Chapter 2 Polynomials (Concepts)

Algebraic Expressions and Polynomials

In your journey through mathematics so far, you have learned about various types of numbers and algebraic expressions. This chapter builds upon that foundation to introduce you to a specific and very important type of algebraic expression called a polynomial. Polynomials are fundamental building blocks in algebra and are used extensively in various areas of mathematics, science, and engineering.

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Review of Algebraic Expressions

Recall from Chapter 9 of Class 8, an algebraic expression is a combination of variables and constants, connected by the fundamental arithmetic operations: addition (+), subtraction (-), multiplication ($\times$), and division ($\div$). Variables are symbols (like $x, y, z$) whose values can vary, while constants are numbers (like 2, -5, $\frac{1}{3}$) whose values are fixed.

Examples of algebraic expressions you've seen include: $x+5$, $2y-3$, $a^2$, $\frac{p}{q}$, $3x^3-4x^2+8x-1$.

The parts of an expression separated by '+' or '-' signs are called terms. In the expression $2x^2 + 5x - 3$, the terms are $2x^2$, $5x$, and $-3$. Each term is a product of factors. For example, the term $2x^2$ has factors 2, $x$, and $x$. The constant factor in a term is the coefficient (e.g., the coefficient of $x^2$ in $2x^2$ is 2, the coefficient of $x$ in $5x$ is 5). A term with no variable is called a constant term (e.g., -3).

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Introduction to Polynomials

While algebraic expressions can involve variables raised to any power (positive, negative, fractional) and variables in denominators, polynomials are a more restricted class of algebraic expressions. A polynomial is an algebraic expression in which the exponents of the variables are restricted to only whole numbers (non-negative integers: 0, 1, 2, 3, ...).

A polynomial in one variable, say $x$, is generally written in the form:

where:

• $x$ is the variable.
• $n$ is a whole number (a non-negative integer). $n$ is the highest power of $x$ with a non-zero coefficient.
• $a_0, a_1, \dots, a_n$ are constants, called the coefficients of the polynomial. Coefficients can be any real numbers (rational or irrational).
• The term $a_n x^n$ is called the leading term (if $a_n \neq 0$ and $n$ is the highest power).
• The term $a_0$ (since $x^0 = 1$) is called the constant term.

For a polynomial of degree $n > 0$, the coefficient $a_n$ is the leading coefficient and must be non-zero ($a_n \neq 0$).

Examples of Polynomials:

• $x+5$: This is a polynomial in one variable $x$. The terms are $x$ (which is $1x^1$) and 5 (which is $5x^0$). The exponents of $x$ are 1 and 0, which are whole numbers. Coefficients are 1 and 5.
• $y^2+2y-5$: This is a polynomial in one variable $y$. The terms are $y^2$, $2y$, and $-5$. The exponents of $y$ are 2, 1, and 0, which are whole numbers. Coefficients are 1, 2, and -5.
• $3x^3-4x^2+8x-1$: A polynomial in $x$ with exponents 3, 2, 1, 0 (all whole numbers). Coefficients are 3, -4, 8, -1.
• $7u^6 - \frac{3}{4} u^4 + \sqrt{2} u - 6$: A polynomial in $u$. Coefficients ($7, -\frac{3}{4}, \sqrt{2}, -6$) can be real numbers. Exponents (6, 4, 1, 0) are whole numbers.
• 5: This is a constant polynomial. It can be written as $5x^0$. The exponent 0 is a whole number. The coefficient is 5.
• 0: This is the zero polynomial. Its terms are all zeros.

Examples of Expressions that are NOT Polynomials:

• $\frac{1}{x} = x^{-1}$: The exponent of the variable $x$ is -1, which is not a whole number.
• $\sqrt{x} = x^{1/2}$: The exponent of the variable $x$ is $\frac{1}{2}$, which is not a whole number.
• $x^2 + \frac{1}{x} = x^2 + x^{-1}$: This expression contains the term $x^{-1}$, which has a negative exponent.
• $\sqrt{y} + y$: This expression contains the term $\sqrt{y} = y^{1/2}$, which has a fractional exponent.
• $3x^{1/2} + 2x + 1$: The term $3x^{1/2}$ has a fractional exponent.

