# 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.