# Factorising quadratic expressions

## Introduction

When **factorising** quadratic expressions you will need two pairs of brackets.
A **quadratic** expression always contains an \(x^2\) term.

Before trying to factorise quadratic expressions you should first make sure you can expand and simplify double brackets and factorise linear expressions.

## How to factorise quadratic expressions

### Example 1 - Factorising a quadratic expression

a) Factorise \(x^2 + 5x + 6\).

We need to find two numbers that multiply together to give \(6\) and add together to give \(5\). In this case \(2 × 3 = 6\) and \(2 + 3 = 5\).

So \(2\) and \(3\) are the numbers we put into double brackets.

\(x^2 + 5x + 6 = (x + 2)(x + 3)\).

You can always check your answer by expanding and simplifying the brackets to make sure you arrive at your original expression.

b) Factorise \(x^2 + 3x - 10\).

This time we need to find two numbers with a product of \(-10\) and a sum of \(3\). One of the numbers must be negative.

\(5 × -2 = -10\) and \(5 + (-2) = 3\).

Putting \(5\) and \(-2\) into double brackets gives:

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

c) Factorise \(x^2 - 4x - 32\).

Find two numbers with a product of \(-32\) and a sum of \(-4\).

\(4 × -8 = -32\) and \(4 + (-8) = -4\).

\(x^2 - 4x - 32 = (x + 4)(x - 8)\).

d) Factorise \(x^2 - 13x + 42\).

The only way to get two numbers with a product of \(42\) and a sum of \(-13\) is if both numbers are negative.

\(-6 × -7 = 42\) and \(-6 + (-7) = -13\).

\(x^2 - 13x + 42 = (x - 6)(x - 7)\).

### Example 2 - Factorising a harder quadratic expression

a) Factorise \(2x^2 + 13x + 15\).

When the coefficient of the \(x^2\) term is not one there are more possibilities to try.

Given a \(2x^2\) term the solution must be of the form \((2x + a)(x + b)\).

\(2x^2 + 13x + 15 = (2x + 3)(x + 5)\).

a) Factorise \(6x^2 + 14x - 12\).

Given a \(6x^2\) term the solution must be of the form \((6x + a)(x + b)\) or \((3x + a)(2x + b)\).

Also, since \(a × b = -12\) we have six possible pairs of values to test:

- \(-12\) and \(1\)
- \(12\) and \(-1\)
- \(-2\) and \(6\)
- \(2\) and \(-6\)
- \(-3\) and \(4\)
- \(3\) and \(-4\)

Testing these gives: \(6x^2 + 14x - 12 = (3x - 2)(2x + 6)\).

### Example 3 - Difference of two squares

a) Factorise \(a^2 - b^2\).

When factorising a difference of two square terms the second bracket will be the same as the first except a minus sign. This is called a conjugate factor.

The linear terms of the expression cancel each other out when expanded.

\(a^2 - b^2 = (a + b)(a - b)\).

b) Factorise \(x^2 - 25\).

The difference of two squares method works for number and letter terms:

\(x^2 - 25 = (x + 5)(x - 5)\).

c) Factorise \(4x^2 - 36y^2\).

For combinations of square numbers and letter terms, the method stays the same:

\(4x^2 - 36y^2 = (2x + 6y)(2x - 6y)\).

## Worksheets to practise factorising linear and quadratic expressions

Try these worksheets to practise your skills.