IGCSEMathsCambridge (CIE)ExtendedRevision NotesAlgebra & SequencesQuadratic EquationsSolving Quadratics by Factorising

Solving Quadratics by Factorising (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

Exam code: 0580 & 0980

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 19 January 2026

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Solving quadratics by factorising

How do I solve a quadratic equation using factorisation?

Rearrange it into the form ax² + bx + c = 0
- Zero must be on one side
- It is easier if you rearrange so that a is positive
Factorise the quadratic and solve each bracket equal to zero
- If (x + 4)(x - 1) = 0, then either x + 4 = 0 or x - 1 = 0
  - Because if two things multiply together to give zero,
    - then one or the other of them must be equal to zero
To solve $(x - 3) (x + 7) = 0$
- …solve first bracket = 0:
  - x – 3 = 0
  - add 3 to both sides: x = 3
- …and solve second bracket = 0
  - x + 7 = 0
  - subtract 7 from both sides: x = -7
- The two solutions are x = 3 or x = -7
  - The solutions in this example are the numbers in the brackets, but with opposite signs

What if there are numbers in front of the x's in the brackets?

The process is the same
- There's a bit more work to find the solutions
- You can't just write down the answers by changing the signs
To solve $(2 x - 3) (3 x + 5) = 0$
- …solve first bracket = 0
  - 2x – 3 = 0
  - add 3 to both sides: 2x = 3
  - divide both sides by 2: x = $\frac{3}{2}$
- …solve second bracket = 0
  - 3x + 5 = 0
  - subtract 5 from both sides: 3x = -5
  - divide both sides by 3: x = $- \frac{5}{3}$
- The two solutions are x = $\frac{3}{2}$ or x = $- \frac{5}{3}$

What if x is a factor?

The process is the same
- Just be sure to handle the x correctly
- That 'x as a factor' gives one of the solutions
To solve $x (x - 4) = 0$
- it may help to think of x as (x – 0) or (x)
- …solve first bracket = 0
  - (x) = 0, so x = 0
- …solve second bracket = 0
  - x – 4 = 0
  - add 4 to both sides: x = 4
- The two solutions are x = 0 or x = 4
It is a common mistake to divide (cancel) both sides by x at the beginning
- If you do this you will lose a solution (the x = 0 solution)

Examiner Tips and Tricks

Your calculator might be able to solve quadratics. If it does, then you can use this to help you to factorise the quadratic.

For example, a calculator gives solutions to $6 x^{2} + x - 2 = 0$ as $x = - \frac{2}{3}$ and $x = \frac{1}{2}$ . You can rearrange these solutions to get zero on one side, $3 x + 2 = 0$ and $2 x - 1 = 0$ . The left-hand side expressions are two factors, $6 x^{2} + x - 2 = (3 x + 2) (2 x - 1)$ .

Worked Example

(a) Solve $x^{2} + 3 x - 10 = 0$ by factorising.

Answer:

Factorise the left-hand side

(x – 2)(x + 5)=0

Set the first bracket equal to zero

x – 2 = 0

Add 2 to both sides

x = 2

Set the second bracket equal to zero

x + 5 = 0

Subtract 5 from both sides

x = -5

Write both solutions together using “or”

x = 2 or x = -5

(b) Solve $5 x^{2} - x = 0$ by factorising.

Answer:

Factorise the left-hand side

x(5x - 1)=0

Do not divide both sides by x (this will lose a solution at the end)
Set the first bracket equal to zero

(x) = 0

Solve this equation to find x

x = 0

Set the second bracket equal to zero

5x - 1 = 0

Add 1 to both sides

5x = 1

Divide both sides by 5

x = $\frac{1}{5}$

Write both solutions together using “or”

x = 0 or x = $\frac{1}{5}$

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Questions students ask about Solving Quadratics by Factorising

Real questions from students revising this topic on Save My Exams

When should you solve a quadratic equation by factorising rather than using the quadratic formula?

Factorise when the question tells you to, or when the quadratic factorises easily; use the formula when no whole numbers or simple fractions work.

Factorise when:

The question says "solve by factorising" — then set out the brackets, because that is the method being asked for
You can spot two numbers that multiply to c and add to b
There is no number term, e.g. 5x² − x = 0 factorises straight to x(5x − 1) = 0, giving x = 0 or x = 1/5

Use the formula when:

No pair of integers works
The answers are not whole numbers or simple fractions, so no pair of brackets will give them exactly

Compare:

x² + 3x − 10 = 0 factorises to (x − 2)(x + 5) = 0, so x = 2 or x = −5
x² + 3x − 5 = 0 does not factorise, so the formula gives x = 1.19 or x = −4.19 (2 d.p.)

Your calculator is a quick test: simple fractions mean it factorises, long decimals mean it does not.

Does it matter which order the brackets are written in?

No — the order of the brackets makes no difference, because multiplication can be done in any order.

Just as 3 × 4 is the same as 4 × 3, (x − 2)(x + 5) and (x + 5)(x − 2) both expand to x² + 3x − 10.
Either way you set each bracket equal to zero, so you get the same pair of solutions.
On this page, x² + 3x − 10 = 0 factorises to (x − 2)(x + 5) = 0: x − 2 = 0 gives x = 2, and x + 5 = 0 gives x = −5. Writing (x + 5)(x − 2) = 0 simply finds x = −5 first.
You can list your two solutions in either order too.

What does matter:

the sign in front of each number — (x + 2)(x − 5) expands to x² − 3x − 10, a different quadratic
keeping each number with the right x term when there are coefficients — (2x − 3)(3x + 5) and (2x + 5)(3x − 3) are not the same
expanding your brackets to check you get back the quadratic you started with