In summary, polynomials are a subset of algebraic expressions characterized by having only whole number exponents for their variables.

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Degree of a Polynomial

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

Example: In $3x^3-4x^2+8x-1$, the exponents are 3, 2, 1, 0. The highest exponent is 3. So, the degree of this polynomial is 3.

Example: The degree of $y^2+2y-5$ is 2.

Example: The degree of $x+5$ (or $x^1+5x^0$) is 1.

Example: The degree of a constant polynomial like 5 (or $5x^0$) is 0, because the highest exponent of the variable (which is implicitly $x^0$) is 0.

Example: The degree of the term $2xy$ in a polynomial of two variables is the sum of the exponents of the variables in that term, which is $1+1=2$. However, for now, we are focusing on polynomials in one variable.

The degree of the zero polynomial (the polynomial 0) is generally considered undefined, as 0 can be written as $0x^n$ for any $n$, or sometimes taken as $-\infty$ by convention in some contexts.

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Classification of Polynomials

Polynomials can be classified based on two main criteria:

1. Based on the number of terms:
   • Monomial: A polynomial that has exactly one non-zero term. Examples: $5x$, $-3y^2$, $10$, $ab$ (if considered as a polynomial in $a$ and $b$), $7u^6$.
   • Binomial: A polynomial that has exactly two non-zero terms. Examples: $x+1$, $y^2-4$, $2p^3+5$, $a^2+b^2$.
   • Trinomial: A polynomial that has exactly three non-zero terms. Examples: $x^2+2x+1$, $y^3-3y+5$, $a^2-ab+b^2$.
   • Polynomials with more than three terms do not have specific names based on the number of terms.
2. Based on the degree:
   • Linear Polynomial: A polynomial of degree 1. Examples: $2x+3$ (degree 1), $y$ (degree 1).
   • Quadratic Polynomial: A polynomial of degree 2. Examples: $x^2-5x+6$ (degree 2), $y^2$ (degree 2), $u^2-4$ (degree 2).
   • Cubic Polynomial: A polynomial of degree 3. Examples: $x^3-1$ (degree 3), $2y^3+y^2-4$ (degree 3), $p^3$.
   • A polynomial of degree 4 is sometimes called a biquadratic polynomial.

Understanding the classification and terminology of polynomials is essential before proceeding to operations and theorems related to them.

Value and Zero of a Polynomial

Think of a polynomial, like $p(x)$, as a machine. You give it an input number (a value for $x$), and it follows a set of instructions (the polynomial expression) to produce a specific output number. Finding this output is called finding the 'value of the polynomial'. A very special input value is one that makes the machine's output exactly zero. This special input is called a 'zero' of the polynomial.

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Value of a Polynomial

If $p(x)$ is a polynomial in the variable $x$, and $k$ is any real number, then the value of the polynomial $p(x)$ at $x=k$ is the result you get when you replace every instance of $x$ in the expression with the number $k$. This value is denoted by $p(k)$.

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Zero of a Polynomial

A zero of a polynomial $p(x)$ is a specific number, let's call it 'k', for which the value of the polynomial is 0. In other words, if you input 'k' into the polynomial machine, the output is 0.

A number 'k' is a zero of $p(x)$ if and only if $p(k) = 0$.

Answer:

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How to Find the Zero of a Polynomial

To find the zero(s) of any polynomial $p(x)$, you must follow one simple rule:

Set the polynomial expression equal to zero, and solve the resulting equation for the variable.

That is, you must find the solution(s) to the equation $p(x) = 0$.

Finding the Zero of a Linear Polynomial

A linear polynomial is of the form $p(x) = ax + b$, where $a \neq 0$. To find its zero, we solve the equation $ax+b=0$:

The zero of the linear polynomial $ax+b$ is always $x = -\frac{b}{a}$.

Using the Zero Product Rule for Higher Degree Polynomials

For polynomials of degree 2 or higher, we often use factoring and the Zero Product Rule. This rule is a fundamental concept in algebra.

