

$(5p)^3 = 5^3 p^3 = 125p^3$

$3(5p)^2(3q) = 3(25p^2)(3q) = 3 \times 25 \times 3 \times p^2 q = 225p^2 q$

$3(5p)(3q)^2 = 3(5p)(9q^2) = 3 \times 5 \times 9 \times p q^2 = 135pq^2$

$(3q)^3 = 3^3 q^3 = 27q^3$

Combine the terms:

$(5p – 3q)^3 = 125p^3 - 225p^2q + 135pq^2 - 27q^3$

Thus, the expanded form of $(5p – 3q)^3$ is $125p^3 - 225p^2q + 135pq^2 - 27q^3$.

We will evaluate the given expressions using suitable algebraic identities.

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(i) $(104)^3$

We can write 104 as $100 + 4$. So, $(104)^3 = (100 + 4)^3$.

This is in the form $(x+y)^3$. We use the identity:

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

Here, $x = 100$ and $y = 4$. Substitute these values into the identity:

$(100 + 4)^3 = (100)^3 + 3(100)^2(4) + 3(100)(4)^2 + (4)^3$

Calculate each term:

$(100)^3 = 100 \times 100 \times 100 = 1,000,000$

$3(100)^2(4) = 3(10000)(4) = 12 \times 10000 = 120,000$

$3(100)(4)^2 = 3(100)(16) = 300 \times 16 = 4800$

$(4)^3 = 4 \times 4 \times 4 = 64$

Now, add the terms:

$(104)^3 = 1,000,000 + 120,000 + 4800 + 64$

$= 1,120,000 + 4800 + 64$

$= 1,124,800 + 64$

$= 1,124,864$

Thus, $(104)^3 = 1,124,864$.

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(ii) $(999)^3$

We can write 999 as $1000 - 1$. So, $(999)^3 = (1000 - 1)^3$.

This is in the form $(x-y)^3$. We use the identity:

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

Here, $x = 1000$ and $y = 1$. Substitute these values into the identity:

$(1000 - 1)^3 = (1000)^3 - 3(1000)^2(1) + 3(1000)(1)^2 - (1)^3$

Calculate each term:

$(1000)^3 = 1000 \times 1000 \times 1000 = 1,000,000,000$

$3(1000)^2(1) = 3(1000000)(1) = 3,000,000$

$3(1000)(1)^2 = 3(1000)(1) = 3000$

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

Combine the terms:

$(999)^3 = 1,000,000,000 - 3,000,000 + 3000 - 1$

$= (1,000,000,000 + 3000) - (3,000,000 + 1)$

$= 1,000,003,000 - 3,000,001$

$= 997,002,999$

Thus, $(999)^3 = 997,002,999$.

We need to factorise the expression $8x^3 + 27y^3 + 36x^2y + 54xy^2$.

We can rewrite the terms to see if this expression fits the expanded form of the sum of cubes of a binomial:

The identity for the cube of a sum of two terms is: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Let's examine the terms in the given expression:

• The first term is $8x^3$. This can be written as $(2x)^3$. So, let $a = 2x$.
• The second term is $27y^3$. This can be written as $(3y)^3$. So, let $b = 3y$.

Now let's check if the remaining terms match $3a^2b$ and $3ab^2$ with $a=2x$ and $b=3y$:

• $3a^2b = 3(2x)^2(3y) = 3(4x^2)(3y) = 3 \times 4 \times 3 \times x^2 \times y = 36x^2y$. This matches the third term.
• $3ab^2 = 3(2x)(3y)^2 = 3(2x)(9y^2) = 3 \times 2 \times 9 \times x \times y^2 = 54xy^2$. This matches the fourth term.

Since all the terms match the expansion of $(a+b)^3$ with $a=2x$ and $b=3y$, the expression can be factorised as $(a+b)^3$.

$8x^3 + 27y^3 + 36x^2y + 54xy^2 = (2x)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 + (3y)^3$

Using the identity, this is equal to $(2x + 3y)^3$.

Thus, the factorisation is $(2x + 3y)^3$.

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Solution:

The given expression is $8x^3 + y^3 + 27z^3 – 18xyz$.

We can rewrite the terms as cubes:

$8x^3 = (2x)^3$

$y^3 = (y)^3$

$27z^3 = (3z)^3$

The expression is $(2x)^3 + y^3 + (3z)^3 - 18xyz$.

This expression is in the form $a^3 + b^3 + c^3 - 3abc$, where $a = 2x$, $b = y$, and $c = 3z$.

Let's verify the term $-3abc$:

We use the algebraic identity:

Substitute $a=2x$, $b=y$, and $c=3z$ into the identity:

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

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

Simplify the second factor:

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

Thus, the factorization of $8x^3 + y^3 + 27z^3 – 18xyz$ is $(2x+y+3z)(4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6zx)$.

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Solution:

We will use suitable algebraic identities to find the products.

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Part (i):

Given expression: $(x + 4) (x + 10)$

This expression is in the form $(x+a)(x+b)$.

