GoalFactorise four-term expressions by grouping them in pairs, regroup when the first try fails, and take out a common factor first.
New words
grouping · splitting the terms into groups that each have a common factorrepeated bracket · the same bracket appearing in both groupsregroup · try a different pairing when the brackets do not matchcomplete factorisation · no factor can be factorised further
Explanation
An expression with four terms can often be factorised by grouping. For x² + 4x + bx + 4b, group the first pair and the second pair: x(x + 4) + b(x + 4). Now (x + 4) is common, so the result is (x + 4)(x + b). The two brackets must be the same. If a pair is subtracted, change the signs inside: ax − ay − bx + by = a(x − y) − b(x − y) = (x − y)(a − b). If the first grouping fails, regroup. For x² − 3x − y² + 3y, the pairs (x² − 3x) and (y² − 3y) fail. Group (x² − y²) − (3x − 3y) = (x − y)(x + y) − 3(x − y) = (x − y)(x + y − 3). Always take out a common factor first, and check by expanding.
Step
x² + 4x + bx + 4b
group in pairs
(x² + 4x) + (bx + 4b)
take out each GCF
x(x + 4) + b(x + 4)
take out the bracket
(x + 4)(x + b)
Worked examples
Factorise ax − ay − bx + by. Group: a(x − y) − b(x − y), because −bx + by = −b(x − y). The bracket is common: (x − y)(a − b). Check: ax − ay − bx + by.
Matching brackets. Write ax + ay + bx + by on a card. Factorise by pairing (ax + ay) with (bx + by). Then pair (ax + bx) with (ay + by). Do you get the same answer? Try it with a = 2, b = 3, x = 4, y = 5 to check the value.
Practice
1
Factorise ax + ay + 3x + 3y.
a(x + y) + 3(x + y) = (x + y)(a + 3).
2
Factorise xy − 3x + 2y − 6.
x(y − 3) + 2(y − 3) = (y − 3)(x + 2).
3
Factorise x³ − 2x² − 9x + 18.
x²(x − 2) − 9(x − 2) = (x − 2)(x − 3)(x + 3).
4
Factorise ab − 2a − 3b + 6.
a(b − 2) − 3(b − 2) = (b − 2)(a − 3).
5
Hana groups x² + 3x − 2x − 6 as x(x + 3) − 2(x − 3). What is wrong?
−2x − 6 = −2(x + 3), not −2(x − 3). The right answer is x(x + 3) − 2(x + 3) = (x + 3)(x − 2).
