Factorising quadratics

In a nutshell

A quadratic in the form $$ax^2+bx+c$$ can be factorised. You have already learnt how to factorise a quadratic when $$a=1$$ by finding two numbers to multiply to give $$c$$ and add to give $$b$$​. This lesson will also discuss how to factorise a quadratic expression where $$a>1$$.

Factorise when $$a=1$$​​

A quadratic is in the form

$$ax^2 + bx +c$$​​

When $$a=1$$ we can factorise a quadratic as follows:

PROCEDURE

Example 1

Factorise 

​$$x^2 + 10x + 24$$​

The quadratic is in the correct form where $$b=10$$ and $$c=24$$.

​$$6 \times 4 = 24 \space and \space 6 + 4 = 10$$, so the numbers are $$6 \space and \space 4$$.

​$$x^2 + 10x+24 = \underline{(x+6)(x+4)}$$

​

Factorise when $$a>1$$

​When $$a>1$$ you can factorise a quadratic as follows:

PROCEDURE

Example 2

Factorise

$$2x^2+13x+15$$​​

The quadratic has $$a=2, b=13$$ and $$c=15$$. Multiply $$a$$ and $$c$$ to give $$2 \times 15 = 30$$.

Think of two numbers which will multiply to give $$30$$ and add to give $$13$$. The two numbers are $$10$$ and $$3$$, as $$10 \times 3 = 30$$ and $$10+3 = 13$$.

Write the quadratic expression using $$10$$ and $$3$$, and factorise.

$$\begin {aligned}&2x^2+13x+15 \\&\underbrace{2x^2 + 10x}_{\text{factorise}} + \underbrace{3x + 15}_{\text{factorise}} \\&2x(x+5) + 3(x+5)\end {aligned}$$​​

The two expressions inside both single brackets should match, in this case $$x+5$$.

The final answer consists of two brackets, the first bracket contains the terms on the outside of each bracket, $$2x$$ and $$+3$$. And the second bracket is the repeated expression inside the single brackets factorised.

$$\underline{(2x+3)(x+5)}$$​​

Example 3

Factorise and solve

​$$3x^2+11x-4=0$$​​

The quadratic has $$a=3, b=11$$ and $$c=-4$$. Multiply $$a$$ and $$c$$ to give $$-12$$.

Think of two numbers which will multiply to give $$-12$$ and add to give $$+11$$. The two numbers are $$+12$$ and $$-1$$, as $$12\times -1 = -12$$ and $$12-1=11$$.

Write the quadratic expression using $$12$$ and $$-1$$, and factorise.

$$\begin {aligned}3x^2+11x-4&=0 \\\underbrace{3x^2+12x}_{\text{factorise}}\space \underbrace{-1x-4}_{\text{factorise}}&=0 \\3x(x+4) -1(x+4)&=0 \\(3x-1)(x+4)&=0\end {aligned}$$​​

Now the quadratic is factorised, it can be solved. Find the solutions by making either bracket equal to $$0$$.

$$\underline{x=\frac 1 3 \space or \space x = -4}$$​​