Factorising: Grouping in Pairs & the Cubes (Grade 10)

A decision guide based on counting the terms, then the two new Grade 10 methods: grouping four terms in pairs, and the sum and difference of two cubes.

By Grade 10 there are five factorising methods, and the exam will not tell you which to use. Fortunately, counting the terms nearly always decides it for you.

1The decision guide

Work down the list. A common factor always comes out first, then let the number of terms choose the method.

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Take out the common factor first, every time. It often turns an ugly expression into a standard one — 2x² − 18 becomes 2(x² − 9) = 2(x − 3)(x + 3).

2Four terms: group in pairs

Worked ExampleFactorise: x³ + 2x² + 3x + 6

1
Split into two pairs: (x³ + 2x²) + (3x + 6).
Four terms → grouping.
2
Factorise each pair: x²(x + 2) + 3(x + 2).
The bracket must come out the SAME — that is the check.
3
Take out the common bracket: (x + 2)(x² + 3).

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If the two brackets do not match, try pairing the terms differently, or take out a negative: ax − ay − bx + by = a(x − y) − b(x − y) = (x − y)(a − b).

3Two terms: difference of squares

a² − b² = (a − b)(a + b)

Note there is no factorisation for a sum of two squares (a² + b²) in this syllabus.

4Two terms: sum and difference of CUBES

a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)

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Memory aid — SOAP: the signs run Same, Opposite, Always Positive. So for a³ − b³ the brackets are (a − b)(a² + ab + b²).

Worked ExampleFactorise: x³ + 8

1
Write both parts as cubes: x³ + 2³.
So a = x and b = 2.
2
Apply the identity: (x + 2)(x² − 2x + 4).
Same, opposite, always positive.
3
Check by expanding — the middle terms cancel and you get x³ + 8 ✓

Worked ExampleFactorise: 8x³ − 27

1
8x³ = (2x)³ and 27 = 3³.
So a = 2x and b = 3.
2
(2x − 3)((2x)² + (2x)(3) + 3²).
Substitute into the identity.
3
= (2x − 3)(4x² + 6x + 9).
Simplify inside the second bracket.

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The second bracket in a cube factorisation does not factorise further, and it is not a perfect square trinomial. Do not try to force (a − b)² on it.

5Three terms: trinomials

Worked ExampleFactorise: x² − 7x + 12

1
Find two numbers multiplying to +12 and adding to −7.
Both must be negative.
2
−3 and −4.
3
= (x − 3)(x − 4).

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Factorise: 3x² − 12
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3(x² − 4) = 3(x − 2)(x + 2)
2
Factorise: x³ + 3x² + 2x + 6
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x²(x+3) + 2(x+3) = (x + 3)(x² + 2)
3
Factorise: ax + ay + bx + by
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a(x+y) + b(x+y) = (x + y)(a + b)
4
Factorise: x³ + 27
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(x + 3)(x² − 3x + 9)
5
Factorise: x³ − 64
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(x − 4)(x² + 4x + 16)
6
Factorise: 8x³ − 1
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(2x − 1)(4x² + 2x + 1)
7
Factorise: 27x³ + 8
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(3x + 2)(9x² − 6x + 4)
8
Factorise: x² − 7x + 12
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(x − 3)(x − 4)
9
Factorise: 2x² + 7x + 3
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(2x + 1)(x + 3)