By first grouping the four terms in pairs, factorise \(2ax-6ay+bx-3by\).
Question 1 Report
By first grouping the four terms in pairs, factorise \(2ax-6ay+bx-3by\).
Answer Details
Factorising by grouping works when four terms split into two pairs that leave the same bracket behind. Take the first two terms together and the last two together.
- From \(2ax-6ay\) the common factor is \(2a\), giving \(2a(x-3y)\) [M1].
- From \(bx-3by\) the common factor is \(b\), giving \(b(x-3y)\) [M1].
- Both pairs now contain the bracket \((x-3y)\), so that bracket is itself a common factor: \(2a(x-3y)+b(x-3y)=(2a+b)(x-3y)\) [A1] oe.
Check by expanding: \((2a+b)(x-3y)=2ax-6ay+bx-3by\), which is the original expression.
The signal that the grouping is right is that both brackets match exactly. If they come out as \((x-3y)\) and \((3y-x)\), the second pair has been factorised with the wrong sign; take out \(-b\) instead of \(b\) to make them agree.
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