1.5.6 Polynomials and rational expressions

What you'll learn

How to expand, collect and factorise polynomial expressions.
How to divide one polynomial by a linear expression.
How the factor theorem helps you find factors of polynomials.
How to simplify rational expressions safely, including by algebraic division.

Polynomial language

A term is a part of an expression separated from the others by addition or subtraction. For example, $4x^3-2x+7$ has the terms $4x^3$ , $- 2 x$ and 7.

The numerical multiplier of a term is its coefficient, so the coefficient of $x^3$ above is 4. The constant term is the term without a variable, which is 7.

Definition

Polynomial

A polynomial in $x$ is a finite sum of terms of the form $ax^n$ , where $a$ is a constant and $n$ is a non-negative integer.

For example, $3x^4-2x^2+x-5$ is a polynomial. Expressions involving $x^{-1}$ , $x\sqrt{x}$ or a variable in a denominator are not polynomials.

The degree of a non-zero polynomial is its highest power of the variable. Therefore, $3x^4-2x^2+x-5$ has degree 4.

Two terms are like terms when they contain exactly the same variable raised to the same power. Only like terms can be collected.

Expanding and collecting terms

To expand brackets, multiply every term in one factor by every term in the other factor. You then collect any like terms.

Example

Expanding a product of polynomials

Expand and simplify $2x-3)(x^2+4x-5)$ .

Multiply $2 x$ by every term in the second bracket:

2x(x^2+4x-5)=2x^3+8x^2-10x.

Multiply $- 3$ by every term in the second bracket:

3(x^2+4x-5)=-3x^2-12x+15.

Add the results and collect like terms:

\begin{aligned} (2x-3)(x^2+4x-5) &=2x^3+8x^2-10x-3x^2-12x+15\\ &=2x^3+5x^2-22x+15. \end{aligned}

Common Mistake

Missing a term

When expanding, a negative multiplier must change the sign of every term in the bracket. Writing the polynomial in descending powers also makes missing terms easier to spot.

Factorising polynomials

To factorise an expression means to rewrite it as a product of factors.

Always check first for a common factor shared by every term. For example,

6x^3-9x^2=3x^2(2x-3).

You should also recognise standard quadratic patterns:

\begin{aligned} x^2+(a+b)x+ab&=(x+a)(x+b),\\ a^2-b^2&=(a-b)(a+b). \end{aligned}

For a quadratic $ax^2+bx+c$ with $a≠1a\neq1$ , you need factors whose products produce both $ax^2$ and $c$ , while the two middle terms combine to give $b x$ .

Example

Factorising a quadratic

Factorise $6x^2+x-2$ .

Multiply the coefficient of $x^2$ by the constant: $6 (- 2) = - 12$ . Find two numbers with product $- 12$ and sum 1: these are 4 and $- 3$ .
Split the middle term using these numbers:

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

Factorise the two pairs:

\begin{aligned} 6x^2+4x-3x-2 &=2x(3x+2)-1(3x+2)\\ &=(2x-1)(3x+2). \end{aligned}

Tip

Check by expanding

After factorising, multiply your factors back out. This quickly detects incorrect signs or coefficients.

Polynomial division

Polynomial division rewrites a polynomial as

\text{dividend}=(\text{divisor})(\text{quotient})+\text{remainder}.

The dividend is the polynomial being divided, the divisor is what you divide by, the quotient is the result, and the remainder is what is left over.

When dividing by a linear expression such as $x - a$ , the remainder is a constant.

Example

Dividing by a linear expression

Divide $2x^3+3x^2-11x-6$ by $x + 3$ .

Divide the leading terms: $2x3÷x=2x22x^3\div x=2x^2$ . Multiply back and subtract:

\begin{aligned} &(2x^3+3x^2)-(2x^3+6x^2)\\ &=-3x^2. \end{aligned}

Bring down the next term to obtain $3x^2-11x$ .

Divide again: $−3x2÷x=−3x-3x^2\div x=-3x$ . Multiply $x + 3$ by $- 3 x$ , then subtract:

3x^2-11x)-(-3x^2-9x)=-2x.

Bring down the constant to obtain $- 2 x - 6$ .

Divide once more: $−2x÷x=−2-2x\div x=-2$ . Since $- 2 (x + 3) = - 2 x - 6$ , the remainder is zero. Therefore,

\frac{2x^3+3x^2-11x-6}{x+3}=2x^2-3x-2.

Key Idea

Keeping powers aligned

Write polynomials in descending powers before dividing. If a power is missing, include a zero term mentally or in your layout, such as $x^3+0x^2-4x+1$ .

The factor theorem

Suppose $f (x)$ is a polynomial.