Zero Product Rule: If the product of two or more factors is zero, then at least one of those factors must be equal to zero.

If $A \times B = 0$, then either $A = 0$ or $B = 0$ (or both are zero).

To use this, we first factor the polynomial expression. Then we set each factor equal to zero and solve.

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Number of Zeros of a Polynomial

The number of zeros a polynomial can have is limited by its degree.

• A non-zero polynomial of degree $n$ can have at most $n$ real zeros.
• A linear polynomial (degree 1) has exactly one zero.
• A quadratic polynomial (degree 2) has at most two zeros.
• A cubic polynomial (degree 3) has at most three zeros.

Special Cases:

A non-zero constant polynomial, like $p(x) = 5$, has a degree of 0. Its value is always 5 and can never be 0. Therefore, a non-zero constant polynomial has no zeros.

The zero polynomial, $p(x) = 0$, is a special case. Its value is always 0, no matter what value of $x$ is substituted. Thus, every real number is a zero of the zero polynomial.

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Polynomials with No Real Zeros

It is possible for a polynomial to have no real zeros. This means there is no real number you can substitute for the variable that will make the polynomial's value equal to zero.

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Zero of a Polynomial vs. Root of an Equation

The terms 'zero' and 'root' are very closely related and often used interchangeably.

• A zero is a property of a polynomial expression. It's the input value that makes the expression equal to 0.
• A root is a property of an equation. It's the solution to the equation.

For example:

We say that 2 is a zero of the polynomial $p(x) = x-2$.

We say that 2 is a root of the equation $x-2 = 0$.

Essentially, they represent the same number but are used in slightly different contexts. The zeros of a polynomial $p(x)$ are the roots (or solutions) of the polynomial equation $p(x)=0$.

Remainder Theorem

When we divide one polynomial by another, the process can sometimes be long and tedious. The Remainder Theorem provides a remarkable shortcut. It allows us to find the remainder of a division, without actually performing the full long division, but only when the divisor is a simple linear polynomial.

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The Division Algorithm for Polynomials

First, let's recall the division algorithm for numbers. When we divide 17 by 5, we get a quotient of 3 and a remainder of 2. We can write this as: $17 = (5 \times 3) + 2$. Notice the remainder (2) is less than the divisor (5).

A similar rule applies to polynomials. If we divide a polynomial $p(x)$ (the dividend) by a non-zero polynomial $g(x)$ (the divisor), we get a quotient polynomial $q(x)$ and a remainder polynomial $r(x)$ such that:

The important condition is that either the remainder is zero ($r(x) = 0$) or the degree of the remainder $r(x)$ is strictly less than the degree of the divisor $g(x)$.

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Statement of the Remainder Theorem

The theorem provides a simple way to find the remainder when our divisor is a linear polynomial of the form $(x-k)$.

Statement: Let $p(x)$ be any polynomial of degree one or more, and let 'k' be any real number. If $p(x)$ is divided by the linear polynomial $(x-k)$, then the remainder is $p(k)$.

In simple terms: to find the remainder, just plug the zero of the divisor into the dividend polynomial.

Proof of the Remainder Theorem

Given: A polynomial $p(x)$ (degree $\ge 1$) and a linear divisor $(x-k)$.

To Prove: The remainder upon dividing $p(x)$ by $(x-k)$ is equal to $p(k)$.

Proof:

From the Division Algorithm for polynomials, we know that when $p(x)$ is divided by $(x-k)$, we can find a quotient $q(x)$ and a remainder $r(x)$ such that:

The degree of the divisor $(x-k)$ is 1. According to the algorithm, the degree of the remainder $r(x)$ must be less than the degree of the divisor. Therefore, the degree of $r(x)$ must be less than 1, which means the degree of $r(x)$ is 0. A polynomial of degree 0 is simply a constant. Let's call this constant remainder 'R'.

So, our equation becomes:

This equation must be true for all values of $x$. To find the value of R, we can choose a convenient value for $x$. The most convenient value is the one that makes the term with $q(x)$ disappear. This happens when $(x-k)=0$, which is at $x=k$.

