Factorising Algebraic Expressions — Grade 9 Maths (CAPS)

The four kinds of factorising the Grade 9 ATP asks for — common factors, the difference of two squares, trinomials of the form x² + bx + c, and trinomials whose x² coefficient comes out as a common factor — each with a worked example and a worksheet, plus a short Grade 10 preview at the end.

Also known as: factoring · factorizing · common factor · difference of two squares

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 a Grade 9 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. (Four terms need grouping, which is Grade 10 work — see the preview at the end.)

1Type 1: Common factor (always check this first)

Factorising is the same picture read backwards — find the side common to both rectangles.

Take out anything every term shares — the highest common number and the lowest power of any shared variable — and write it outside a bracket.

Worked ExampleExample 1a: numbers and one variable: 6x² + 9x

1
Highest common number of 6 and 9 is 3.
3 divides into both 6 and 9.
2
Both terms contain x, so x also comes out.
Take the lowest power of x present, which is x¹.
3
Factor out 3x: 6x² + 9x = 3x(2x + 3).
6x² ÷ 3x = 2x and 9x ÷ 3x = 3.
4
Check by expanding: 3x × 2x = 6x² and 3x × 3 = 9x. ✓
Expanding back must return the original.

Worked ExampleExample 1b: two variables: 12a²b − 8ab²

1
Common number of 12 and 8 is 4.
4 is the HCF of 12 and 8.
2
Both terms share a and b; the lowest powers are a¹ and b¹, so ab comes out.
Take the smallest power of each shared variable.
3
Factor 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.

a² − b² = (a + b)(a − b)

Worked ExampleExample 2a: basic: x² − 25

1
Recognise both parts are perfect squares: x² and 25 = 5².
The first sign must be minus for this method.
2
Take the square root of each: √x² = x and √25 = 5.
These become the a and b in the brackets.
3
Write (x + 5)(x − 5).
One bracket with +, one with −.

Worked ExampleExample 2b: with a coefficient: 9x² − 16

1
9x² = (3x)² and 16 = 4², so both are squares.
Square-root the coefficient too: √9 = 3.
2
Roots are 3x and 4, giving (3x + 4)(3x − 4).
Apply a² − b² with a = 3x, b = 4.

Worked ExampleExample 2c: common factor FIRST, then squares: 2x² − 50

1
Take out the common factor 2: 2x² − 50 = 2(x² − 25).
Always do Type 1 first — it simplifies what's left.
2
Now 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.

Worked ExampleExample 3a: both signs positive: x² + 7x + 12

1
Need two numbers: product 12, sum 7.
Both positive because c and b are positive.
2
3 and 4: 3 × 4 = 12 and 3 + 4 = 7.
Test factor pairs of 12 against the sum you need.
3
Answer: (x + 3)(x + 4).
Both brackets have +.

Worked ExampleExample 3b: c positive, b negative: x² − 7x + 12

1
Need 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.
3
Answer: (x − 3)(x − 4).
Both brackets have −.

Worked ExampleExample 3c: c negative (bigger number positive): x² + x − 12

1
Need 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.
3
Answer: (x + 4)(x − 3).
Different signs in the brackets.

Worked ExampleExample 3d: c negative (bigger number negative): x² − x − 12

1
Need 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.
3
Answer: (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: ax² + bx + c when a is a common factor

If the x² has a coefficient, look for a common factor first. In Grade 9 the coefficient always comes out as a common factor, and what is left is an ordinary Type 3 trinomial.

Worked ExampleExample 4: 3x² + 15x + 18

1
Take out the common factor 3: 3(x² + 5x + 6).
3 divides every term, so it comes out first.
2
Factorise the trinomial: two numbers with a product of 6 and a sum of 5 → 2 and 3.
This is Type 3 again.
3
3(x + 2)(x + 3)
Keep the 3 in front — it is part of the answer.

5How 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). If the x² has a coefficient, take it out as a common factor first (Type 4).
4 terms → grouping — Grade 10 work (see the preview below).

💡

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.

Types 1 to 4 are exactly what the Grade 9 ATP asks for. 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.

6Looking ahead: Grade 10 factorising

Not in the Grade 9 ATP. Grade 10 extends factorising to trinomials whose x² coefficient is NOT a common factor, and to expressions with four terms. You will not be asked these in a Grade 9 test; they are here so that the next step is not a surprise.

When the x² has a coefficient other than 1, use the ac-method: multiply a × c, split the middle term, then factorise by grouping.

Worked ExampleGrade 10 preview — the ac-method: 2x² + 7x + 3

1
Multiply 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.
2
Split the middle term: 2x² + 6x + 1x + 3.
Rewrite 7x as 6x + 1x using the two numbers.
3
Group in pairs: 2x(x + 3) + 1(x + 3).
Factor each pair separately.
4
Factor out (x + 3): (x + 3)(2x + 1).
Both pieces share the bracket (x + 3).

Grouping (Grade 10). With four terms, group them into two pairs, factor each pair, then take out the common bracket.

Worked ExampleGrade 10 preview — grouping: x³ + 2x² + 3x + 6

1
Group: (x³ + 2x²) + (3x + 6).
Pair terms that share a factor.
2
Factor each pair: x²(x + 2) + 3(x + 2).
x² comes out of the first pair, 3 out of the second.
3
Factor out (x + 2): (x + 2)(x² + 3).
The shared bracket is the common factor.

Practice exercises

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

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Factorise 5x + 15 (common factor)
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HCF is 5 → 5(x + 3)
2
Factorise 8a²b − 12ab (common factor)
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HCF is 4ab → 4ab(2a − 3)
3
Factorise x² − 49 (difference of squares)
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√x² = x, √49 = 7 → (x + 7)(x − 7)
4
Factorise 25x² − 9 (difference of squares)
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√25x² = 5x, √9 = 3 → (5x + 3)(5x − 3)
5
Factorise 3x² − 27 (common factor first)
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= 3(x² − 9) = 3(x + 3)(x − 3)
6
Factorise x² + 9x + 20 (trinomial, both +)
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Product 20, sum 9 → 4 and 5 → (x + 4)(x + 5)
7
Factorise x² − 8x + 15 (trinomial, c + b −)
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Product 15, sum −8 → −3 and −5 → (x − 3)(x − 5)
8
Factorise x² + 2x − 15 (trinomial, c −)
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Product −15, sum +2 → +5 and −3 → (x + 5)(x − 3)
9
Factorise x² − 2x − 15 (trinomial, c −)
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Product −15, sum −2 → −5 and +3 → (x − 5)(x + 3)
10
Factorise 2x² + 14x + 20 (common factor, then trinomial)
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= 2(x² + 7x + 10); product 10, sum 7 → 2 and 5 → 2(x + 2)(x + 5)
11
Factorise 5x² − 5x − 30 (common factor, then trinomial)
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= 5(x² − x − 6); product −6, sum −1 → −3 and +2 → 5(x − 3)(x + 2)
12
Factorise x²y − 9y (common factor, then difference of squares)
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= y(x² − 9) = y(x + 3)(x − 3)
13
Factorise a² − 6a + 9 (trinomial)
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Product 9, sum −6 → −3 and −3 → (a − 3)(a − 3) = (a − 3)²
14
Factorise 4x² − 100 (mixed)
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= 4(x² − 25) = 4(x + 5)(x − 5)
15
Factorise 2x² − 2x − 24 (mixed)
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= 2(x² − x − 12) = 2(x − 4)(x + 3)