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.
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)
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.
After trying it yourself, choose your next review. This is your self-assessment.
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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.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources