Factorising Quadratics

If there is no constant term then just factorise out (a multiple of)
- $x^{2} - 9 x = x (x - 9)$
- $5 x^{2} + 30 x = 5 x (x + 6)$
Factorise quadratics of the form by inspection
- Find a pair of numbers, and , that multiply to give and add to give
  - E.g. for $x^{2} - 21 x - 100$ the numbers would be $4$ and $- 25$
- The quadratic will factorise as
  - So $x^{2} - 21 x - 100 = (x + 4) (x - 25)$

A harder quadratic is of the form where is not equal to 1 (or 0)
- E.g. $12 x^{2} - 11 x - 5$
These can also be factorised by inspection
- This requires a lot of practice and there are no simple rules to follow
They can be factorised reliably by grouping
- Find a pair of numbers that multiply to and add to
  - For $12 x^{2} - 11 x - 5$ , $a c = - 60$ and $b = - 11$
  - So the two numbers are $4$ and $- 15$
- Rewrite the middle term using those two numbers
  - $12 x^{2} + 4 x - 15 x - 5$
- Group and factorise the first two terms and the last two terms by pulling out common factors
  - $4 x (3 x + 1) - 5 (3 x + 1)$
- Those two terms now have a common factor (in brackets) that can be factorised out
  - $(4 x - 5) (3 x + 1)$

As a check, expand your answer and make sure you get the same expression as the one you were trying to factorise.
You should be able to recognise a difference of squares in both factorised and unfactorised form

(a)

Factorise

x^{2} - 4 x - 21

We will factorise by inspection

We need two numbers that multiply to $- 21$ and add to $- 4$

$+ 3$ and $- 7$ satisfy this

Write down the brackets

$(x + 3) (x - 7)$

(b)

Factorise

6 x^{2} - 7 x - 3

We will factorise by splitting the middle term and grouping

We need two numbers that multiply to $6 \times (- 3) = - 18$ and add to $- 7$

$+ 2$ and $- 9$ satisfy this

Split the middle term

$6 x^{2} + 2 x - 9 x - 3$

Factorise $2 x$ out of the first two terms, and $- 3$ out of the last two terms

$2 x (3 x + 1) - 3 (3 x + 1)$

These have a common factor of $(3 x + 1)$ which can be factored out

$(2 x - 3) (3 x + 1)$

(c)

Factorise

9 x^{2} - 16

Recognise that this is a difference of two squares, because $9 x^{2} - 16 = {(3 x)}^{2} - {(4)}^{2}$

Use the relation $a^{2} - b^{2} = (a + b) (a - b)$

$(3 x + 4) (3 x - 4)$

Factorising Quadratics (Edexcel IGCSE Further Maths)