0580

Expanding and Factorising Brackets

Algebra and Sequences

Reversing the expansion

  • Factorising writes an expression as a product, which is the reverse of expanding
  • Take the largest number dividing every coefficient, together with the highest power of each letter present in all terms
  • Dividing each term by that combined factor leaves what belongs inside the bracket
    • 20p³ − 15p² has highest common factor 5p², giving 5p²(4p − 3)
  • Expanding the answer back is a reliable check, and it costs one line

When the common factor is a bracket

  • The common factor taken out need not be a number or a letter: a whole bracket can be shared
    • 3(x + 1) + y(x + 1) has (x + 1) in both terms, giving (x + 1)(3 + y)
  • This is the step that finishes a factorisation by grouping, and it is worth spotting on its own
Exam tip

Take out the highest common factor, letters included. 6x² + 9x written as 3(2x² + 3x) has been factorised but not fully, and the mark is for 3x(2x + 3). Expanding your own answer back is a two-second check that catches both a missed factor and a sign slip.

Worked example

Taking out a common factor

Factorise 12x³ − 18x², and factorise 5(x − 2) + y(x − 2).

Solution:

  • For the first, find the highest common factor of the numbers and of the letters separately
  • The highest common factor of 12 and 18 is 6
  • Both terms contain x², and that is the highest power common to both
  • So the common factor is 6x²
  • 12x³ ÷ 6x² = 2x, and −18x² ÷ 6x² = −3
  • 12x³ − 18x² = 6x²(2x − 3)
  • For the second, the common factor is the whole bracket (x − 2)
  • Taking it out leaves 5 and y, so the answer is (x − 2)(5 + y)
  • Check either by expanding back