The suitable identity is:

Here, $a = 4$ and $b = 10$.

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

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

$= x^2 + 14x + 40$

Thus, the product is $x^2 + 14x + 40$.

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Part (ii):

Given expression: $(x + 8) (x – 10)$

This expression is in the form $(x+a)(x+b)$.

The suitable identity is:

Here, $a = 8$ and $b = -10$.

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

$(x+8)(x-10) = x^2 + (8+(-10))x + (8)(-10)$

$= x^2 + (8-10)x - 80$

$= x^2 - 2x - 80$

Thus, the product is $x^2 - 2x - 80$.

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Part (iii):

Given expression: $(3x + 4) (3x – 5)$

This expression is in the form $(y+a)(y+b)$, where $y = 3x$.

The suitable identity is:

Here, $y = 3x$, $a = 4$, and $b = -5$.

Substituting the values of $y$, $a$, and $b$ into the identity:

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

$= (3x)^2 + (-1)(3x) - 20$

$= 9x^2 - 3x - 20$

Thus, the product is $9x^2 - 3x - 20$.

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Part (iv):

Given expression: $(y^2 + \frac{3}{2}) (y^2 - \frac{3}{2})$

This expression is in the form $(a+b)(a-b)$.

The suitable identity is:

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

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

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

$= y^{2 \times 2} - \frac{3^2}{2^2}$

$= y^4 - \frac{9}{4}$

Thus, the product is $y^4 - \frac{9}{4}$.

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Part (v):

Given expression: $(3 – 2x) (3 + 2x)$

This expression is in the form $(a-b)(a+b)$.

The suitable identity is:

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

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

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

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

$= 9 - 4x^2$

Thus, the product is $9 - 4x^2$.

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Solution:

We will evaluate the products using suitable algebraic identities without direct multiplication.

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Part (i):

Given product: $103 \times 107$

We can write $103$ as $100 + 3$ and $107$ as $100 + 7$.

So the product is $(100 + 3)(100 + 7)$.

This expression is in the form $(x+a)(x+b)$.

The suitable identity is:

Here, $x = 100$, $a = 3$, and $b = 7$.

Substituting the values into the identity:

$103 \times 107 = (100 + 3)(100 + 7)$

$= (100)^2 + (3+7)(100) + (3)(7)$

$= 10000 + (10)(100) + 21$

$= 10000 + 1000 + 21$

$= 11000 + 21$

$= 11021$

Thus, $103 \times 107 = 11021$.

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Part (ii):

Given product: $95 \times 96$

We can write $95$ as $100 - 5$ and $96$ as $100 - 4$.

So the product is $(100 - 5)(100 - 4)$.

This expression is in the form $(x+a)(x+b)$.

The suitable identity is:

Here, $x = 100$, $a = -5$, and $b = -4$.

Substituting the values into the identity:

$95 \times 96 = (100 - 5)(100 - 4)$

$= (100)^2 + ((-5)+(-4))(100) + (-5)(-4)$

$= 10000 + (-9)(100) + 20$

$= 10000 - 900 + 20$

$= 9100 + 20$

$= 9120$

Thus, $95 \times 96 = 9120$.

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Part (iii):

Given product: $104 \times 96$

We can write $104$ as $100 + 4$ and $96$ as $100 - 4$.

So the product is $(100 + 4)(100 - 4)$.

This expression is in the form $(a+b)(a-b)$.

The suitable identity is:

Here, $a = 100$ and $b = 4$.

Substituting the values into the identity:

$104 \times 96 = (100 + 4)(100 - 4)$

$= (100)^2 - (4)^2$

$= 10000 - 16$

$= 9984$

Thus, $104 \times 96 = 9984$.

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Solution:

We will factorise the given expressions using appropriate algebraic identities.

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Part (i):

Given expression: $9x^2 + 6xy + y^2$

We can rewrite the terms as squares and check for the middle term:

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

$y^2 = (y)^2$

The middle term is $6xy$. Let's check if it fits the pattern $2ab$ for a perfect square trinomial.

$2(3x)(y) = 6xy$.

The expression matches the form $a^2 + 2ab + b^2$, where $a = 3x$ and $b = y$.

The suitable identity is:

Substituting $a = 3x$ and $b = y$ into the identity:

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

Thus, the factorization is $(3x+y)^2$.

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Part (ii):

Given expression: $4y^2 – 4y + 1$

We can rewrite the terms as squares and check for the middle term:

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

$1 = (1)^2$

The middle term is $-4y$. Let's check if it fits the pattern $-2ab$ for a perfect square trinomial.

$-2(2y)(1) = -4y$.

The expression matches the form $a^2 - 2ab + b^2$, where $a = 2y$ and $b = 1$.

The suitable identity is:

Substituting $a = 2y$ and $b = 1$ into the identity:

$4y^2 – 4y + 1 = (2y)^2 - 2(2y)(1) + (1)^2 = (2y-1)^2$

Thus, the factorization is $(2y-1)^2$.