Definition

Factor theorem

The linear expression $x - a$ is a factor of $f (x)$ if and only if $f (a) = 0$ .

This works because division by $x - a$ leaves the remainder $f (a)$ . Therefore, obtaining zero shows that the division is exact.

Be careful with signs: to test whether $x + 2$ is a factor, write it as $x - (- 2)$ and calculate $f (- 2)$ .

Example

Factorising a cubic using the factor theorem

Factorise $f(x)=x^3-2x^2-5x+6$ fully.

Possible integer values to test come from the factors of the constant term 6. Testing $x = 1$ gives

f (1) = 1 - 2 - 5 + 6 = 0,

so $x - 1$ is a factor.

Divide $f (x)$ by $x - 1$ . Polynomial division gives

f(x)=(x-1)(x^2-x-6).

Factorise the quadratic:

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

Hence,

f (x) = (x - 1) (x - 3) (x + 2) .

Tip

Choosing values to test

For a polynomial with integer coefficients and leading coefficient 1, integer roots must be factors of the constant term. This narrows the search, although you must still verify each candidate by substitution.

Rational expressions

Definition

Rational expression

A rational expression is a fraction whose numerator and denominator are polynomials, such as $x2−9x2+x−6\frac{x^2-9}{x^2+x-6}$ .

A rational expression is undefined whenever its original denominator equals zero. These excluded values must remain excluded even if a factor later cancels.

Simplifying by factorising

You may cancel a common factor from the numerator and denominator. You cannot cancel individual terms joined by addition or subtraction.

Example

Simplifying a rational expression

Simplify $x2−9x2+x−6\frac{x^2-9}{x^2+x-6}$ .

Factorise both polynomials:

\frac{x^2-9}{x^2+x-6} =\frac{(x-3)(x+3)}{(x+3)(x-2)}.

Cancel the common factor $x + 3$ :

\frac{x^2-9}{x^2+x-6}=\frac{x-3}{x-2}.

Use the original denominator to identify excluded values. Since $(x+3)(x−2)≠0(x+3)(x-2)\neq0$ , the restrictions are $x≠−3x\neq-3$ and $x≠2x\neq2$ .

Common Mistake

Cancelling terms instead of factors

In $x+3x\frac{x+3}{x}$ , the $x$ cannot be cancelled because $x + 3$ is a sum, not a product containing an $x$ factor.

Algebraic division of rational expressions

A rational expression is improper when the degree of its numerator is at least as large as the degree of its denominator. Divide to rewrite it as a polynomial plus a proper rational expression.

Example

Rewriting an improper rational expression

Express $2x2+5x−1x+2\frac{2x^2+5x-1}{x+2}$ as a polynomial plus a proper fraction.

Divide the leading terms: $2x2÷x=2x2x^2\div x=2x$ . Subtract $2x(x+2)=2x^2+4x$ from the numerator, leaving $x - 1$ .
Divide again: $x÷x=1x\div x=1$ . Subtract $x + 2$ , leaving the remainder $- 3$ .
Combine the quotient and remainder:

\frac{2x^2+5x-1}{x+2}=2x+1-\frac{3}{x+2},\qquad x\neq-2.

Key Idea

Division identity

Your final result can always be checked using

\text{numerator}=(\text{denominator})(\text{quotient})+\text{remainder}.

The remainder must have lower degree than the denominator.

Exam technique

In the exam

Put polynomials in descending powers, and account for any missing powers before dividing.
Factorise fully before cancelling, and cancel only complete factors.
When using the factor theorem, match signs carefully: the factor $x - a$ corresponds to testing $f (a)$ .
State values excluded by the original denominator, even when the corresponding factor cancels.
Check factorisations by expanding and divisions by multiplying the divisor and quotient, then adding the remainder.

Self review

Check yourself

Can you factorise $2x^3-5x^2-18x+45$ fully?
How would you test whether $x + 3$ is a factor of a polynomial $f (x)$ ?
Can you simplify $x2−4x2−x−6\frac{x^2-4}{x^2-x-6}$ and state every excluded value?

What you'll learn

How to expand, collect and factorise polynomial expressions.
How to divide one polynomial by a linear expression.
How the factor theorem helps you find factors of polynomials.
How to simplify rational expressions safely, including by algebraic division.

Polynomial language

A term is a part of an expression separated from the others by addition or subtraction. For example, $4x^3-2x+7$ has the terms $4x^3$ , $- 2 x$ and 7.

The numerical multiplier of a term is its coefficient, so the coefficient of $x^3$ above is 4. The constant term is the term without a variable, which is 7.

