IGCSEFurther MathsEdexcelRevision NotesEquations, Identities & InequalitiesQuadratic FunctionsFactorising Quadratics

Factorising Quadratics (Edexcel IGCSE Further Pure Maths): Revision Note

Exam code: 4PM1

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 19 October 2024

Factorising Quadratics

How can I factorise simple quadratics?

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

How can I factorise harder quadratics?

A harder quadratic is of the form $a x^{2} + b x + c$ where $a$ 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 $a c$ and add to $b$
  - For $12 x^{2} - 11 x - 5$ , $a c = - 60$ and $b = - 11$
  - So the two numbers are $4$ and $- 15$
- Rewrite the middle $b x$ 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
  - $(4 x - 5) (3 x + 1)$

How do I factorise a difference of two squares

A difference of two squares refers to any expression of the form $a^{2} - b^{2}$
- I.e. 'something squared subtracted from something else squared'
- For example,
  - $x^{2} - 36$
  - $9^{2} - 5^{2}$
  - ${(x + 1)}^{2} - {(x - 4)}^{2}$
  - $4 m^{2} - 25 n^{2}$ which is equal to ${(2 m)}^{2} - {(5 n)}^{2}$
Such expressions will factorise as $(a + b) (a - b)$
- This is because $(a + b) (a - b) = a^{2} - a b + a b - b^{2} = a^{2} - b^{2}$
- So
  - $x^{2} - 36 = (x + 6) (x - 6)$
  - $9^{2} - 5^{2} = (9 + 5) (9 - 5)$ which is equal to $14 \times 4 = 56$
  - ${(x + 1)}^{2} - {(x - 4)}^{2} = ((x + 1) + (x - 4)) ((x + 1) - (x - 4))$ which is equal to $(2 x - 3) (5) = 10 x - 15$
  - $4 m^{2} - 25 n^{2} = (2 m + 5 n) (2 m - 5 n)$

Examiner Tips and Tricks

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

Worked Example

(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

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

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

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

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

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