Factorization
Factorising four terms by grouping
In this note
- 2.2Factorise four terms by grouping into pairs, e.g. \(ax+bx+ay+by\).
- 2.2Choose the sign and the order of the terms so the brackets match.
- 2.2Finish with a difference of two squares when one appears.
Before you start: Common factors. Next: Solving quadratics by factorising.
Formula bank
Group into pairs
Four terms rarely share one factor. But the first two often share one factor and the last two another, and after taking them out the same bracket appears twice. That bracket is then a common factor of everything.
General methodHow to factorise four terms
- Pair the terms: 1st with 2nd, 3rd with 4th.
- Take out the common factor of each pair. If the second pair starts with a minus, take out a NEGATIVE factor.
- The two brackets must be identical. If they are not, re-order the terms and try again.
- Take out the bracket: (outside factors added)(bracket). Expand to check.
- Look again: can any bracket be factorised further (for example a difference of two squares)?
Example Factorise \(xy-4x-3y+12\).
Pairs \(xy-4x\) and \(-3y+12\); take \(-3\) out of the second so the brackets match.
- \(x(y-4)-3(y-4)\)
- \((x-3)(y-4)\)
Answer \((x-3)(y-4)\)
Example 2 Factorise \(6pq+5-15q-2p\).
\(6pq\) and \(5\) share nothing, so re-order: \(6pq-2p-15q+5\).
- \(2p(3q-1)-5(3q-1)\)
- \((2p-5)(3q-1)\)
Answer \((2p-5)(3q-1)\)
Try it yourself
Show answer
Answer \((x+2)(y+3)\)
Show answer
Answer \((x+4)(y-5)\)
Hint
Show answer
Answer \((a-1)(b-1)\)
Ways they test it
Type 1 In order
"Factorise \(5ax+10ay+2bx+4by\)." (2 marks)
Spot it four terms, the pairs work as written.
- Factor each pair.
- Take out the matching bracket.
For example \(5ax+10ay+2bx+4by\)
- \(5a(x+2y)+2b(x+2y)\)
Answer \((5a+2b)(x+2y)\)
Type 2 A minus in the middle
"Factorise \(mr-2ms-3r+6s\)." (2 marks)
Spot it the 3rd term is negative.
- Take out a negative factor from the second pair.
- Check the signs inside match.
For example \(mr-2ms-3r+6s\)
- \(m(r-2s)-3(r-2s)\)
Answer \((m-3)(r-2s)\)
Type 3 Out of order
"Factorise \(ay+bx+by+ax\)."
Spot it the first two terms share nothing useful.
- Re-order so each pair shares a letter.
- Then group.
For example \(ay+ax+by+bx\)
- \(a(y+x)+b(y+x)\)
Answer \((a+b)(x+y)\)
Type 4 Group, then DOTS
"Factorise completely \(x^{3}+2x^{2}-9x-18\)." (3 marks)
Spot it a cubic with four terms; 'completely'.
- Group.
- One bracket is a difference of two squares: factorise it.
For example \(x^{3}+2x^{2}-9x-18\)
- \(x^{2}(x+2)-9(x+2)\)
- \((x+2)(x^{2}-9)\)
Answer \((x+2)(x+3)(x-3)\)
Mistakes to avoid
- ✗ \(x(y-4)+3(-y+4)\) and stop✓ \(x(y-4)-3(y-4)=(x-3)(y-4)\)Choose the sign that makes the brackets match.
- ✗ \((x+2)(2y+3)(2y+3)\)✓ \((x+2)(2y+3)\)The common bracket appears once.
- ✗ \((x+2)(x^{2}-9)\) for 'completely'✓ \((x+2)(x+3)(x-3)\)\(x^{2}-9\) is a difference of two squares.
Summary
- Pair the terms, factor each pair, then take out the matching bracket.
- A minus before the second pair: take out a negative factor.
- No match? Re-order the terms.
- Check every bracket again at the end.
If you see …
- four terms, nothing common to allgroup into pairs.
- brackets like (y − 4) and (−y + 4)take out −1 more: they become the same.
Practice
Practice: easy, medium, hard
Easy
Show answer
Answer \((x+5)(y+2)\)
Hint
Show answer
Answer \((x+2)(2y+3)\)
Hint
Show working
- \(ay+ax+by+bx\)
- \(a(y+x)+b(y+x)\)
- \((a+b)(x+y)\)
Answer \((a+b)(x+y)\)
Medium
Guide me
What is it asking? a product
What do we know? pairs share \(3a\) and \(b\)
Plan bracket \((x-2y)\)
Show working
- \(3a(x-2y)+b(x-2y)\)
- \((3a+b)(x-2y)\)
Answer \((3a+b)(x-2y)\)
Guide me
What is it asking? a product
What do we know? 4xy and −15 share nothing
Plan re-order: 4xy − 6x + 10y − 15
Show working
- \(2x(2y-3)+5(2y-3)\)
- \((2x+5)(2y-3)\)
Answer \((2x+5)(2y-3)\)
Guide me
What is it asking? a product
What do we know? pairs: \(3(2x-3y)\) and \(y(2x-3y)\)
Plan take out the bracket
Show working
- \(3(2x-3y)+y(2x-3y)\)
- \((2x-3y)(3+y)\)
Answer \((2x-3y)(3+y)\)
Hard
Guide me
What is it asking? a product
What do we know? first pair is \(1(1-a)\)
Plan second \(b(1-a)\)
Show working
- \(1(1-a)+b(1-a)\)
- \((1+b)(1-a)\)
Answer \((1+b)(1-a)\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(1(1-a)+b(1-a)\) | M1 | ||
| \((1+b)(1-a)\) | A1 | ||
| Total: 2 marks | |||
Guide me
What is it asking? a product
What do we know? group: \(x^{2}(x+2)-9(x+2)\)
Plan then \(x^{2}-9\) is DOTS
Show working
- \((x+2)(x^{2}-9)\)
- \((x+2)(x+3)(x-3)\)
Answer \((x+2)(x+3)(x-3)\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(x^{2}(x+2)-9(x+2)\) | M1 | ||
| \((x+2)(x^{2}-9)\) | M1 | ||
| \((x+2)(x+3)(x-3)\) | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? a product
What do we know? \(x^{2}(2x-1)-4(2x-1)\)
Plan then \(x^{2}-4\)
Show working
- \((2x-1)(x^{2}-4)\)
- \((2x-1)(x+2)(x-2)\)
Answer \((2x-1)(x+2)(x-2)\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(x^{2}(2x-1)-4(2x-1)\) | M1 | ||
| \((2x-1)(x^{2}-4)\) | M1 | ||
| \((2x-1)(x+2)(x-2)\) | A1 | ||
| Total: 3 marks | |||
Can I do this?
- I can factorise four terms by grouping.
- I can handle minus signs and re-order terms.
- I can finish with a difference of two squares where possible.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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