Definition

Polynomial

A polynomial in $x$ is a finite sum of terms of the form $ax^n$ , where $a$ is a constant and $n$ is a non-negative integer.

For example, $3x^4-2x^2+x-5$ is a polynomial. Expressions involving $x^{-1}$ , $x\sqrt{x}$ or a variable in a denominator are not polynomials.

The degree of a non-zero polynomial is its highest power of the variable. Therefore, $3x^4-2x^2+x-5$ has degree 4.

Two terms are like terms when they contain exactly the same variable raised to the same power. Only like terms can be collected.

Expanding and collecting terms

To expand brackets, multiply every term in one factor by every term in the other factor. You then collect any like terms.

Example

Expanding a product of polynomials

Expand and simplify $2x-3)(x^2+4x-5)$ .

Multiply $2 x$ by every term in the second bracket:

2x(x^2+4x-5)=2x^3+8x^2-10x.

Multiply $- 3$ by every term in the second bracket:

3(x^2+4x-5)=-3x^2-12x+15.

Add the results and collect like terms:

\begin{aligned} (2x-3)(x^2+4x-5) &=2x^3+8x^2-10x-3x^2-12x+15\\ &=2x^3+5x^2-22x+15. \end{aligned}

Common Mistake

Missing a term

When expanding, a negative multiplier must change the sign of every term in the bracket. Writing the polynomial in descending powers also makes missing terms easier to spot.

Factorising polynomials

To factorise an expression means to rewrite it as a product of factors.

Always check first for a common factor shared by every term. For example,

6x^3-9x^2=3x^2(2x-3).

You should also recognise standard quadratic patterns:

\begin{aligned} x^2+(a+b)x+ab&=(x+a)(x+b),\\ a^2-b^2&=(a-b)(a+b). \end{aligned}

For a quadratic $ax^2+bx+c$ with $a≠1a\neq1$ , you need factors whose products produce both $ax^2$ and $c$ , while the two middle terms combine to give $b x$ .

Example

Factorising a quadratic

Factorise $6x^2+x-2$ .

Multiply the coefficient of $x^2$ by the constant: $6 (- 2) = - 12$ . Find two numbers with product $- 12$ and sum 1: these are 4 and $- 3$ .
Split the middle term using these numbers:

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

Factorise the two pairs:

\begin{aligned} 6x^2+4x-3x-2 &=2x(3x+2)-1(3x+2)\\ &=(2x-1)(3x+2). \end{aligned}

Tip

Check by expanding

After factorising, multiply your factors back out. This quickly detects incorrect signs or coefficients.

Polynomial division

Polynomial division rewrites a polynomial as

\text{dividend}=(\text{divisor})(\text{quotient})+\text{remainder}.

The dividend is the polynomial being divided, the divisor is what you divide by, the quotient is the result, and the remainder is what is left over.

When dividing by a linear expression such as $x - a$ , the remainder is a constant.

Example

Dividing by a linear expression

Divide $2x^3+3x^2-11x-6$ by $x + 3$ .

Divide the leading terms: $2x3÷x=2x22x^3\div x=2x^2$ . Multiply back and subtract:

\begin{aligned} &(2x^3+3x^2)-(2x^3+6x^2)\\ &=-3x^2. \end{aligned}

Bring down the next term to obtain $3x^2-11x$ .

Divide again: $−3x2÷x=−3x-3x^2\div x=-3x$ . Multiply $x + 3$ by $- 3 x$ , then subtract:

3x^2-11x)-(-3x^2-9x)=-2x.

Bring down the constant to obtain $- 2 x - 6$ .

Divide once more: $−2x÷x=−2-2x\div x=-2$ . Since $- 2 (x + 3) = - 2 x - 6$ , the remainder is zero. Therefore,

\frac{2x^3+3x^2-11x-6}{x+3}=2x^2-3x-2.

Key Idea

Keeping powers aligned

Write polynomials in descending powers before dividing. If a power is missing, include a zero term mentally or in your layout, such as $x^3+0x^2-4x+1$ .

The factor theorem

Suppose $f (x)$ is a polynomial.

Definition

Factor theorem

The linear expression $x - a$ is a factor of $f (x)$ if and only if $f (a) = 0$ .

This works because division by $x - a$ leaves the remainder $f (a)$ . Therefore, obtaining zero shows that the division is exact.

Be careful with signs: to test whether $x + 2$ is a factor, write it as $x - (- 2)$ and calculate $f (- 2)$ .

Example

Factorising a cubic using the factor theorem

Factorise $f(x)=x^3-2x^2-5x+6$ fully.

