A full walkthrough of every kind of factorising you can be asked in an exam — common factors, difference of squares, all four trinomial sign patterns, leading coefficients and grouping — with a worked example for each and a worksheet to practise.
Factorising means writing an expression as a product (things multiplied) instead of a sum. It is the exact reverse of expanding brackets: expanding turns x(x + 3) into x2 + 3x, so factorising turns x2 + 3x back into x(x + 3). This lesson covers every type of factorising an exam can ask for, in the order you should try them.
The one habit that never fails: before anything else, look for a common factor. Then count the terms — that tells you which method comes next: 2 terms → difference of squares, 3 terms → trinomial, 4 terms → grouping.
1Type 1: Common factor (always check this first)
Take out anything every term shares — the highest common number and the lowest power of any shared variable — and write it outside a bracket.
- 1Highest common number of 6 and 9 is 3.3 divides into both 6 and 9.
- 2Both terms contain x, so x also comes out.Take the lowest power of x present, which is x1.
- 3Factor out 3x: 6x2 + 9x = 3x(2x + 3).6x2 ÷ 3x = 2x and 9x ÷ 3x = 3.
- 4Check by expanding: 3x × 2x = 6x2 and 3x × 3 = 9x. ✓Expanding back must return the original.
- 1Common number of 12 and 8 is 4.4 is the HCF of 12 and 8.
- 2Both terms share a and b; the lowest powers are a1 and b1, so ab comes out.Take the smallest power of each shared variable.
- 3Factor out 4ab: 12a2b − 8ab2 = 4ab(3a − 2b).12a2b ÷ 4ab = 3a and 8ab2 ÷ 4ab = 2b.
2Type 2: Difference of two squares (exactly 2 terms)
When you see a square minus a square, it splits into two brackets that are identical except for the sign — one plus, one minus.
- 1Recognise both parts are perfect squares: x2 and 25 = 52.The first sign must be minus for this method.
- 2Take the square root of each: x2 = x and 25 = 5.These become the a and b in the brackets.
- 3Write (x + 5)(x − 5).One bracket with +, one with −.
- 19x2 = (3x)2 and 16 = 42, so both are squares.Square-root the coefficient too: 9 = 3.
- 2Roots are 3x and 4, giving (3x + 4)(3x − 4).Apply a2 − b2 with a = 3x, b = 4.
- 1Take out the common factor 2: 2x2 − 50 = 2(x2 − 25).Always do Type 1 first — it simplifies what's left.
- 2Now x2 − 25 is a difference of squares: 2(x + 5)(x − 5).Factorise the bracket fully.
A SUM of squares does not factorise. x2 + 25 stays as it is — the difference-of-squares rule only works with a minus sign.
3Type 3: Trinomials x2 + bx + c (3 terms)
Find two numbers that multiply to give c and add to give b. The signs of b and c tell you what kind of numbers to look for — here are all four cases.
- 1Need two numbers: product 12, sum 7.Both positive because c and b are positive.
- 23 and 4: 3 × 4 = 12 and 3 + 4 = 7.Test factor pairs of 12 against the sum you need.
- 3Answer: (x + 3)(x + 4).Both brackets have +.
- 1Need product +12, sum −7.A positive product with a negative sum means BOTH numbers are negative.
- 2−3 and −4: (−3)(−4) = 12 and −3 + (−4) = −7.Two negatives multiply to positive, add to negative.
- 3Answer: (x − 3)(x − 4).Both brackets have −.
- 1Need product −12, sum +1.A negative product means the two numbers have DIFFERENT signs.
- 2+4 and −3: (4)(−3) = −12 and 4 + (−3) = +1.The larger number (4) is positive because the sum is positive.
- 3Answer: (x + 4)(x − 3).Different signs in the brackets.
- 1Need product −12, sum −1.Different signs again (negative product).
- 2−4 and +3: (−4)(3) = −12 and −4 + 3 = −1.The larger number (4) is now negative because the sum is negative.
- 3Answer: (x − 4)(x + 3).Compare with 3c — only the signs swapped.