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Part (iii):

Given expression: $x^2 – \frac{y^2}{100}$

We can rewrite the terms as squares:

$x^2 = (x)^2$

$\frac{y^2}{100} = \frac{y^2}{10^2} = (\frac{y}{10})^2$

The expression matches the form $a^2 - b^2$, where $a = x$ and $b = \frac{y}{10}$.

The suitable identity is:

Substituting $a = x$ and $b = \frac{y}{10}$ into the identity:

$x^2 – \frac{y^2}{100} = (x)^2 - (\frac{y}{10})^2 = (x + \frac{y}{10})(x - \frac{y}{10})$

Thus, the factorization is $(x + \frac{y}{10})(x - \frac{y}{10})$.

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Solution:

We will expand each expression using the identity for the square of a trinomial:

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Part (i):

Given expression: $(x + 2y + 4z)^2$

Here, $a = x$, $b = 2y$, and $c = 4z$.

Using the identity:

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

$= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$

Thus, the expansion is $x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$.

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Part (ii):

Given expression: $(2x – y + z)^2$

Here, $a = 2x$, $b = -y$, and $c = z$.

Using the identity:

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

$= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$

Thus, the expansion is $4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$.

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Part (iii):

Given expression: $(–2x + 3y + 2z)^2$

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

Using the identity:

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

$= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx$

Thus, the expansion is $4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx$.

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Part (iv):

Given expression: $(3a – 7b – c)^2$

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

Using the identity:

$(3a – 7b – c)^2 = (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$

$= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$

Thus, the expansion is $9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$.

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Part (v):

Given expression: $(–2x + 5y – 3z)^2$

Here, $a = -2x$, $b = 5y$, and $c = -3z$.

Using the identity:

$(–2x + 5y – 3z)^2 = (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)$

$= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx$

Thus, the expansion is $4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx$.

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Part (vi):

Given expression: $\left[ \frac{1}{4}a \;-\; \frac{1}{2}b \;+\; 1 \right]^2$

Here, $a = \frac{1}{4}a$, $b = -\frac{1}{2}b$, and $c = 1$.

Using the identity:

$\left[ \frac{1}{4}a \;-\; \frac{1}{2}b \;+\; 1 \right]^2 = \left( \frac{1}{4}a \right)^2 + \left( -\frac{1}{2}b \right)^2 + (1)^2 + 2\left( \frac{1}{4}a \right)\left( -\frac{1}{2}b \right) + 2\left( -\frac{1}{2}b \right)(1) + 2(1)\left( \frac{1}{4}a \right)$

$= \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a$

Thus, the expansion is $\frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a$.

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Solution:

We will factorise the given expressions using the algebraic identity for the square of a trinomial:

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Part (i):

Given expression: $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$

We can write the terms as squares:

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

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

$16z^2 = (4z)^2$

The expression is in the form $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Observing the signs of the cross terms ($12xy$ is positive, $-24yz$ is negative, $-16xz$ is negative), we can deduce the signs of $a, b, c$.

Let $a = 2x$, $b = 3y$, and $c = 4z$.

$2ab = 2(2x)(3y) = 12xy$ (Positive, matches)

$2bc = 2(3y)(4z) = 24yz$ (Should be negative $-24yz$, so one of $3y$ or $4z$ must be negative)

$2ca = 2(4z)(2x) = 16xz$ (Should be negative $-16xz$, so one of $4z$ or $2x$ must be negative)

Since $12xy$ is positive, $2x$ and $3y$ must have the same sign. Let's assume they are positive, so $a=2x$ and $b=3y$.

Now consider the negative terms $-24yz$ and $-16xz$. Since $3y$ and $2x$ are assumed positive, the term $4z$ must be negative to make these products negative.

Let's choose $a = 2x$, $b = 3y$, and $c = -4z$.

Check the cross terms with these values:

$2ab = 2(2x)(3y) = 12xy$

$2bc = 2(3y)(-4z) = -24yz$

$2ca = 2(-4z)(2x) = -16xz$

These match the terms in the given expression.

So, $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz = (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$

Using the identity, this is equal to $(2x + 3y - 4z)^2$.

Thus, the factorization is $(2x + 3y - 4z)^2$.

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Part (ii):

Given expression: $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$

We can write the terms as squares:

$2x^2 = (\sqrt{2}x)^2$

$y^2 = (y)^2$

$8z^2 = (\sqrt{8}z)^2 = (2\sqrt{2}z)^2$

The expression is in the form $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Observing the signs of the cross terms ($-2\sqrt{2}xy$ is negative, $4\sqrt{2}yz$ is positive, $-8xz$ is negative), we can deduce the signs of $a, b, c$.

Let $a = \sqrt{2}x$, $b = y$, and $c = 2\sqrt{2}z$ initially (ignoring signs).

