Revision Hub / Algebra / Factorising Quadratics

Algebra

Grade 5-7

Factorising Quadratics

Factorising a quadratic is the reverse of expanding double brackets, and the method depends on whether the coefficient of x² is 1 or greater than 1. This lesson covers factorising when the coefficient of x² is 1, factorising with mixed or negative signs, factorising when the coefficient of x² is greater than 1, and factorising fully by removing a common factor first.

What you need to know

  • For x squared plus b x plus c, find two numbers that multiply to c and add to b.
  • Those two numbers go straight into the brackets: (x + p)(x + q).
  • If c is positive both numbers share the sign of b; if c is negative they have opposite signs.
  • When the x squared coefficient is not 1, split the middle term or use the grid method.
Asad, co-founder of Teachably

Written by Asad, co-founder of Teachably

Video walkthrough coming soon

The written lesson below covers everything you need in the meantime.

Factorising when the coefficient of x² is 1

Find two numbers that multiply together to give the constant term and add together to give the coefficient of x, then write these as the second terms in two brackets.

To factorise x² + 7x + 12, find two numbers that multiply to 12 and add to 7: these are 3 and 4. This gives (x + 3)(x + 4).

Factorising with mixed or negative signs

The same method applies when the constant term or coefficient of x is negative, but the two numbers found may include negative values.

To factorise x² − 3x − 10, find two numbers that multiply to −10 and add to −3: these are −5 and 2. This gives (x − 5)(x + 2).

Factorising when the coefficient of x² is greater than 1

Multiply the coefficient of x² by the constant term to find a new target product, find two numbers that multiply to this and add to the coefficient of x, then split the middle term and factorise in pairs.

To factorise 2x² + 7x + 3, multiply 2 × 3 = 6, then find two numbers that multiply to 6 and add to 7: these are 6 and 1. Splitting the middle term gives 2x² + 6x + x + 3, which factorises in pairs as 2x(x + 3) + 1(x + 3), giving (2x + 1)(x + 3).

Factorising fully

If every term in the quadratic shares a common factor, take this out first, then factorise the simpler quadratic that remains inside the bracket.

To factorise fully 3x² + 21x + 30, take out the common factor 3 to get 3(x² + 7x + 10), then factorise the quadratic inside using two numbers that multiply to 10 and add to 7: 2 and 5, giving 3(x + 2)(x + 5).

Worked examples

Three exam-style questions, fully solved

Factorise x² + 7x + 12.

Easy
  1. 1.Find two numbers that multiply to 12 and add to 7: 3 and 4

Answer: (x + 3)(x + 4)

Factorise x² − 3x − 10.

Medium
  1. 1.Find two numbers that multiply to −10 and add to −3: −5 and 2

Answer: (x − 5)(x + 2)

Factorise fully 3x² + 21x + 30.

Hard
  1. 1.Take out the common factor: 3(x² + 7x + 10)
  2. 2.Find two numbers that multiply to 10 and add to 7: 2 and 5

Answer: 3(x + 2)(x + 5)

Practice

6 questions, marked instantly

Type an answer and check it. The worked solution appears once you have had a go. No login needed, and your progress saves in this browser.

Practice

Now try these yourself.

Type your answer and check it. The worked solution appears once you have had a go.

1

Factorise x² + 7x + 10.

2

Factorise x² + 2x − 15.

3

Factorise x² − 9x + 20.

4

Factorise x² − x − 12.

5

Factorise fully 2x² + 10x + 12.

6

Factorise 3x² + 5x − 2.

Avoid these

The mistakes students make most often

01

Finding two numbers that multiply to the coefficient of x instead of the constant term, or add to the constant term instead of the coefficient of x.

02

Getting a sign wrong when the two numbers include a negative value, especially when the constant term is negative and the two numbers must have different signs.

03

When the coefficient of x² is greater than 1, trying to find two numbers that multiply to the constant term alone, instead of multiplying the coefficient of x² by the constant term first.

04

Forgetting to check for a common factor before factorising, which means the quadratic is only partly factorised, not fully factorised.

05

Not checking the answer by expanding the brackets back out to confirm it matches the original quadratic.

FAQ

Questions students and parents ask

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