MathematicsExam code: 0580
Expanding and Factorising Brackets
Algebra and Sequences
- Four terms with no factor common to all of them may still factorise in pairs
- Factorise the first two terms and the last two separately, aiming to produce the same bracket in both
- That shared bracket is then taken out as a common factor
- ab + 4a + 3b + 12 becomes a(b + 4) + 3(b + 4), which factorises to (b + 4)(a + 3)
- If the two brackets do not match, try pairing the terms differently
Worked example
Factorising by grouping
Factorise xy + 3x + 2y + 6.
Solution:
- Four terms with nothing common to all of them, which is the signal to group them in pairs
- Take the first two together and the last two together, keeping the order
- First pair: xy + 3x, whose common factor is x, giving x(y + 3)
- Second pair: 2y + 6, whose common factor is 2, giving 2(y + 3)
- The two brackets are now identical, and that is the check that the pairing worked: different brackets mean the terms should be paired the other way round
- The expression reads x(y + 3) + 2(y + 3), which is (y + 3) multiplied by x and by 2
- Take the common bracket out: (y + 3)(x + 2)
- Check by expanding: xy + 3x + 2y + 6, which is the expression we started from
Worked example
Grouping when a pair starts with a minus
Factorise 6xy − 9x − 4y + 6.
Solution:
- Nothing is common to all four terms, so factorise them in pairs
- First pair: 6xy − 9x = 3x(2y − 3)
- The second pair starts with a minus, so take the minus out with its factor: −4y + 6 = −2(2y − 3)
- Dividing by −2 flips both signs inside: −4y ÷ (−2) = 2y and 6 ÷ (−2) = −3. Taking out +2 instead gives 2(3 − 2y), a bracket that does not match
- The expression is now 3x(2y − 3) − 2(2y − 3)
- Take the common bracket out: (2y − 3)(3x − 2)
- Check by expanding: 6xy − 4y − 9x + 6, the four terms we started with