$2ab = 2(\sqrt{2}x)(y) = 2\sqrt{2}xy$ (Should be negative $-2\sqrt{2}xy$, so one of $\sqrt{2}x$ or $y$ must be negative)

$2bc = 2(y)(2\sqrt{2}z) = 4\sqrt{2}yz$ (Positive, matches $4\sqrt{2}yz$, so $y$ and $2\sqrt{2}z$ must have the same sign)

$2ca = 2(2\sqrt{2}z)(\sqrt{2}x) = 8xz$ (Should be negative $-8xz$, so one of $2\sqrt{2}z$ or $\sqrt{2}x$ must be negative)

From the second term $4\sqrt{2}yz$ being positive, $y$ and $2\sqrt{2}z$ have the same sign. From the first ($-2\sqrt{2}xy$) and third ($-8xz$) terms being negative, $\sqrt{2}x$ must have the opposite sign to both $y$ and $2\sqrt{2}z$.

Let's assume $y$ and $2\sqrt{2}z$ are positive, so $b=y$ and $c=2\sqrt{2}z$. Then $\sqrt{2}x$ must be negative.

Let's choose $a = -\sqrt{2}x$, $b = y$, and $c = 2\sqrt{2}z$.

Check the cross terms with these values:

$2ab = 2(-\sqrt{2}x)(y) = -2\sqrt{2}xy$

$2bc = 2(y)(2\sqrt{2}z) = 4\sqrt{2}yz$

$2ca = 2(2\sqrt{2}z)(-\sqrt{2}x) = -8xz$

These match the terms in the given expression.

So, $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz = (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2(-\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) + 2(2\sqrt{2}z)(-\sqrt{2}x)$

Using the identity, this is equal to $(-\sqrt{2}x + y + 2\sqrt{2}z)^2$.

Thus, the factorization is $(-\sqrt{2}x + y + 2\sqrt{2}z)^2$.

Alternate Solution:

Another valid factorization is $(\sqrt{2}x - y - 2\sqrt{2}z)^2$, which is equivalent because $(P)^2 = (-P)^2$. In this case, $a = \sqrt{2}x$, $b = -y$, and $c = -2\sqrt{2}z$.

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Solution:

We will write the given cubes in expanded form using the following identities:

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Part (i):

Given expression: $(2x + 1)^3$

This is in the form $(a+b)^3$ where $a=2x$ and $b=1$.

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

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

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

$= 8x^3 + 12x^2 + 6x + 1$

The expanded form is $8x^3 + 12x^2 + 6x + 1$.

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Part (ii):

Given expression: $(2a – 3b)^3$

This is in the form $(a-b)^3$ where $a=2a$ and $b=3b$.

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

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

$= 8a^3 - 3(4a^2)(3b) + 3(2a)(9b^2) - 27b^3$

$= 8a^3 - 36a^2b + 54ab^2 - 27b^3$

The expanded form is $8a^3 - 36a^2b + 54ab^2 - 27b^3$.

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Part (iii):

Given expression: $\left[ \frac{3}{2} x \;+\; 1\right]^3$

This is in the form $(a+b)^3$ where $a=\frac{3}{2}x$ and $b=1$.

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

$\left[ \frac{3}{2} x \;+\; 1\right]^3 = \left(\frac{3}{2}x\right)^3 + 3\left(\frac{3}{2}x\right)^2(1) + 3\left(\frac{3}{2}x\right)(1)^2 + (1)^3$

$= \frac{27}{8}x^3 + 3\left(\frac{9}{4}x^2\right)(1) + 3\left(\frac{3}{2}x\right)(1) + 1$

$= \frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$

The expanded form is $\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$.

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Part (iv):

Given expression: $\left[ x \;-\; \frac{2}{3}y \right]^3$

This is in the form $(a-b)^3$ where $a=x$ and $b=\frac{2}{3}y$.

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

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

$= x^3 - \cancel{3}x^2\left(\frac{2}{\cancel{3}}y\right) + \cancel{3}x\left(\frac{4}{\cancel{9}_{3}}y^2\right) - \frac{8}{27}y^3$

$= x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$

The expanded form is $x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$.

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Solution:

We will evaluate the following using suitable algebraic identities.

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Part (i):

Given expression: $(99)^3$

We can write $99$ as $100 - 1$. So, $(99)^3 = (100-1)^3$.

This is in the form $(a-b)^3$, where $a=100$ and $b=1$.

Using the identity:

Substituting the values of $a$ and $b$:

$(100-1)^3 = (100)^3 - 3(100)^2(1) + 3(100)(1)^2 - (1)^3$

$= 1000000 - 3(10000)(1) + 3(100)(1) - 1$

$= 1000000 - 30000 + 300 - 1$

$= 970000 + 300 - 1$

$= 970300 - 1$

$= 970299$

Thus, $(99)^3 = 970299$.

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Part (ii):

Given expression: $(102)^3$

We can write $102$ as $100 + 2$. So, $(102)^3 = (100+2)^3$.

