A decision guide based on counting the terms, then the three new Grade 10 methods: a trinomial with a number in front of x² (splitting the middle term), grouping four terms in pairs, and the sum and difference of two cubes.
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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
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
- 1Split into two pairs: (x³ + 2x²) + (3x + 6).Four terms → grouping.
- 2Factorise each pair: x²(x + 2) + 3(x + 2).The bracket must come out the SAME — that is the check.
- 3Take out the common bracket: (x + 2)(x² + 3).
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
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)
Memory aid — SOAP: the signs run Same, Opposite, Always Positive. So for a³ − b³ the brackets are (a − b)(a² + ab + b²).
- 1Write both parts as cubes: x³ + 2³.So a = x and b = 2.
- 2Apply the identity: (x + 2)(x² − 2x + 4).Same, opposite, always positive.
- 3Check by expanding — the middle terms cancel and you get x³ + 8 ✓
- 18x³ = (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.
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
- 1Find two numbers multiplying to +12 and adding to −7.Both must be negative.
- 2−3 and −4.
- 3= (x − 3)(x − 4).
6Three terms with a number in front of x²: split the middle term
In Grade 9 the x² either stood alone, or the number in front came out as a common factor first. Grade 10 is where you meet the real case: 2x² + 7x + 3, where nothing divides out. Guessing the brackets works sometimes; the ac-method works every time, and it turns the trinomial into a grouping question you can already do.
- 1a × c = 2 × 3 = 6.a is the number in front of x², c is the last number.
- 2Two numbers with product 6 and sum 7: 6 and 1.The sum must be the middle coefficient, +7.
- 3Split the middle term: 2x² + 6x + x + 3.7x becomes 6x + x — nothing has changed in value, but there are four terms now.
- 4Group in pairs: 2x(x + 3) + 1(x + 3).The same bracket both times — that is the check that the split was right.
- 5= (2x + 1)(x + 3).Take out the common bracket, exactly as in the four-term method above.
- 1a × c = 3 × 8 = 24, and the sum must be −10.The product is positive and the sum negative, so both numbers are negative.
- 2−6 and −4.−6 × −4 = 24 ✓ and −6 + (−4) = −10 ✓
- 33x² − 6x − 4x + 8 = 3x(x − 2) − 4(x − 2).Take out −4, not +4, so the second bracket matches the first.
- 4= (x − 2)(3x − 4).Expand to check: 3x² − 4x − 6x + 8 ✓
Check the common factor first, as always. 6x² + 21x + 9 is not a new problem — take out the 3 and it is 3(2x² + 7x + 3), the example above.
Splitting the middle term and then stopping. 2x² + 6x + x + 3 is not an answer — it is four terms, not a product. Group them and take out the common bracket. The other slip is dropping the 1: 2x(x + 3) + (x + 3) still has a 1 in front of the second bracket, so the answer is (2x + 1)(x + 3).
Practice exercises
Work each one out, then click to reveal the answer.
- 1Factorise: 3x² − 12
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3(x² − 4) = 3(x − 2)(x + 2) - 2Factorise: x³ + 3x² + 2x + 6
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x²(x + 3) + 2(x + 3) = (x + 3)(x² + 2) - 3Factorise: ax + ay + bx + by
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a(x + y) + b(x + y) = (x + y)(a + b) - 4Factorise: x³ + 27
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(x + 3)(x² − 3x + 9) - 5Factorise: x³ − 64
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(x − 4)(x² + 4x + 16) - 6Factorise: 8x³ − 1
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(2x − 1)(4x² + 2x + 1) - 7Factorise: 27x³ + 8
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(3x + 2)(9x² − 6x + 4) - 8Factorise: x² − 7x + 12
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(x − 3)(x − 4) - 9Factorise: 2x² + 7x + 3
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a × c = 6; 6 and 1; 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) - 10Factorise: 2x² − 5x − 3
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a × c = −6; −6 and +1; 2x² − 6x + x − 3 = 2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3)
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