Let's substitute $x=k$ into equation (ii):

Since $(k-k) = 0$, the expression simplifies:

This shows that the constant remainder $R$ is exactly equal to the value of the polynomial at $x=k$. This proves the theorem.

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How to Apply the Remainder Theorem

To find the remainder when a polynomial $p(x)$ is divided by a linear polynomial:

1. Take the divisor and set it equal to zero to find its zero.
2. Substitute this zero value into the dividend polynomial $p(x)$.
3. The result of the calculation is the remainder.

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Factor Theorem

The Factor Theorem is a powerful tool in polynomial algebra that directly relates the zeros of a polynomial to its linear factors. It is a direct consequence of the Remainder Theorem and provides a convenient way to determine if a linear expression is a factor of a polynomial or to find possible zeros of a polynomial.

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Statement of the Factor Theorem

Let $p(x)$ be a polynomial of degree greater than or equal to 1, and let $a$ be any real number.

The Factor Theorem states two interconnected conditions:

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

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

These two parts mean that the statements "$p(a) = 0$" and "$(x-a)$ is a factor of $p(x)$" are equivalent for a linear factor $(x-a)$.

Proof of the Factor Theorem

The proof relies directly on the Remainder Theorem, which states that when a polynomial $p(x)$ is divided by $(x-a)$, the remainder is $p(a)$. According to the Division Algorithm, we can write:

By the Remainder Theorem, $r(x) = p(a)$. So, we have:

where $q(x)$ is the quotient.

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

Assume that $p(a) = 0$. Substitute $p(a)=0$ into equation (i):

This equation shows that $p(x)$ can be written as the product of $(x-a)$ and the polynomial $q(x)$. By definition of factors, if a polynomial can be expressed as a product of other polynomials, those polynomials are its factors. Therefore, $(x-a)$ is a factor of $p(x)$.

Part (i) is proven.

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

Assume that $(x-a)$ is a factor of $p(x)$. By the definition of factors, this means that $p(x)$ can be expressed as the product of $(x-a)$ and some other polynomial, say $q(x)$.

This equation holds true for all real values of $x$. Let's substitute $x=a$ into this equation:

Since $(a-a) = 0$, the right side of the equation becomes 0:

Thus, if $(x-a)$ is a factor of $p(x)$, then $p(a) = 0$.

Part (ii) is proven.

Both parts together complete the proof of the Factor Theorem.

Relationship between Zero and Factor

The Factor Theorem establishes a direct link between the zeros of a polynomial and its linear factors. A real number 'a' is a zero of a polynomial $p(x)$ if and only if $(x-a)$ is a factor of $p(x)$.

• Finding the zeros of a polynomial helps us find its linear factors.
• Finding the linear factors of a polynomial helps us find its zeros.

For example, if we find that $p(5)=0$, then by the Factor Theorem, $(x-5)$ is a factor of $p(x)$. If we know that $(x+3)$ is a factor of $p(x)$, then since $x+3 = x-(-3)$, by the Factor Theorem, $p(-3)=0$, meaning -3 is a zero of $p(x)$.

Applications of the Factor Theorem

The Factor Theorem is a powerful tool for:

• Determining if a given linear expression $(x-a)$ or $(x+a)$ is a factor of a polynomial $p(x)$ by simply checking if $p(a)$ or $p(-a)$ is equal to 0.
• Finding the zeros of a polynomial if one or more linear factors are known or can be guessed.
• Factorising polynomials, especially when combined with other techniques like polynomial division (if a factor is found, division by that factor yields a polynomial of lower degree, which might be easier to factorise).

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Algebraic Identities

An algebraic identity is an equality that holds true for all possible values of the variables involved. Think of them as the "golden rules" of algebra. This is a key distinction from an algebraic equation, which is only true for specific values. Identities are powerful tools used for two main purposes: to expand products quickly and to factorise complex expressions into simpler ones.

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Distinction between Identity and Equation

• An Equation is true for only certain values. For example, the equation $x+3=10$ is true only when $x=7$.
• An Identity is true for all values. For example, $(a+b)^2 = a^2 + 2ab + b^2$. No matter what numbers you choose for $a$ and $b$, this statement will always be true.

