Quadratics Factorising Methods (AQA GCSE Further Maths)

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Written by: Jamie Wood

Reviewed by: Dan Finlay

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Quadratics Factorising Methods

How do I know if it factorises?

Method 1: Use a calculator to solve the quadratic expression equal to 0
- If the solutions are integers or fractions (without square roots), then the quadratic expression factorises
Method 2: Find the value under the square root in the quadratic formula, b² – 4ac (called the discriminant)
- If this number is a perfect square number, then the quadratic expression factorises

Which factorisation method should I use for a quadratic expression?

Does it have 2 terms only?
- Yes, like $x^{2} - 7 x$
  - Use "basic factorisation" to take out the highest common factor
  - $x (x - 7)$
- 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 use "basic factorisation" to factorise it out first
  - $3 (x^{2} + 5 x + 6)$
  - 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
  - $(3 x + 1) (x - 2)$

Worked Example

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

Spot the common factor of -4 and put outside a set of brackets, work out the terms inside the brackets by dividing the terms in the original expression by -4.

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

Check the discriminant for the expression inside the brackets, $(b^{2} - 4 a c)$ , to see if it will factorise.

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

$529 = 13^{2}$ , it is a perfect square so the expression will factorise.

Proceed with factorising $2 x^{2} - 25 x + 12$ as you would for a harder quadratic, where $a \neq 1$ .
"+12" means the signs will be the same.
"-25" means that both signs will be negative.

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

The only numbers which multiply to give 24 and follow the rules for the signs above are:
$(- 1) \times (- 24)$ and $(- 2) \times (- 12)$ and $(- 3) \times (- 8)$ and $(- 4) \times (- 6)$
but only the first pair add to give $- 25$ .

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 highest common factor and group and factorise the last two terms using $1$ as the highest 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 now be factorised out.

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

Put it all together.

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

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

Last updated: 7 October 2024

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