This is in the form $(a+b)^3$, where $a=100$ and $b=2$.

Using the identity:

Substituting the values of $a$ and $b$:

$(100+2)^3 = (100)^3 + 3(100)^2(2) + 3(100)(2)^2 + (2)^3$

$= 1000000 + 3(10000)(2) + 3(100)(4) + 8$

$= 1000000 + 60000 + 1200 + 8$

$= 1060000 + 1200 + 8$

$= 1061200 + 8$

$= 1061208$

Thus, $(102)^3 = 1061208$.

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Part (iii):

Given expression: $(998)^3$

We can write $998$ as $1000 - 2$. So, $(998)^3 = (1000-2)^3$.

This is in the form $(a-b)^3$, where $a=1000$ and $b=2$.

Using the identity:

Substituting the values of $a$ and $b$:

$(1000-2)^3 = (1000)^3 - 3(1000)^2(2) + 3(1000)(2)^2 - (2)^3$

$= 1000000000 - 3(1000000)(2) + 3(1000)(4) - 8$

$= 1000000000 - 6000000 + 12000 - 8$

$= 994000000 + 12000 - 8$

$= 994012000 - 8$

$= 994011992$

Thus, $(998)^3 = 994011992$.

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Solution:

We will factorise the given expressions using the identities for the cube of a binomial:

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Part (i):

Given expression: $8a^3 + b^3 + 12a^2b + 6ab^2$

We can rewrite the first two terms as cubes:

$8a^3 = (2a)^3$

$b^3 = (b)^3$

The expression looks like the expansion of $(a+b)^3$. Let's check if the middle terms match $3a^2b + 3ab^2$ with $a=2a$ and $b=b$.

$3a^2b = 3(2a)^2(b) = 3(4a^2)(b) = 12a^2b$

$3ab^2 = 3(2a)(b)^2 = 6ab^2$

The terms match the identity $a^3 + b^3 + 3a^2b + 3ab^2 = (a+b)^3$ where $a=2a$ and $b=b$.

So, $8a^3 + b^3 + 12a^2b + 6ab^2 = (2a)^3 + (b)^3 + 3(2a)^2(b) + 3(2a)(b)^2 = (2a+b)^3$.

Thus, the factorization is $(2a+b)^3$.

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Part (ii):

Given expression: $8a^3 – b^3 – 12a^2b + 6ab^2$

Rearranging the terms: $8a^3 – 12a^2b + 6ab^2 – b^3$.

We can rewrite the first and last terms as cubes:

$8a^3 = (2a)^3$

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

The expression looks like the expansion of $(a-b)^3$. Let's check if the middle terms match $-3a^2b + 3ab^2$ with $a=2a$ and $b=b$.

$-3a^2b = -3(2a)^2(b) = -3(4a^2)(b) = -12a^2b$

$3ab^2 = 3(2a)(b)^2 = 6ab^2$

The terms match the identity $a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3$ where $a=2a$ and $b=b$.

So, $8a^3 – b^3 – 12a^2b + 6ab^2 = (2a)^3 - 3(2a)^2(b) + 3(2a)(b)^2 - (b)^3 = (2a-b)^3$.

Thus, the factorization is $(2a-b)^3$.

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Part (iii):

Given expression: $27 – 125a^3 – 135a + 225a^2$

Rearranging the terms in descending powers of $a$: $-125a^3 + 225a^2 - 135a + 27$.

We can rewrite the first and last terms as cubes:

$-125a^3 = (-5a)^3$

$27 = (3)^3$

Let's consider the form $(a-b)^3$ with $a=3$ and $b=5a$.

Check the terms: $a^3 = (3)^3 = 27$.

$-b^3 = -(5a)^3 = -125a^3$.

$-3a^2b = -3(3)^2(5a) = -3(9)(5a) = -135a$.

$3ab^2 = 3(3)(5a)^2 = 3(3)(25a^2) = 225a^2$.

The terms match the identity $a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3$ where $a=3$ and $b=5a$.

So, $27 – 125a^3 – 135a + 225a^2 = (3)^3 - 3(3)^2(5a) + 3(3)(5a)^2 - (5a)^3 = (3-5a)^3$.

Thus, the factorization is $(3-5a)^3$.

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Part (iv):

Given expression: $64a^3 – 27b^3 – 144a^2b + 108ab^2$

Rearranging the terms: $64a^3 – 144a^2b + 108ab^2 – 27b^3$.

We can rewrite the first and last terms as cubes:

$64a^3 = (4a)^3$

$-27b^3 = -(3b)^3$

The expression looks like the expansion of $(a-b)^3$. Let's check if the middle terms match $-3a^2b + 3ab^2$ with $a=4a$ and $b=3b$.

$-3a^2b = -3(4a)^2(3b) = -3(16a^2)(3b) = -144a^2b$

$3ab^2 = 3(4a)(3b)^2 = 3(4a)(9b^2) = 108ab^2$

The terms match the identity $a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3$ where $a=4a$ and $b=3b$.