Standard Algebraic Identities

Here are the fundamental algebraic identities. Understanding their proofs helps in remembering and applying them correctly. Let the variables $x, y, z$ represent any real numbers.

1. Identity I: Square of a Sum

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

   ... (I)

   Proof (Algebraic):

   We expand the left side by multiplying $(x+y)$ by itself.

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

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

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

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

   (Since $xy=yx$)

   Proof (Geometric):

   Consider a square with side length $(x+y)$. Its total area is $(x+y)^2$. This square can be divided into four parts: a square of side $x$ (area $x^2$), a square of side $y$ (area $y^2$), and two rectangles of sides $x$ and $y$ (each with area $xy$). The total area is the sum of these parts: $x^2 + y^2 + xy + xy = x^2 + 2xy + y^2$.

   Example 1. Use an identity to calculate $103^2$.

   Answer:

   $103^2 = (100+3)^2$

   This is in the form $(x+y)^2$ with $x=100$ and $y=3$.

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

   $= 10000 + 600 + 9 = 10609$

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2. Identity II: Square of a Difference

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

   ... (II)

   Proof:

   We expand the left side.

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

   $= x(x-y) - y(x-y)$

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

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

   Example 2. Expand $(3p-q)^2$.

   Answer:

   This is in the form $(x-y)^2$ with $x=3p$ and $y=q$.

   $(3p-q)^2 = (3p)^2 - 2(3p)(q) + (q)^2 = 9p^2 - 6pq + q^2$

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3. Identity III: Difference of Squares

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

   ... (III)

   Proof:

   We expand the right side.

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

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

   $= x^2 - y^2$

   (Since $-xy+yx=0$)

   Example 3. Use an identity to calculate $102 \times 98$.

   Answer:

   $102 \times 98 = (100+2)(100-2)$

   This is in the form $(x+y)(x-y)$ with $x=100$ and $y=2$.

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

   $= 10000 - 4 = 9996$

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4. Identity IV: Product of Binomials

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

   ... (IV)

   Proof:

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

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

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

   (Factoring $x$ from middle terms)

   Example 4. Find the product of $(m+3)(m-5)$.

   Answer:

   Here, $x=m, a=3, b=-5$.

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

   $= m^2 - 2m - 15$

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5. Identity V: Square of a Trinomial

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

   ... (V)

   Proof:

   Let $(x+y) = A$. Then $(x+y+z)^2 = (A+z)^2$.

   Using Identity I: $(A+z)^2 = A^2 + 2Az + z^2$.

   Substitute back $A=(x+y)$:

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

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

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

   Example 5. Expand $(2a - 3b + c)^2$.

   Answer:

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

   Here $x=2a, y=-3b, z=c$.

   $(2a)^2 + (-3b)^2 + c^2 + 2(2a)(-3b) + 2(-3b)(c) + 2(c)(2a)$

   $= 4a^2 + 9b^2 + c^2 - 12ab - 6bc + 4ca$

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6. Identity VI: Cube of a Sum

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

   ... (VI)

   This is often also written as $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$.

   Proof:

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

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

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

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

   Example 6. Expand $(2a+b)^3$.

   Answer:

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

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

   $= 8a^3 + 3(4a^2)b + 6ab^2 + b^3$

   $= 8a^3 + 12a^2b + 6ab^2 + b^3$

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7. Identity VII: Cube of a Difference

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

   ... (VII)

   This is often also written as $(x-y)^3 = x^3 - y^3 - 3xy(x-y)$.

   Proof:

   We can prove this identity in two ways. One is by direct expansion, and the other is by using the identity for $(x+y)^3$.

   Method 1: Direct Expansion

   We expand the left side by writing it as a product and using the identity for $(x-y)^2$.

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

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

   Now, multiply the binomial by the trinomial:

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

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

   Distribute the negative sign and combine like terms:

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

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

   Method 2: Using Identity VI

   We can prove this by replacing $y$ with $-y$ in the identity for the cube of a sum, $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$.

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

   Now, simplify each term:

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

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

   Example 7. Use an identity to calculate $99^3$.

