Solving Quadratic Equations by Factorising: GCSE Guide

A quadratic equation has the form ax² + bx + c = 0. To solve it by factorising, you rewrite the left-hand side as a product of two brackets, then use the fact that if two things multiply to give zero, at least one of them must be zero. This gives you two solutions.

What is a quadratic equation?

A quadratic equation contains an x² term (and no higher powers). The standard form is:

ax² + bx + c = 0

where a, b, and c are numbers and a ≠ 0. Examples: x² + 5x + 6 = 0, 2x² − x − 3 = 0, x² − 9 = 0.

The solutions (also called roots) are the values of x that make the equation true. Most quadratics have two roots; some have one repeated root; a few have none (no real solutions).

How do you factorise a quadratic with a = 1?

When the coefficient of x² is 1, the equation looks like x² + bx + c = 0. You need two numbers that:

multiply to give c

add to give b

Step 1 — List factor pairs of c.
Step 2 — Identify the pair that adds to b.
Step 3 — Write the factorised form: (x + p)(x + q) = 0.
Step 4 — Set each bracket equal to zero: x + p = 0 or x + q = 0.
Step 5 — Solve each: x = −p or x = −q.

Worked example 1: solve x² + 7x + 12 = 0

Factor pairs of 12: (1 × 12), (2 × 6), (3 × 4).
Which pair adds to 7? 3 and 4 (3 + 4 = 7, 3 × 4 = 12).
Factorised: (x + 3)(x + 4) = 0.
Set each to zero: x + 3 = 0 → x = −3 or x + 4 = 0 → x = −4.

Solutions: x = −3 or x = −4

Worked example 2: solve x² − 5x + 6 = 0

Factor pairs of 6: (1 × 6), (2 × 3).
Need pair adding to −5 and multiplying to +6: both must be negative. −2 and −3 (−2 + −3 = −5, −2 × −3 = +6).
Factorised: (x − 2)(x − 3) = 0.

Solutions: x = 2 or x = 3

How do you factorise when a ≠ 1?

When a > 1, use the ac method (also called splitting the middle term).

Step 1 — Multiply a × c.
Step 2 — Find two numbers that multiply to ac and add to b.
Step 3 — Split the middle term using these two numbers.
Step 4 — Factorise by grouping.
Step 5 — Set each bracket to zero and solve.

Worked example 3: solve 2x² + 5x + 3 = 0

a × c = 2 × 3 = 6. Find two numbers multiplying to 6 and adding to 5: 2 and 3.
Split: 2x² + 2x + 3x + 3 = 0.
Group: 2x(x + 1) + 3(x + 1) = 0.
Factor out (x + 1): (2x + 3)(x + 1) = 0.
Set to zero: 2x + 3 = 0 → x = −3/2, or x + 1 = 0 → x = −1.

Solutions: x = −3/2 or x = −1

What is the difference of two squares?

A special case arises when the quadratic has no middle term and is of the form a² − b²:

a² − b² = (a + b)(a − b)

Worked example 4: solve x² − 25 = 0

Recognise x² − 25 = x² − 5². Apply the rule: (x + 5)(x − 5) = 0.
Set to zero: x + 5 = 0 → x = −5, or x − 5 = 0 → x = 5.

Solutions: x = −5 or x = 5

Worked example 5: solve 4x² − 9 = 0

Rewrite as (2x)² − 3². Factorise: (2x + 3)(2x − 3) = 0.
2x + 3 = 0 → x = −3/2, or 2x − 3 = 0 → x = 3/2.

Solutions: x = −3/2 or x = 3/2

Why must the equation equal zero before factorising?

This is the most important rule: the equation must equal zero before you factorise and apply the zero-product property. If you have x² + 3x = 10, you cannot say "x(x + 3) = 10, therefore x = 10 or x + 3 = 10." That reasoning is wrong. You must rearrange first:

x² + 3x − 10 = 0 → (x + 5)(x − 2) = 0 → x = −5 or x = 2.

Quick reference table

Type | Example | Factorised form | Solutions
a = 1, both positive | x² + 5x + 4 = 0 | (x + 1)(x + 4) = 0 | x = −1 or x = −4
a = 1, mixed signs | x² − x − 6 = 0 | (x − 3)(x + 2) = 0 | x = 3 or x = −2
a > 1 | 3x² + 7x + 2 = 0 | (3x + 1)(x + 2) = 0 | x = −1/3 or x = −2
Difference of squares | x² − 16 = 0 | (x + 4)(x − 4) = 0 | x = ±4

Frequently asked questions

What if I cannot find two numbers that work?

If no integer factor pair exists, the quadratic may not factorise neatly. In that case, use the quadratic formula or complete the square — both are alternative GCSE methods. Always check whether the question specifies "by factorising" or allows any method.

How do I check my solutions?

Substitute each solution back into the original equation. If both sides balance, the solution is correct. For example, for x² + 7x + 12 = 0 with x = −3: (−3)² + 7(−3) + 12 = 9 − 21 + 12 = 0 ✓.

Can a quadratic have only one solution?

Yes. When the factorised form is (x + p)² = 0, both brackets are identical and the only solution is x = −p. This is called a repeated root. For example, x² − 6x + 9 = (x − 3)² = 0, giving x = 3 only.

Is factorising always the quickest method?

Factorising is fastest when the quadratic factors neatly. If the numbers are awkward or you cannot spot the factor pair quickly, the quadratic formula is more reliable and works every time. In an exam, if a question says "solve by factorising," you must use factorising — not the formula.

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