Expanding and Factorising Brackets

Algebra and Sequences

Where one square is subtracted from another, the expression factorises into two brackets differing only in sign
a² − b² factorises to (a + b)(a − b)
- 25x² − 49 is (5x)² − 7², giving (5x + 7)(5x − 7)
Both terms must be perfect squares and the operation must be a subtraction
Recognising this pattern is faster than any general method, and it also appears when rationalising a denominator

Exam tip

Look for it before trying anything else: two terms, both squares, with a minus between them. x² − 49 factorises straight to (x + 7)(x − 7). Spotting it first saves the whole trial-and-error route, and it is the one factorisation that has no middle term to work from.

Worked example

Difference of two squares, factorised fully

Factorise 9x² − 25, and factorise 2x² − 8.

Solution:

9x² − 25 is one square subtracted from another: 9x² = (3x)² and 25 = 5²
The difference of two squares factorises as (first − second)(first + second)
9x² − 25 = (3x − 5)(3x + 5)
2x² − 8 is not yet a difference of two squares, because 2 and 8 are not squares
Take out the common factor of 2 first: 2(x² − 4)
Now x² − 4 is a difference of two squares: x² − 2²
2x² − 8 = 2(x − 2)(x + 2)
Stopping at 2(x² − 4) is not fully factorised, and the syllabus defines "factorise" to mean factorise fully