So, $64a^3 – 27b^3 – 144a^2b + 108ab^2 = (4a)^3 - 3(4a)^2(3b) + 3(4a)(3b)^2 - (3b)^3 = (4a-3b)^3$.

Thus, the factorization is $(4a-3b)^3$.

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Part (v):

Given expression: $27p^3 – \frac{1}{216} – \frac{9}{2} p^2 + \frac{1}{4} p$

Rearranging the terms in descending powers of $p$: $27p^3 – \frac{9}{2} p^2 + \frac{1}{4} p – \frac{1}{216}$.

We can rewrite the first and last terms as cubes:

$27p^3 = (3p)^3$

$-\frac{1}{216} = -(\frac{1}{6})^3$

The expression looks like the expansion of $(a-b)^3$. Let's check if the middle terms match $-3a^2b + 3ab^2$ with $a=3p$ and $b=\frac{1}{6}$.

$-3a^2b = -3(3p)^2(\frac{1}{6}) = -3(9p^2)(\frac{1}{6}) = -\cancel{3} \times \frac{9p^2}{\cancel{6}_{2}} = -\frac{9}{2}p^2$

$3ab^2 = 3(3p)(\frac{1}{6})^2 = 3(3p)(\frac{1}{36}) = \cancel{9}p \times \frac{1}{\cancel{36}_{4}} = \frac{1}{4}p$

The terms match the identity $a^3 - 3a^2b + 3ab^2 - b^3 = (a-b)^3$ where $a=3p$ and $b=\frac{1}{6}$.

So, $27p^3 – \frac{9}{2} p^2 + \frac{1}{4} p – \frac{1}{216} = (3p)^3 - 3(3p)^2(\frac{1}{6}) + 3(3p)(\frac{1}{6})^2 - (\frac{1}{6})^3 = (3p - \frac{1}{6})^3$.

Thus, the factorization is $(3p - \frac{1}{6})^3$.

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Solution:

We will verify the given identities by expanding the Right Hand Side (RHS) and showing that it equals the Left Hand Side (LHS).

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Part (i):

Identity to verify: $x^3 + y^3 = (x + y)(x^2 – xy + y^2)$

Consider the RHS:

Expand the product:

RHS = $x(x^2 – xy + y^2) + y(x^2 – xy + y^2)$

RHS = $(x \cdot x^2) + (x \cdot (-xy)) + (x \cdot y^2) + (y \cdot x^2) + (y \cdot (-xy)) + (y \cdot y^2)$

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

Combine like terms:

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

RHS = $x^3 + 0 + 0 + y^3$

RHS = $x^3 + y^3$

This is equal to the LHS ($x^3 + y^3$).

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Part (ii):

Identity to verify: $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$

Consider the RHS:

Expand the product:

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

RHS = $(x \cdot x^2) + (x \cdot xy) + (x \cdot y^2) + (-y \cdot x^2) + (-y \cdot xy) + (-y \cdot y^2)$

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

Combine like terms:

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

RHS = $x^3 + 0 + 0 - y^3$

RHS = $x^3 - y^3$

This is equal to the LHS ($x^3 - y^3$).

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Solution:

We will factorise the given expressions using the identities for the sum and difference of cubes.

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Part (i):

Given expression: $27y^3 + 125z^3$

We can rewrite the terms as cubes:

$27y^3 = (3y)^3$

$125z^3 = (5z)^3$

The expression is in the form $a^3 + b^3$, where $a=3y$ and $b=5z$.

The suitable identity is:

Substitute $a=3y$ and $b=5z$ into the identity:

$27y^3 + 125z^3 = (3y)^3 + (5z)^3$

$= (3y + 5z)((3y)^2 - (3y)(5z) + (5z)^2)$

Simplify the second factor:

$= (3y + 5z)(9y^2 - 15yz + 25z^2)$

Thus, the factorization is $(3y + 5z)(9y^2 - 15yz + 25z^2)$.

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Part (ii):

Given expression: $64m^3 – 343n^3$

We can rewrite the terms as cubes:

$64m^3 = (4m)^3$

$343n^3 = (7n)^3$

The expression is in the form $a^3 - b^3$, where $a=4m$ and $b=7n$.

The suitable identity is:

Substitute $a=4m$ and $b=7n$ into the identity:

$64m^3 – 343n^3 = (4m)^3 - (7n)^3$

$= (4m - 7n)((4m)^2 + (4m)(7n) + (7n)^2)$

Simplify the second factor:

$= (4m - 7n)(16m^2 + 28mn + 49n^2)$

Thus, the factorization is $(4m - 7n)(16m^2 + 28mn + 49n^2)$.

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Solution:

The given expression is $27^3 + y^3 + z^3 – 9xyz$.

It is highly likely that the question intended the first term to be $27x^3$ instead of $27^3$ to fit a standard algebraic identity involving multiple variables. Assuming the intended expression is $27x^3 + y^3 + z^3 – 9xyz$, we proceed as follows:

The assumed expression is $27x^3 + y^3 + z^3 – 9xyz$.

