Factorising quadratics and grouped expressions
Reverse expansion by finding a common structure, including quadratics whose leading coefficient is not one.
Extended only: quadratic patterns, grouping and higher-power expressions. Extracting a common factor is also taught in the preceding Core lesson sequence below. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Expand two brackets and collect terms.
- Find factor pairs of positive and negative integers.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Expand (x + 2)(x − 2).
What you will learn
- Extract the greatest common factor first.
- Factorise quadratics, difference of squares and perfect squares.
- Use grouping and factorise cubic expressions with a common variable.
Remove the common factor first
Factorising expresses a sum as a product. In 12x² + 18x, both terms contain 6x, giving 6x(2x + 3). This is fully factorised because the bracket has no further common factor. Extracting only 3x leaves another factor of two inside. Check by multiplying the factors back out.
A cubic such as 2x³ − 8x² + 6x first becomes 2x(x² − 4x + 3), then 2x(x − 1)(x − 3). Extracting x exposes a quadratic. Factorising is not the same as solving: an expression by itself gives no values of x until an equation is supplied.
Find the middle coefficient through factor pairs
For x² + bx + c, seek p and q with p + q = b and pq = c. Then the factors are (x + p)(x + q). For x² − x − 12, the pair 3 and −4 has product −12 and sum −1, so the factors are (x + 3)(x − 4). The constant's sign tells whether the pair has equal or different signs.
For 2x² + 7x + 3, the product of the leading coefficient and constant is 6. Split 7x as 6x + x: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Grouping works because the same binomial is now common. Expanding the result checks every coefficient.
Expand several brackets or several variables
Multiply three brackets in stages. For (x + 1)(x − 1)(x + 2), the first two give x² − 1. Distribute the third to obtain x³ + 2x² − x − 2. Keep the intermediate expression visible so every term is multiplied; the answer's highest degree is three.
For (2x + y)(x − 3y), the four products are 2x², −6xy, xy and −3y². Collect only the matching xy terms, giving 2x² − 5xy − 3y². The variables are distinct, so x², xy and y² cannot be combined. Expanding a perfect square such as (2x + 3y)² gives 4x² + 12xy + 9y², including its middle product.
Recognise patterns and group a shared bracket
The difference u² − v² equals (u − v)(u + v), so 9x² − 16y² = (3x − 4y)(3x + 4y). A sum of squares is not the same pattern. A perfect square u² + 2uv + v² equals (u + v)²; the middle term must match twice the product, not merely look even.
For px + qx + 2py + 2qy, collect as x(p + q) + 2y(p + q), then (p + q)(x + 2y). Grouping may need the signs of an entire bracket reversed. In ax − ay − bx + by, the second group is −b(x − y), giving (a − b)(x − y).
Pause and explain
Which pair factorises x² + 2x − 15?
Worked example
Factorise 3x³ − 12x fully.
Show the worked solution
- Extract the greatest common factor: 3x(x² − 4).
- Recognise x² − 4 as a difference of squares.
- The complete product is 3x(x − 2)(x + 2); expanding returns the original.
Answer 3x(x − 2)(x + 2)
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Factorise 15a²b − 10ab² fully.
Give me a hint
Find the common numerical factor and minimum powers.
Compare my reasoning
- The numerical common factor is 5; both terms contain ab.
- Divide each term by 5ab to obtain 3a and −2b.
- Write 5ab(3a − 2b) and check its expansion.
5ab(3a − 2b)
Look for these in your work
- Extracted the greatest common monomial.
- Preserved the subtraction sign.
Factorise 6x² − x − 2.
Give me a hint
Split the middle term using a pair with product −12 and sum −1.
Compare my reasoning
- The pair is 3 and −4, so write 6x² + 3x − 4x − 2.
- Group as 3x(2x + 1) − 2(2x + 1).
- Take out the common bracket to get (3x − 2)(2x + 1).
(3x − 2)(2x + 1)
Look for these in your work
- Matched both product and sum.
- Checked all three coefficients by expansion.
A frame has outer sides (x + 4) cm and inner sides x cm, both squares. Factorise its area difference and find the frame area at x = 6.
Give me a hint
Use a difference of squares for the two areas.
Compare my reasoning
- Frame area is (x + 4)² − x².
- Factorise as [(x + 4) − x][(x + 4) + x] = 4(2x + 4) = 8(x + 2).
- For x = 6, the area is 8 × 8 = 64 cm², agreeing with 10² − 6².
8(x + 2) cm²; 64 cm²
Look for these in your work
- Subtracted areas rather than side lengths.
- Used and checked the factorised expression.
Expand (x + 1)(x − 1)(x + 2) and factorise px + qx + 2py + 2qy.
Give me a hint
For the expansion, multiply two brackets first. For factorisation, create two groups with the same bracket.
Compare my reasoning
- (x + 1)(x − 1) = x² − 1, so multiplying by (x + 2) gives x³ + 2x² − x − 2.
- The four-term expression groups as x(p + q) + 2y(p + q).
- Extract the common bracket to get (p + q)(x + 2y); expansion checks it.
x³ + 2x² − x − 2; (p + q)(x + 2y)
Look for these in your work
- Distributed the third bracket to every intermediate term.
- Extracted an entire shared binomial.
Common mistakes
- Stopping before removing every common factor.
- Using a factor pair that has the right product but wrong sum.
- Treating a sum of squares as a difference of squares.
What can you explain now?
Factorise x² − 6x + 9.
Compare with the explanation
(x − 3)²
The numbers −3 and −3 sum to −6 and multiply to 9.
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Make it stick
Factorise one common-factor expression, one non-monic quadratic and one difference of squares. Expand each result to check the identity.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.