Possible integer values to test come from the factors of the constant term 6. Testing $x = 1$ gives

f (1) = 1 - 2 - 5 + 6 = 0,

so $x - 1$ is a factor.

Divide $f (x)$ by $x - 1$ . Polynomial division gives

f(x)=(x-1)(x^2-x-6).

Factorise the quadratic:

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

Hence,

f (x) = (x - 1) (x - 3) (x + 2) .

Tip

Choosing values to test

Rational expressions

Definition

Rational expression

A rational expression is a fraction whose numerator and denominator are polynomials, such as $x2−9x2+x−6\frac{x^2-9}{x^2+x-6}$ .

A rational expression is undefined whenever its original denominator equals zero. These excluded values must remain excluded even if a factor later cancels.

Simplifying by factorising

You may cancel a common factor from the numerator and denominator. You cannot cancel individual terms joined by addition or subtraction.

Example

Simplifying a rational expression

Simplify $x2−9x2+x−6\frac{x^2-9}{x^2+x-6}$ .

Factorise both polynomials:

\frac{x^2-9}{x^2+x-6} =\frac{(x-3)(x+3)}{(x+3)(x-2)}.

Cancel the common factor $x + 3$ :

\frac{x^2-9}{x^2+x-6}=\frac{x-3}{x-2}.

Use the original denominator to identify excluded values. Since $(x+3)(x−2)≠0(x+3)(x-2)\neq0$ , the restrictions are $x≠−3x\neq-3$ and $x≠2x\neq2$ .

Common Mistake

Cancelling terms instead of factors

In $x+3x\frac{x+3}{x}$ , the $x$ cannot be cancelled because $x + 3$ is a sum, not a product containing an $x$ factor.

Algebraic division of rational expressions

A rational expression is improper when the degree of its numerator is at least as large as the degree of its denominator. Divide to rewrite it as a polynomial plus a proper rational expression.

Example

Rewriting an improper rational expression

Express $2x2+5x−1x+2\frac{2x^2+5x-1}{x+2}$ as a polynomial plus a proper fraction.

Divide the leading terms: $2x2÷x=2x2x^2\div x=2x$ . Subtract $2x(x+2)=2x^2+4x$ from the numerator, leaving $x - 1$ .
Divide again: $x÷x=1x\div x=1$ . Subtract $x + 2$ , leaving the remainder $- 3$ .
Combine the quotient and remainder:

\frac{2x^2+5x-1}{x+2}=2x+1-\frac{3}{x+2},\qquad x\neq-2.

Key Idea

Division identity

Your final result can always be checked using

\text{numerator}=(\text{denominator})(\text{quotient})+\text{remainder}.

The remainder must have lower degree than the denominator.

Exam technique

In the exam

Put polynomials in descending powers, and account for any missing powers before dividing.
Factorise fully before cancelling, and cancel only complete factors.
When using the factor theorem, match signs carefully: the factor $x - a$ corresponds to testing $f (a)$ .
State values excluded by the original denominator, even when the corresponding factor cancels.
Check factorisations by expanding and divisions by multiplying the divisor and quotient, then adding the remainder.

Self review

Check yourself

Can you factorise $2x^3-5x^2-18x+45$ fully?
How would you test whether $x + 3$ is a factor of a polynomial $f (x)$ ?
Can you simplify $x2−4x2−x−6\frac{x^2-4}{x^2-x-6}$ and state every excluded value?

What you'll learn

Polynomial language

Expanding and collecting terms

Factorising polynomials

Polynomial division

The factor theorem

Rational expressions

Simplifying by factorising

Algebraic division of rational expressions

Review 1.5.6 Polynomials and rational expressions by teaching Genie

1.5.6 Polynomials and rational expressions

What you'll learn

Polynomial language

Expanding and collecting terms

Factorising polynomials

Polynomial division

The factor theorem

Rational expressions

Simplifying by factorising

Algebraic division of rational expressions

Review 1.5.6 Polynomials and rational expressions by teaching Genie

1.5.6 Polynomials and rational expressions

What you'll learn

Polynomial language

Expanding and collecting terms

Factorising polynomials

Polynomial division

The factor theorem

Rational expressions

Simplifying by factorising

Algebraic division of rational expressions

Review 1.5.6 Polynomials and rational expressions by teaching Genie

1.5.6 Polynomials and rational expressions Revision Guide

1.5.6 Polynomials and rational expressions

What you'll learn

Polynomial language

Expanding and collecting terms

Factorising polynomials

Polynomial division

The factor theorem

Rational expressions

Simplifying by factorising

Algebraic division of rational expressions

Review 1.5.6 Polynomials and rational expressions by teaching Genie

1.5.6 Polynomials and rational expressions Revision Guide