Factorisation by grouping

Some four-term expressions have no factor common to all four terms, but they split into two pairs that share a bracket: \(ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)\).

Why: after taking a common factor out of each pair, the same bracket appears twice, so the bracket is itself a common factor. In a multiplication frame the four terms fill the four cells of the product of two brackets.

Multiplication frame for \(6xy-8x+9y-12\)

Method

1. Arrange the four terms into two pairs that each have a common factor.
2. Factorise each pair.
3. If the two brackets differ only in sign, take a negative factor out of the second pair so that they match.
4. Take out the common bracket, then check by expanding. If no common bracket appears, rearrange the terms and try again.

Example

Factorise \(10ab-15a-4b+6\).

\(\begin{aligned} 10ab-15a-4b+6 &= 5a(2b-3)-2(2b-3) \\ &= (5a-2)(2b-3) \end{aligned}\)

Check: \((5a-2)(2b-3)=10ab-15a-4b+6\). A common error is to write \(-4b+6\) as \(-2(2b+3)\); taking out \(-2\) changes both signs, so \(-4b+6=-2(2b-3)\).