Algebra · GCSE Maths
Factorising quadratics
GCSE Maths factorising quadratics: highest common factor first, then two brackets for x² + bx + c, and grouping when a ≠ 1 — expand every time as a check.
Factorising is expanding in reverse. HCF first. For x² + bx + c, hunt two numbers that multiply to c and add to b. Expand to check.
The important bits
What you need to know
- 1
Factorising writes a sum as a product. 6x + 9 = 3(2x + 3) is HCF factorisation. Quadratics need two brackets when they factorise over the integers.
- 2
For x² + bx + c = 0, find two numbers p and q with pq = c and p + q = b. Then (x + p)(x + q) = 0. For x² + 7x + 10, p = 2 and q = 5.
- 3
Signs matter. x² − 5x + 6 needs numbers that multiply to +6 and add to −5: −2 and −3, so (x − 2)(x − 3). x² − x − 6 needs +2 and −3.
- 4
Rearrange to ax² + bx + c = 0 before factorising. x² + 5x = 6 becomes x² + 5x − 6 = 0, then (x + 6)(x − 1) = 0.
- 5
When a ≠ 1, such as 2x² + 7x + 3, look for two numbers that multiply to a × c = 6 and add to b = 7: 6 and 1. Split: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
- 6
The zero-product property: if (x + 2)(x − 3) = 0, then x + 2 = 0 or x − 3 = 0. Each bracket gives a root. Do not divide away a factor and lose a root.
- 7
Difference of two squares: x² − 9 = (x + 3)(x − 3). Perfect square: x² + 6x + 9 = (x + 3)². Recognise these before you hunt arbitrary pairs.
- 8
If no integer pair works, the quadratic may not factorise over the integers. Use the quadratic formula instead. Five minutes of false pairs wastes the paper.
Quotations worth analysing
Short evidence. Real method.
“(x + p)(x + q) where pq = c and p + q = b”
The hunt is multiply to c, add to b. Writing (x + 10)(x + 1) for x² + 11x + 10 is correct; for x² + 9x + 10 you need 5 and 4, not 10 and 1.
“2x² + 7x + 3 = (2x + 1)(x + 3)”
Split the middle term using numbers that multiply to ac and add to b, then factor by grouping. Skipping the split and guessing brackets rarely expands back to the original.
“Expand immediately as a check”
A neat bracket with a wrong sign inside fails the check in ten seconds. (x − 2)(x − 3) expands to x² − 5x + 6, not x² − 5x − 6.
Go deeper
HCF before brackets
Always scan for a common factor first. 4x² − 12x = 4x(x − 3). 6x² + 9x − 15 = 3(2x² + 3x − 5), then factorise the quadratic inside if it factorises. Students who jump to two brackets on 2x² + 10x miss that 2x is common: 2x(x + 5). The HCF line is often a method mark. For a difference of two squares, x² − 16 = (x + 4)(x − 4) with no middle term. For a perfect square, x² − 10x + 25 = (x − 5)² because half of −10 is −5 and (−5)² = 25. Spot these before the pair hunt; they are faster and less error-prone.
Go deeper
The pair hunt when a = 1
x² + bx + c: list factor pairs of c, pick the pair that sums to b. For c = 12 and b = 7, pairs of 12 are 1×12, 2×6, 3×4; 3 + 4 = 7, so (x + 3)(x + 4). For c = −12 and b = −1, you need −4 and 3 because (−4)(3) = −12 and −4 + 3 = −1: (x − 4)(x + 3). If c is positive and b is negative, both numbers in the pair are negative. If c is negative, one positive and one negative. After writing brackets, expand: first terms, outside, inside, last (FOIL). The middle term is the cross terms added. If the middle does not match, the pair was wrong, not the method.
Go deeper
Grouping when a ≠ 1
2x² + 7x + 3: ac = 6, b = 7, so split 7x as 6x + x. Rewrite 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). The common bracket (x + 3) is the signal that grouping worked. If you get different brackets, recheck the split. 6x² − 7x − 3: ac = −18, b = −7, split as −9x and 2x because (−9)(2) = −18 and −9 + 2 = −7: 6x² − 9x + 2x − 3 = 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Solving: set each factor zero. (2x + 1)(x + 3) = 0 gives x = −½ or x = −3. Write both roots; a single answer from one bracket loses a mark when the quadratic had two distinct factors.
See the idea in action
Factorise fully 2x² + 7x + 3. Step 1: No HCF across all three terms. a = 2, b = 7, c = 3, so ac = 6. Step 2: Numbers that multiply to 6 and add to 7 are 6 and 1. Step 3: Split: 2x² + 6x + x + 3. Step 4: Group: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Step 5: Check: (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3. If solving 2x² + 7x + 3 = 0: x = −½ or x = −3.
Exam technique
Turn knowledge into marks
Write “ac = …, need × to ac and + to b” before you split. Expand your final brackets; if the middle term is wrong, the pair was wrong.
Common mistakes
Do not give these marks away
- 01
Hunting bracket pairs before taking out an HCF, or forgetting to factorise the HCF in the final answer.
- 02
Choosing a pair that multiplies to c but does not add to b, especially with negative c.
- 03
Giving only one root after factorising into two different linear factors.
x² − 5x + 6 factorises as
A(x − 2)(x − 3)
B(x + 2)(x + 3)
C(x − 6)(x + 1)
D(x − 1)(x + 6)
Show the answer
(x − 2)(x − 3). Need numbers that multiply to +6 and add to −5: −2 and −3. (x + 2)(x + 3) sums to +5. (x − 6)(x + 1) sums to −5 but multiplies to −6.
Quick questions
If this is the bit you searched
What if I cannot find two numbers?
The quadratic may not factorise over the integers. Use the quadratic formula or completing the square. Do not invent decimals inside brackets.
Do I need = 0 to factorise?
To solve, rearrange to = 0 first. To factorise an expression such as x² + 7x + 10, the brackets are the same: (x + 2)(x + 5).
How is this different from expanding?
Expanding multiplies brackets to a sum. Factorising is the reverse. Every factorisation should expand back to the start.
What is factorising by grouping?
Split the middle term, group into two pairs, factor each pair, then factor the common bracket. It is the standard method when a ≠ 1.