   Answer:

   $99^3 = (100-1)^3$. Here, $x=100, y=1$.

   Using the identity $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$:

   $(100-1)^3 = (100)^3 - 3(100)^2(1) + 3(100)(1)^2 - (1)^3$

   $ = 1000000 - 3(10000) + 3(100) - 1$

   $ = 1000000 - 30000 + 300 - 1$

   $ = 970000 + 300 - 1$

   $ = 970300 - 1 = 970299$

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8. Identity VIII & IX: Sum and Difference of Cubes

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

   ... (VIII)

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

   ... (IX)

   These are crucial for factorising. A helpful mnemonic for the signs is SOAP: Same sign (in the binomial), Opposite sign (in the trinomial), Always Positive (the last term).

   Proof of Identity VIII ($x^3+y^3$):

   We start by multiplying the right-hand side (RHS) and show that it simplifies to the left-hand side (LHS).

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

   Distribute each term from the first bracket over the second bracket:

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

   Expand the products:

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

   Group the like terms together:

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

   The middle terms cancel out:

   $= x^3 + 0 + 0 + y^3 = x^3 + y^3 =$ LHS

   Proof of Identity IX ($x^3-y^3$):

   Similarly, we expand the right-hand side.

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

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

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

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

   $= x^3 + 0 + 0 - y^3 = x^3 - y^3 =$ LHS

   Example 8. Factorise $8a^3 + 27b^3$.

   Answer:

   This is a sum of cubes: $(2a)^3 + (3b)^3$. Here, $x=2a, y=3b$.

   Using Identity VIII: $(x+y)(x^2 - xy + y^2)$

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

   $= (2a+3b)(4a^2 - 6ab + 9b^2)$

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9. Identity X: The Sum of Three Cubes

   $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$

   ... (X)

   Proof:

   We prove this identity by expanding the more complex right-hand side (RHS).

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

   First, distribute $x$ over the second bracket:

   $x(x^2 + y^2 + z^2 - xy - yz - zx) = x^3 + xy^2 + xz^2 - x^2y - xyz - x^2z$

   Next, distribute $y$ over the second bracket:

   $y(x^2 + y^2 + z^2 - xy - yz - zx) = x^2y + y^3 + yz^2 - xy^2 - y^2z - xyz$

   Finally, distribute $z$ over the second bracket:

   $z(x^2 + y^2 + z^2 - xy - yz - zx) = x^2z + y^2z + z^3 - xyz - yz^2 - xz^2$

   Now, we add all these results together and cancel out pairs of terms:

   $x^3 + y^3 + z^3$

   $+ xy^2 - xy^2$ (cancels)

   $+ xz^2 - xz^2$ (cancels)

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

   $- x^2z + x^2z$ (cancels)

   $+ yz^2 - yz^2$ (cancels)

   $- y^2z + y^2z$ (cancels)

   $- xyz - xyz - xyz = -3xyz$

   Combining the remaining terms, we get:

   $x^3 + y^3 + z^3 - 3xyz$, which is the LHS.

   Special Case: Notice from Identity X that if the first factor on the right is zero, the whole expression is zero. That is, if $x+y+z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.

   Example 9. Without calculating the cubes, find the value of $(-12)^3 + (7)^3 + (5)^3$.

   Answer:

   This is in the form $x^3+y^3+z^3$ with $x=-12, y=7, z=5$.

   First, check the sum of the bases: $x+y+z = -12 + 7 + 5 = -12 + 12 = 0$.

   Since the sum is 0, we can use the special case identity: $x^3 + y^3 + z^3 = 3xyz$.

   $(-12)^3 + (7)^3 + (5)^3 = 3(-12)(7)(5)$

   $= -36 \times 35$

   $= -1260$

Factorising Polynomials

In the previous sections, we learned how to multiply polynomials to get a single, expanded expression. For example, we know that $(x+2)(x+3) = x^2+5x+6$. Factorisation is the exact reverse process. It is the art of taking an expanded polynomial and breaking it down into a product of simpler polynomials (its factors).

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What is Factorisation?

Think about the number 12. We can write 12 as a product of its factors, for example, $2 \times 6$ or $3 \times 4$. Similarly, factorising a polynomial means rewriting it as a product of polynomials of lower degree.