We can rewrite the terms as cubes:

$27x^3 = (3x)^3$

$y^3 = (y)^3$

$z^3 = (z)^3$

The expression is in the form $a^3 + b^3 + c^3 - 3abc$, where $a = 3x$, $b = y$, and $c = z$.

Let's verify the term $-3abc$:

We use the algebraic identity:

Substitute $a=3x$, $b=y$, and $c=z$ into the identity:

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

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

Simplify the second factor:

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

Thus, assuming the expression is $27x^3 + y^3 + z^3 – 9xyz$, the factorization is $(3x+y+z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$.

Given:

The identity to verify is:

$x^3 + y^3 + z^3 - 3xyz = \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$

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To Verify:

Verify that the given identity is true for all values of $x$, $y$, and $z$.

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Solution:

We will start with the Right Hand Side (RHS) of the identity and simplify it to obtain the Left Hand Side (LHS).

RHS $= \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$

First, let's expand the squared terms inside the brackets using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

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

$(y - z)^2 = y^2 - 2yz + z^2$

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

Now, sum these expanded terms:

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

Combine like terms:

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

$= 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx$

Factor out 2 from the expression:

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

Substitute this back into the RHS expression:

RHS $= \frac{1}{2} (x + y + z)[2(x^2 + y^2 + z^2 - xy - yz - zx)]$

The $\frac{1}{2}$ and 2 cancel out:

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

Now, expand this product by multiplying each term in the first parenthesis by each term in the second parenthesis:

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

RHS $= (x^3 + xy^2 + xz^2 - x^2y - xyz - x^2z) + (yx^2 + y^3 + yz^2 - xy^2 - y^2z - xyz) + (zx^2 + zy^2 + z^3 - xyz - yz^2 - z^2x)$

Rearrange the terms and group them:

RHS $= x^3 + y^3 + z^3$

$+ (xy^2 - xy^2)$

$+ (xz^2 - z^2x)$

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

$+ (yz^2 - yz^2)$

$+ (-y^2z + zy^2)$

$+ (zx^2 - x^2z)$

$+ (-xyz - xyz - xyz)$

Simplify the grouped terms. Note that $xy^2 = y^2x$, $xz^2 = z^2x$, $x^2y = yx^2$, etc.:

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

$(xz^2 - z^2x) = 0$

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

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

$(-y^2z + zy^2) = 0$

$(zx^2 - x^2z) = 0$

$(-xyz - xyz - xyz) = -3xyz$

Substitute these simplified terms back into the expression for RHS:

RHS $= x^3 + y^3 + z^3 + 0 + 0 + 0 + 0 + 0 + 0 - 3xyz$

RHS $= x^3 + y^3 + z^3 - 3xyz$

This is the Left Hand Side (LHS) of the identity.

Since LHS $=$ RHS, the identity is verified.

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Verification Complete.

Given:

$x + y + z = 0$

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To Show:

$x^3 + y^3 + z^3 = 3xyz$

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Solution:

We know the algebraic identity:

$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

We are given that $x + y + z = 0$.

Substitute $x + y + z = 0$ into the identity:

$x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)$

Since anything multiplied by zero is zero, the right side becomes 0:

$x^3 + y^3 + z^3 - 3xyz = 0$

Now, transpose the term $-3xyz$ to the right side of the equation:

$x^3 + y^3 + z^3 = 3xyz$

Thus, it is shown that if $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.

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Alternate Method:

Given $x + y + z = 0$.

From this, we can write $x + y = -z$.

Cube both sides of this equation:

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

Using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$, we expand the left side:

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

We know that $x + y = -z$. Substitute this into the equation:

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

$x^3 + y^3 - 3xyz = -z^3$

Now, transpose $-z^3$ to the left side and $-3xyz$ to the right side:

$x^3 + y^3 + z^3 = 3xyz$

This also shows the required result.

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Statement Shown.

We know the identity: If $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$. We will use this identity to find the values without calculating the cubes directly.

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(i) $(-12)^3 + (7)^3 + (5)^3$

Let $x = -12$, $y = 7$, and $z = 5$.

Check the sum $x + y + z$:

$x + y + z = (-12) + 7 + 5 = -12 + 12 = 0$

Since $x + y + z = 0$, we can use the identity $x^3 + y^3 + z^3 = 3xyz$.

Therefore,

$(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12) \times 7 \times 5$

Calculate the product:

$3 \times (-12) = -36$

$-36 \times 7 = -252$

$-252 \times 5 = -1260$

So, $(-12)^3 + (7)^3 + (5)^3 = -1260$.

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(ii) $(28)^3 + (-15)^3 + (-13)^3$

Let $x = 28$, $y = -15$, and $z = -13$.

Check the sum $x + y + z$:

$x + y + z = 28 + (-15) + (-13) = 28 - 15 - 13 = 28 - (15 + 13) = 28 - 28 = 0$

Since $x + y + z = 0$, we can use the identity $x^3 + y^3 + z^3 = 3xyz$.