- c positive, b positive → both numbers positive: (x + _)(x + _)
- c positive, b negative → both numbers negative: (x − _)(x − _)
- c negative → one of each sign; the bigger number takes the sign of b
4Type 4: Trinomials with a leading coefficient, ax2 + bx + c (Grade 10–11)
When the x2 has a coefficient other than 1, use the ac-method: multiply a × c, split the middle term, then factorise by grouping.
- 1Multiply a × c = 2 × 3 = 6. Find two numbers: product 6, sum 7 → 6 and 1.Same two-number idea, but the product is now ac, not just c.
- 2Split the middle term: 2x2 + 6x + 1x + 3.Rewrite 7x as 6x + 1x using the two numbers.
- 3Group in pairs: 2x(x + 3) + 1(x + 3).Factor each pair separately.
- 4Factor out (x + 3): (x + 3)(2x + 1).Both pieces share the bracket (x + 3).
5Type 5: Grouping (4 terms)
With four terms, group them into two pairs, factor each pair, then take out the common bracket.
- 1Group: (x3 + 2x2) + (3x + 6).Pair terms that share a factor.
- 2Factor each pair: x2(x + 2) + 3(x + 2).x2 comes out of the first pair, 3 out of the second.
- 3Factor out (x + 2): (x + 2)(x2 + 3).The shared bracket is the common factor.
6How to choose the right method
- Step 1 — Always take out a common factor first (Type 1).
- Step 2 — Then count the terms of what's left.
- 2 terms and a minus between squares → difference of squares (Type 2).
- 3 terms → trinomial (Type 3, or Type 4 if the x2 has a coefficient).
- 4 terms → grouping (Type 5).
Factorise fully. After using any method, check whether a bracket can still be factorised further — e.g. 2(x2 − 9) is not finished until it becomes 2(x + 3)(x − 3).
Always check by expanding. Multiply your brackets back out; if you don't get the original expression exactly, a sign or number is wrong. This one habit catches most exam errors.
Every case above is drawn straight from the Grade 9–11 CAPS syllabus. Work through the worksheet below until you can name the correct method the moment you see the expression — that instant recognition is what earns full marks under exam time pressure.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Factorise 5x + 15 (common factor)
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HCF is 5 → 5(x + 3) - 2Factorise 8a2b − 12ab (common factor)
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HCF is 4ab → 4ab(2a − 3) - 3Factorise x2 − 49 (difference of squares)
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x2 = x, 49 = 7 → (x + 7)(x − 7) - 4Factorise 25x2 − 9 (difference of squares)
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25x2 = 5x, 9 = 3 → (5x + 3)(5x − 3) - 5Factorise 3x2 − 27 (common factor first)
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= 3(x2 − 9) = 3(x + 3)(x − 3) - 6Factorise x2 + 9x + 20 (trinomial, both +)
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Product 20, sum 9 → 4 and 5 → (x + 4)(x + 5) - 7Factorise x2 − 8x + 15 (trinomial, c + b −)
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Product 15, sum −8 → −3 and −5 → (x − 3)(x − 5) - 8Factorise x2 + 2x − 15 (trinomial, c −)
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Product −15, sum +2 → +5 and −3 → (x + 5)(x − 3) - 9Factorise x2 − 2x − 15 (trinomial, c −)
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Product −15, sum −2 → −5 and +3 → (x − 5)(x + 3) - 10Factorise 2x2 + 7x + 6 (leading coefficient)
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ac = 12, numbers 4 and 3 → 2x2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3) - 11Factorise 3x2 − 10x + 8 (leading coefficient)
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ac = 24, numbers −6 and −4 → 3x2 − 6x − 4x + 8 = 3x(x − 2) − 4(x − 2) = (x − 2)(3x − 4) - 12Factorise ax + ay + bx + by (grouping)
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a(x + y) + b(x + y) = (x + y)(a + b) - 13Factorise x3 + 4x2 + 2x + 8 (grouping)
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x2(x + 4) + 2(x + 4) = (x + 4)(x2 + 2) - 14Factorise 4x2 − 100 (mixed)
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= 4(x2 − 25) = 4(x + 5)(x − 5) - 15Factorise 2x2 − 2x − 24 (mixed)
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= 2(x2 − x − 12) = 2(x − 4)(x + 3)
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