Deciding the Factorisation Method (Edexcel IGCSE Maths A (Modular))

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Written by: Jamie Wood

Reviewed by: Dan Finlay

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Quadratics Factorising Methods

How do I know if an expression factorises?

The easiest way to check if ax² + bx + c factorises is to check if you can find a pair of integers which:
- Multiply to give ac
- Sum to give b
- If you can find integers to satisfy this, the expression must factorise
There are some alternate methods to check:
- Method 1: Use a calculator to solve the quadratic expression equal to 0
  - Only some calculators have this functionality
  - If the solutions are integers or fractions (without square roots), then the quadratic expression will factorise
- Method 2: Find the value under the square root in the quadratic formula
  - b² – 4ac
  - If this number is a square number, then the quadratic expression will factorise

Which factorisation method should I use for a quadratic expression?

Does it have 2 terms only?
- Yes, like $x^{2} - 7 x$
  - Factorise out the highest common factor, x
  - $x (x - 7)$
- Yes, like $x^{2} - 9$
  - Use the "difference of two squares" to factorise
  - $(x + 3) (x - 3)$

Does it have 3 terms?

Yes, starting with x² like $x^{2} - 3 x - 10$
- Use "factorising simple quadratics" by finding two numbers that add to -3 and multiply to -10
- $(x + 2) (x - 5)$
Yes, starting with ax² like $3 x^{2} + 15 x + 18$
- Check to see if the 3 in front of x² is a common factor for all three terms (which it is in this case), then factorise it out of all three terms
- $3 (x^{2} + 5 x + 6)$
- The quadratic expression inside the brackets is now x² +... , which factorises more easily
- $3 (x + 2) (x + 3)$
Yes, starting with ax² like $3 x^{2} - 5 x - 2$
- The 3 in front of x² is not a common factor for all three term
- Use "factorising harder quadratics", for example factorising by grouping or factorising using a grid
- $(3 x + 1) (x - 2)$

What other expressions should I be able to factorise?

You may have a cubed term like $x^{3} - 3 x^{2} - 10 x$
- Check to see if x is a common factor for all three terms (which it is in this case), so factorise it out of all three terms
- $x (x^{2} - 3 x - 10)$
- The remaining quadratic can then be factorised
- $x (x + 2) (x - 5)$
It can also be useful to spot a quadratic in the form $x^{2} + 2 a x + a^{2}$
- This factorises to ${(x + a)}^{2}$
- E.g. $x^{2} + 6 x + 9 = {(x + 3)}^{2}$

Examiner Tips and Tricks

A common mistake in the exam is to divide expressions by numbers, e.g. $2 x^{2} + 4 x + 2$ becomes $x^{2} + 2 x + 1$ (which is incorrect)
- This can only be done with equations
- e.g. $2 x^{2} + 4 x + 2 = 0$ becomes $x^{2} + 2 x + 1 = 0$ (dividing "both sides" by 2)

Worked Example

Factorise $- 8 x^{2} + 100 x - 48$ .

Spot the common factor of -4 and factorise it out

$- 8 x^{2} + 100 x - 48 = - 4 (2 x^{2} - 25 x + 12)$

Check to see if the quadratic in the bracket will factorise using $b^{2} - 4 a c$

$\begin{array}{rcl} {(- 25)}^{2} - (4 \times 2 \times 12) \\ = & 625 - 96 \\ = & 529 \end{array}$

529 is a square number (23²) so the expression will factorise

Factorise $2 x^{2} - 25 x + 12$

We require a pair of numbers which multiply to ac, and sum to b

$a \times c = 2 \times 12 = 24$

The only numbers which multiply to 24 and sum to -25 are

-24 and -1

Split the $- 25 x$ term into $- 24 x - x$

$\begin{array}{rcl} 2 x^{2} - 24 x - x + 12 \end{array}$

Group and factorise the first two terms, using $2 x$ as the common factor
Group and factorise the last two terms using $- 1$ as the common factor

$\begin{array}{rcl} 2 x (x - 12) - 1 (x - 12) \end{array}$

These factorised terms now have a common term of $(x - 12)$ , so this can be factorised out

$(2 x - 1) (x - 12)$

Recall that -4 was factorised out at the start

$- 8 x^{2} + 100 x - 48 = - 4 (2 x^{2} - 25 x + 12) = - 4 (2 x - 1) (x - 12)$

$- 4 (2 x - 1) (x - 12)$

Last updated: 15 November 2024

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