Therefore,

$(28)^3 + (-15)^3 + (-13)^3 = 3 \times 28 \times (-15) \times (-13)$

Calculate the product:

$3 \times 28 = 84$

$(-15) \times (-13) = 15 \times 13 = 195$

Now calculate $84 \times 195$:

So, $84 \times 195 = 16380$.

Thus, $(28)^3 + (-15)^3 + (-13)^3 = 16380$.

To find possible expressions for the length and breadth of a rectangle given its area, we need to factorize the quadratic expression representing the area. The area of a rectangle is given by the product of its length and breadth.

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(i) Area: $25a^2 - 35a + 12$

We need to factorize the quadratic trinomial $25a^2 - 35a + 12$.

We look for two numbers whose product is $(25) \times (12) = 300$ and whose sum is $-35$.

The numbers are $-15$ and $-20$, since $(-15) \times (-20) = 300$ and $(-15) + (-20) = -35$.

Now we rewrite the middle term $-35a$ as the sum of these two numbers times $a$:

$25a^2 - 35a + 12 = 25a^2 - 15a - 20a + 12$

Next, we group the terms and factor by grouping:

$= (25a^2 - 15a) + (-20a + 12)$

Factor out the greatest common factor from each group:

$= 5a(5a - 3) - 4(5a - 3)$

Now, factor out the common binomial $(5a - 3)$:

$= (5a - 3)(5a - 4)$

Thus, the factored form of the area is $(5a - 3)(5a - 4)$.

Possible expressions for the length and breadth are the two factors.

One possible set of expressions is:

Length $= 5a - 3$

Breadth $= 5a - 4$

Alternatively, the length could be $5a - 4$ and the breadth $5a - 3$.

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(ii) Area: $35y^2 + 13y - 12$

We need to factorize the quadratic trinomial $35y^2 + 13y - 12$.

We look for two numbers whose product is $(35) \times (-12) = -420$ and whose sum is $13$.

The numbers are $28$ and $-15$, since $(28) \times (-15) = -420$ and $28 + (-15) = 13$.

Now we rewrite the middle term $13y$ as the sum of these two numbers times $y$:

$35y^2 + 13y - 12 = 35y^2 + 28y - 15y - 12$

Next, we group the terms and factor by grouping:

$= (35y^2 + 28y) + (-15y - 12)$

Factor out the greatest common factor from each group:

$= 7y(5y + 4) - 3(5y + 4)$

Now, factor out the common binomial $(5y + 4)$:

$= (5y + 4)(7y - 3)$

Thus, the factored form of the area is $(5y + 4)(7y - 3)$.

Possible expressions for the length and breadth are the two factors.

One possible set of expressions is:

Length $= 5y + 4$

Breadth $= 7y - 3$

Alternatively, the length could be $7y - 3$ and the breadth $5y + 4$.

To find the possible expressions for the dimensions (length, breadth, and height) of a cuboid given its volume, we need to factorize the polynomial representing the volume into three factors. The volume of a cuboid is given by the product of its length, breadth, and height.

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(i) Volume: $3x^2 - 12x$

We need to factorize the expression $3x^2 - 12x$.

We can factor out the common factor $3x$ from both terms:

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

The factors are $3$, $x$, and $(x - 4)$.

Therefore, the possible expressions for the dimensions of the cuboid are $3$, $x$, and $(x - 4)$.

Possible dimensions: Length $= 3$, Breadth $= x$, Height $= x - 4$ (or any permutation of these three expressions).

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(ii) Volume: $12ky^2 + 8ky - 20k$

We need to factorize the expression $12ky^2 + 8ky - 20k$.

First, factor out the common factor $4k$ from all terms:

$12ky^2 + 8ky - 20k = 4k(3y^2 + 2y - 5)$

Now, we need to factorize the quadratic trinomial $3y^2 + 2y - 5$.

We look for two numbers whose product is $(3) \times (-5) = -15$ and whose sum is $2$.

The numbers are $5$ and $-3$, since $5 \times (-3) = -15$ and $5 + (-3) = 2$.

Rewrite the middle term $2y$ as the sum of these two numbers times $y$:

$3y^2 + 2y - 5 = 3y^2 + 5y - 3y - 5$

Group the terms and factor by grouping:

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

$= y(3y + 5) - 1(3y + 5)$

Factor out the common binomial $(3y + 5)$:

$= (3y + 5)(y - 1)$

So, the complete factorization of the volume expression is $4k(3y + 5)(y - 1)$.

The three factors are $4k$, $(3y + 5)$, and $(y - 1)$.

Therefore, the possible expressions for the dimensions of the cuboid are $4k$, $(3y + 5)$, and $(y - 1)$.

Possible dimensions: Length $= 4k$, Breadth $= 3y + 5$, Height $= y - 1$ (or any permutation of these three expressions).