Expanding and Factorising Brackets

Algebra and Sequences

The expression	Method
Every term shares a factor	Take out the common factor first
Two square terms subtracted	Difference of two squares
Three terms, x² coefficient 1	Find a pair multiplying to c, adding to b
Three terms, x² coefficient not 1	Find a pair multiplying to ac, adding to b, then group
Four terms	Factorise in pairs and look for a common bracket

Always take out a common factor before anything else, since it makes what remains simpler
A quadratic factorises with integers only when a pair multiplying to ac and adding to b exists, so a short search settles it

Exam tip

Write out all four products when expanding a pair of brackets before collecting — the unsimplified expansion earns a mark of its own. The papers say only "Factorise", never "factorise fully", because the syllabus defines factorise to mean fully: 2x² − 8 must reach 2(x + 2)(x − 2). Check any factorisation by expanding it back.

Worked example

Choosing the method

Factorise each expression: (a) 6x² + 9x (b) x² − 49 (c) x² − 7x + 12

Solution:

(a) Two terms, and both share a factor. 3x divides both, so take it out
6x² + 9x = 3x(2x + 3)
(b) Two terms, one square subtracted from another, and no common factor
That is the difference of two squares: x² − 7² = (x − 7)(x + 7)
(c) Three terms with no number in front of x², so look for two numbers multiplying to +12 and adding to −7
Both must be negative to give a positive product and a negative sum: −3 and −4
x² − 7x + 12 = (x − 3)(x − 4)
Always test for a common factor first, because taking one out can turn what is left into one of the standard forms