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Factorization

Factorization

Factorising four terms by grouping

Four Terms Paper 1 + 2 5 min read

On this page
  1. In this note
  2. Formula bank
  3. Group into pairs
  4. Ways they test it
  5. Mistakes to avoid
  6. Summary
  7. Practice
  8. Can I do this?
  9. Quiz

In this note

Syllabus 2.2: factorising four terms
  • 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

GroupingLearn it\(ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)\)the matching bracket is the common factor.Use it when four terms with no factor common to all four.

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

  1. Pair the terms: 1st with 2nd, 3rd with 4th.
  2. Take out the common factor of each pair. If the second pair starts with a minus, take out a NEGATIVE factor.
  3. The two brackets must be identical. If they are not, re-order the terms and try again.
  4. Take out the bracket: (outside factors added)(bracket). Expand to check.
  5. 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.

  1. \(x(y-4)-3(y-4)\)
  2. \((x-3)(y-4)\)

Answer \((x-3)(y-4)\)

Example 2 Factorise \(6pq+5-15q-2p\).

\(6pq\) and \(5\) share nothing, so re-order: \(6pq-2p-15q+5\).

  1. \(2p(3q-1)-5(3q-1)\)
  2. \((2p-5)(3q-1)\)

Answer \((2p-5)(3q-1)\)

Try it yourself

1
Factorise \(xy+3x+2y+6\).
Show answer

Answer \((x+2)(y+3)\)

2
Factorise \(xy-5x+4y-20\).
Show answer

Answer \((x+4)(y-5)\)

3
Factorise \(ab-a-b+1\).
Hint
Second pair: take out \(-1\).
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.

  1. Factor each pair.
  2. Take out the matching bracket.

For example \(5ax+10ay+2bx+4by\)

  1. \(5a(x+2y)+2b(x+2y)\)

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.

  1. Take out a negative factor from the second pair.
  2. Check the signs inside match.

For example \(mr-2ms-3r+6s\)

  1. \(m(r-2s)-3(r-2s)\)

Answer \((m-3)(r-2s)\)

Type 3 Out of order

"Factorise \(ay+bx+by+ax\)."

Spot it the first two terms share nothing useful.

  1. Re-order so each pair shares a letter.
  2. Then group.

For example \(ay+ax+by+bx\)

  1. \(a(y+x)+b(y+x)\)

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'.

  1. Group.
  2. One bracket is a difference of two squares: factorise it.

For example \(x^{3}+2x^{2}-9x-18\)

  1. \(x^{2}(x+2)-9(x+2)\)
  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

1
Factorise \(xy+2x+5y+10\).
Show answer

Answer \((x+5)(y+2)\)

2
Factorise \(2xy+3x+4y+6\).
Hint
\(x(2y+3)+2(2y+3)\).
Show answer

Answer \((x+2)(2y+3)\)

3
Factorise \(ay+bx+by+ax\).
Hint
Put the a-terms together first.
Show working
  1. \(ay+ax+by+bx\)
  2. \(a(y+x)+b(y+x)\)
  3. \((a+b)(x+y)\)

Answer \((a+b)(x+y)\)

Medium

1
Factorise \(3ax-6ay+bx-2by\).
Guide me

What is it asking? a product

What do we know? pairs share \(3a\) and \(b\)

Plan bracket \((x-2y)\)

Show working
  1. \(3a(x-2y)+b(x-2y)\)
  2. \((3a+b)(x-2y)\)

Answer \((3a+b)(x-2y)\)

2
Factorise \(4xy-15-6x+10y\).
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
  1. \(2x(2y-3)+5(2y-3)\)
  2. \((2x+5)(2y-3)\)

Answer \((2x+5)(2y-3)\)

3
Factorise \(6x-9y+2xy-3y^{2}\).
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
  1. \(3(2x-3y)+y(2x-3y)\)
  2. \((2x-3y)(3+y)\)

Answer \((2x-3y)(3+y)\)

Hard

1
Factorise \(1-a+b-ab\).
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. \(1(1-a)+b(1-a)\)
  2. \((1+b)(1-a)\)

Answer \((1+b)(1-a)\)

PartAnswer / workingMarksGuidance
\(1(1-a)+b(1-a)\)M1
\((1+b)(1-a)\)A1
Total: 2 marks
2
Factorise completely \(x^{3}+2x^{2}-9x-18\).
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
  1. \((x+2)(x^{2}-9)\)
  2. \((x+2)(x+3)(x-3)\)

Answer \((x+2)(x+3)(x-3)\)

PartAnswer / workingMarksGuidance
\(x^{2}(x+2)-9(x+2)\)M1
\((x+2)(x^{2}-9)\)M1
\((x+2)(x+3)(x-3)\)A1
Total: 3 marks
3
Factorise completely \(2x^{3}-x^{2}-8x+4\).
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
  1. \((2x-1)(x^{2}-4)\)
  2. \((2x-1)(x+2)(x-2)\)

Answer \((2x-1)(x+2)(x-2)\)

PartAnswer / workingMarksGuidance
\(x^{2}(2x-1)-4(2x-1)\)M1
\((2x-1)(x^{2}-4)\)M1
\((2x-1)(x+2)(x-2)\)A1
Total: 3 marks

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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