Factorise quadratics where the x2 coefficient is not 1, and use the factors to solve the equation. It is Higher only, aimed at grade 7. Free questions with answers are on this page. Worked examples show each line before you try the questions.
When the x2 coefficient is not 1, look for two numbers that multiply to ac and add to b. Split the middle term, factor in pairs, and the two brackets that appear must match. 6x2+7x−3: ac=−18, numbers 9 and −2, so 6x2+9x−2x−3=3x(2x+3)−1(2x+3)=(3x−1)(2x+3).
Difference of two squares with coefficients: 9x2−16=(3x−4)(3x+4). Take out a numerical common factor first if you can. Expand to check. Solving is then each bracket =0.
For ax2+bx+c with a=1, find two numbers that multiply to ac and add to b, split the middle, factor by grouping. Difference of two squares still applies to 4x2−9.
Objectives
Factorise ax² + bx + c with a greater than 1.
Factorise a difference of squares.
Factorise 2a + 2b as 2(a + b).
Key points
Find two numbers that multiply to ac and add to b.
Split the middle term and factorise in pairs (grouping).
Spot a difference of two squares such as 4x2−9.
Take out a common factor first if every term shares one.
Worked examples
Example 1
Factorise 2x²+7x+3.
2 marks
Identify factors of 2: 2 and 1, so brackets start (2x + ?)(x + ?).
Find factors of 3: 1 and 3, testing cross terms 2x × 3 + 1 × x = 7x.
Write the factorisation: (2x+1)(x+3).
Answer: (2x+1)(x+3)
Common slips
Slip: Assume brackets start (x + ?)(x + ?). The leading coefficient 2 must be split into factors.
Slip: Use (2x + 3)(x + 1). (2x + 3)(x + 1) gives 5x instead of 7x.
Slip: Write (x+1)(x+3). (x+1)(x+3) expands to x² + 4x + 3, which has no 2x² term.
Example 2
Factorise 6x²+5x+1.
2 marks
Identify potential splits for 6x²: (6x, x) or (3x, 2x).
Test factors of 1: only 1 × 1, giving cross terms 3x × 1 + 2x × 1 = 5x.
State the factorisation: (3x+1)(2x+1).
Answer: (3x+1)(2x+1)
Common slips
Slip: Always assume (6x, x). Composite coefficients can split as 3 and 2.
Slip: Test (6x + 1)(x + 1). (6x + 1)(x + 1) produces middle term 7x.
Slip: Write (6x+1)(x+1). (3x+1)(2x+1) gives the correct 5x.
Example 3
Solve 2x²+7x+3=0.
3 marks
Factorise the quadratic: (2x+1)(x+3) = 0.
Set each factor to zero: 2x + 1 = 0 or x + 3 = 0.
Solve each linear equation: x = -1/2 or x = -3.
Answer: -3 or -1/2
Common slips
Slip: Attempt to divide by 2. Factorising is the exact algebraic method.
Slip: Set only x + 3 = 0. A product is zero if either factor equals zero.
Slip: Give only integer root x = -3. Include both solutions.
Questions and answers
GCSE Maths harder factorising (ax² + bx + c) questions and answers. Try each one, then open the answer. Easy questions come first.
Question 1 Easy2 marks
Factorise 2a+2b completely.
Show answer
Answer: 2(a+b)
Take out the common factor 2.
Divide terms by 2: 2(a+b).
M1 — Take out the common factor 2.
A1 — Divide terms by 2: 2(a+b).
Question 2 Easy2 marks
Factorise 3x²+7x+2.
Show answer
Answer: (3x+1)(x+2)
Test splits of 3 and 2.
Cross terms 6x + x = 7x: (3x+1)(x+2).
M1 — Test splits of 3 and 2.
A1 — Cross terms 6x + x = 7x: (3x+1)(x+2).
Question 3 Medium2 marks
Factorise 6x²+7x+2.
Show answer
Answer: (3x+2)(2x+1)
Split 6 as 3 and 2, and 2 as 2 and 1.
Cross terms 4x + 3x = 7x: (3x+2)(2x+1).
M1 — Split 6 as 3 and 2, and 2 as 2 and 1.
A1 — Cross terms 4x + 3x = 7x: (3x+2)(2x+1).
Question 4 Medium2 marks
Factorise 4x²+8x+3.
Show answer
Answer: (2x+1)(2x+3)
Split 4 as 2 and 2, and 3 as 1 and 3.
Cross terms 6x + 2x = 8x: (2x+1)(2x+3).
M1 — Split 4 as 2 and 2, and 3 as 1 and 3.
A1 — Cross terms 6x + 2x = 8x: (2x+1)(2x+3).
Question 5 Hard2 marks
Factorise 4x²-9.
Show answer
Answer: (2x-3)(2x+3)
Square root 4x² is 2x and square root 9 is 3.
Difference of squares: (2x-3)(2x+3).
M1 — Square root 4x² is 2x and square root 9 is 3.
A1 — Difference of squares: (2x-3)(2x+3).
Question 6 Hard2 marks
Factorise 9x²-16.
Show answer
Answer: (3x-4)(3x+4)
Square root 9x² is 3x and square root 16 is 4.
Difference of squares: (3x-4)(3x+4).
M1 — Square root 9x² is 3x and square root 16 is 4.
A1 — Difference of squares: (3x-4)(3x+4).
Question 7 Exam-style4 marks
Solve 2x² + 9x + 4 = 0.
Show answer
Answer: -4 or -1/2
Factorise quadratic: 2x² + 9x + 4.
Factors are (2x + 1)(x + 4) = 0.
Set each factor to zero: 2x + 1 = 0 or x + 4 = 0.
Solve 2x + 1 = 0: x = -1/2.
Solve x + 4 = 0: x = -4.
M1 — (2x + a)(x + b) with ab = 4
A1 — (2x+1)(x+4)
M1 — Set a factor equal to zero
A1 — Both roots, x = -4 and x = -1/2
Common mistakes
Most lost marks on this topic are one slip.
A pair that multiplies to c but adds to the wrong middle
Test the cross terms by expanding
Forgetting to divide by the leading coefficient when solving
2x + 1 = 0 gives x = -1/2, not -1
Dropping the minus sign in difference of squares
One bracket has minus and one has plus
Spot the slip
Factorise 2x²+7x+3. Dan writes (2x+3)(x+1). What is the correct factorisation?
(2x+1)(x+3)
(2x+3)(x+1)
(2x+7)(x+3)
Show the slip
(2x+1)(x+3) (2x+1)(x+3) gives middle term 6x + x = 7x.
Exam tips
Be ready to do this without a calculator. Paper 1 is the non-calculator paper.
If the question says “show that”, the result is given. The marks are for the steps that reach it.
The method is the same on AQA-style and Edexcel-style papers.
Write the pair of numbers you found (e.g. '−9 and 2') — it shows method even if the grouping goes wrong.
Expand your brackets mentally before moving on; a two-second check saves the accuracy mark.
Printable worksheet
The same questions are in a PDF you can print. Answers start on a later page.