Example: The polynomial $x^2 - 4$ can be written as the product $(x-2)(x+2)$. So, $(x-2)$ and $(x+2)$ are the factors of $x^2 - 4$.

Factorisation is a crucial skill because it helps us to:

• Simplify complex algebraic expressions.
• Solve polynomial equations by finding their zeros.
• Work with algebraic fractions.

Methods of Factorisation

The method you choose depends on the structure of the polynomial you are trying to factorise. Here are the most common methods.

1. Factorising by Taking Out a Common Factor

   This is the simplest method and should always be the first thing you check. If every term in the polynomial shares a common factor, you can "pull it out" using the distributive property in reverse.

   Steps:

   1. Find the Greatest Common Factor (GCF) of the numerical coefficients.
   2. Find the highest power of each variable that is common to all terms.
   3. The common factor is the product of these.
   4. Write the common factor outside a set of brackets, and inside the brackets, write what is left from each term after dividing by the common factor.

   Example 1. Factorise $5x^2 + 20x$.

   Answer:

   a) The GCF of the coefficients 5 and 20 is 5.

   b) The terms are $x^2$ and $x$. The highest power of $x$ common to both is $x^1$ or $x$.

   c) The common factor is $5x$.

   d) We write $5x(\dots)$. To find what goes inside, we divide each original term by $5x$:

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

   $\frac{20x}{5x} = 4$

   So, $5x^2 + 20x = 5x(x+4)$. The factors are $5x$ and $(x+4)$.

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2. Factorising by Grouping Terms

   This method is typically used for polynomials with four terms when there is no single factor common to all four. The strategy is to create groups of terms that have their own common factors.

   Steps:

   1. Arrange the four terms into two pairs.
   2. Factor out the common factor from the first pair.
   3. Factor out the common factor from the second pair.
   4. If the binomials in the brackets are identical, factor out this common binomial.

   Example 2. Factorise $ax + by + bx + ay$.

   Answer:

   a) First, let's rearrange the terms so that terms with common factors are next to each other: $(ax+bx) + (ay+by)$.

   b) Factor out $x$ from the first pair: $x(a+b)$.

   c) Factor out $y$ from the second pair: $y(a+b)$.

   The expression is now $x(a+b) + y(a+b)$.

   d) The binomial $(a+b)$ is now a common factor. We can factor it out:

   $(a+b)(x+y)$.

   The factors are $(a+b)$ and $(x+y)$.

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3. Factorising using Algebraic Identities

   If a polynomial has a special form that matches the expanded side of one of the algebraic identities, you can use that identity in reverse to factorise it instantly.

   Difference of Squares: $x^2 - y^2 = (x+y)(x-y)$

   Example 3. Factorise $p^2 - 49$.

   Answer:

   Recognise this as a difference between two perfect squares: $(p)^2 - (7)^2$.

   This matches the form $x^2 - y^2$ with $x=p$ and $y=7$.

   Using the identity, $p^2 - 49 = (p+7)(p-7)$.

   Perfect Square Trinomials: $x^2 + 2xy + y^2 = (x+y)^2$ and $x^2 - 2xy + y^2 = (x-y)^2$

   Example 4. Factorise $9a^2 - 30ab + 25b^2$.

   Answer:

   Check if it fits the perfect square trinomial pattern.

   The first term is a perfect square: $9a^2 = (3a)^2$.

   The last term is a perfect square: $25b^2 = (5b)^2$.

   The middle term has a minus sign, so we check if it matches $-2xy$.

   Let $x=3a$ and $y=5b$. Then $-2xy = -2(3a)(5b) = -30ab$. This matches the middle term.

   So, the polynomial is a perfect square trinomial of the form $(x-y)^2$.

   $9a^2 - 30ab + 25b^2 = (3a - 5b)^2$.

Other important factorisation methods include "Splitting the Middle Term" for quadratic trinomials and using the "Factor Theorem" for higher-degree polynomials. These are powerful techniques that will be covered in the next sections.

Factorisation of Quadratic Polynomials

A quadratic polynomial is a polynomial of degree 2. In one variable, say $x$, its general form is $ax^2 + bx + c$, where $a, b,$ and $c$ are constants and $a \neq 0$. To factorise a quadratic polynomial means to break it down into a product of its factors, which are typically two linear (degree 1) polynomials.

While some quadratics can be factorised using identities (like $x^2-9 = (x-3)(x+3)$), we will focus on two systematic methods that work for general trinomials: splitting the middle term and using the Factor Theorem.

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Method 1: Factorising by Splitting the Middle Term

This powerful technique involves rewriting the middle term ($bx$) as a sum or difference of two new terms. This transforms the three-term polynomial into a four-term polynomial, which can then be factorised by grouping.

The goal is to find two special numbers, let's call them $p$ and $q$, that help us "split" the middle term. These numbers must satisfy two conditions:

• Their sum must be equal to $b$ (the coefficient of the middle term).
• Their product must be equal to $ac$ (the product of the first and last coefficients).

$p + q = b$ and $p \times q = ac$

Quick Guide to Finding the Numbers (p and q)

Finding the two numbers is the key step. This table will help you determine the signs of the numbers you are looking for:

If the product `ac` is...	And the sum `b` is...	Then the two numbers `p` and `q` are...
Positive (+)	Positive (+)	Both positive
	Negative (−)	Both negative
Negative (−)	Positive (+)	One positive, one negative (the larger number is positive)
	Negative (−)	One positive, one negative (the larger number is negative)

Answer:

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Method 2: Factorising using the Factor Theorem

The Factor Theorem provides an alternative method that relies on finding the zeros of the polynomial. The core principle is: If $k$ is a zero of a polynomial $p(x)$, then $(x-k)$ is a factor of $p(x)$.

To find possible zeros, we use the Rational Root Theorem. For a polynomial $ax^2+bx+c$, any rational zero must be of the form $\frac{p}{q}$, where:

• $p$ is a factor of the constant term, $c$.
• $q$ is a factor of the leading coefficient, $a$.

Answer:

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While the Factor Theorem is very powerful, especially for cubic and higher-degree polynomials, the "splitting the middle term" method is often quicker and more direct for factorising quadratic trinomials.

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Factorisation of Cubic Polynomials

A cubic polynomial is a polynomial of degree 3, with the general form $ax^3 + bx^2 + cx + d$. Factorising a cubic polynomial means rewriting it as a product of simpler polynomials, which are typically one linear factor and one quadratic factor, or sometimes three linear factors.

While some special cubic polynomials can be factorised using identities (like $x^3+y^3$), the most reliable and general method involves using the Factor Theorem combined with polynomial long division.

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Method: Factorising using the Factor Theorem and Division

This method is a systematic, step-by-step process. The core idea is to find one linear factor by "guessing" a zero, and then use division to find the remaining quadratic factor, which is easier to handle.

Steps to factorise a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$:

1. Find one linear factor using the Factor Theorem.
   • Identify possible zeros: According to the Rational Root Theorem, any rational zero of the polynomial must be of the form $\frac{\text{factor of constant term } d}{\text{factor of leading coefficient } a}$.
   • Test the possibilities: Start by testing simple integer factors of the constant term (like $\pm 1, \pm 2$, etc.) by substituting them into $p(x)$.
   • Find a zero: If you find a number 'k' such that $p(k) = 0$, then you have found a zero. By the Factor Theorem, $(x-k)$ is a factor of $p(x)$.
2. Divide the cubic polynomial by the linear factor.
   • Use polynomial long division to divide $p(x)$ by the factor $(x-k)$ you found in Step 1.
   • Since $(x-k)$ is a factor, the remainder of this division must be 0.
   • The result of the division (the quotient) will be a quadratic polynomial.
3. Factorise the resulting quadratic polynomial.
   • Take the quadratic quotient from Step 2 and factorise it completely using any appropriate method (e.g., splitting the middle term or using identities).
4. Write the final factorisation.
   • The complete factorisation of the cubic polynomial is the product of the linear factor from Step 1 and the factors of the quadratic from Step 3.

Answer:

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