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A4bFactorising

Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Algebra

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Revision notes

Factorising is the reverse of expanding: taking out common factors, factorising quadratics \(x^2 + bx + c\) and differences of two squares (both tiers) and, on Higher, quadratics \(ax^2 + bx + c\). You need it to solve quadratics, simplify algebraic fractions and write proofs.

Grade by grade

What you need to be able to do, from the first marks up to the top grade.

  1. 4
    Take out a single common factorFind a number or letter that divides every term, e.g. \(6x + 15 = 3(2x + 5)\).
  2. 5
    Factorise fully using the HCFTake out the highest common factor of the numbers and letters, e.g. \(8x^2y - 12xy = 4xy(2x - 3)\).
  3. 5
    Factorise quadratics with x² coefficient 1Find two numbers that multiply to c and add to b, e.g. \(x^2 + 2x - 15 = (x + 5)(x - 3)\).
  4. 5
    Factorise a difference of two squaresUse \(a^2 - b^2 = (a + b)(a - b)\), e.g. \(x^2 - 49 = (x + 7)(x - 7)\).

Notes

Common factors

  • Find the HCF of the numbers and the lowest power of each letter that is in every term.
  • Write the HCF outside the bracket and divide each term by it to find what goes inside: \(12a^2 - 18ab = 6a(2a - 3b)\).
  • 'Factorise fully' means nothing else can be taken out: \(3a(4a - 6b)\) is correct but not fully factorised.
  • Check by expanding your answer.

Quadratics \(x^2 + bx + c\)

  • Find two numbers that multiply to give c and add to give b.
  • \(x^2 + 7x + 12\): \(3 \times 4 = 12\) and \(3 + 4 = 7\), so it is \((x + 3)(x + 4)\).
  • If c is negative, the numbers have different signs: \(x^2 - 2x - 24 = (x - 6)(x + 4)\).
  • If c is positive and b is negative, both numbers are negative: \(x^2 - 9x + 20 = (x - 4)(x - 5)\).

Difference of two squares

  • \(a^2 - b^2 = (a + b)(a - b)\). Look for two square terms with a minus sign between them.
  • \(x^2 - 36 = (x + 6)(x - 6)\) and \(25y^2 - 4 = (5y + 2)(5y - 2)\).
  • A sum of two squares, such as \(x^2 + 36\), does not factorise.

Cheatsheet

  • Factorising is the reverse of expanding
  • Common factor: HCF outside, divide each term by it to fill the bracket
  • \(x^2 + bx + c = (x + p)(x + q)\) where \(pq = c\) and \(p + q = b\)
  • \(a^2 - b^2 = (a + b)(a - b)\)
  • Check any factorisation by expanding it

How to answer each type of question

Factorise fully (common factor)

2 marks5
  1. Find the HCF of the numbers.
  2. Find the lowest power of each letter that appears in every term.
  3. Divide each term by the HCF to fill the bracket, then check nothing else divides both terms.

Example. Factorise fully \(10p^2q + 15pq^2\)

Show the model answer
\(5pq(2p + 3q)\) B2 (B1 for a correct but incomplete factorisation, e.g. \(5p(2pq + 3q^2)\))

Factorise a quadratic \(x^2 + bx + c\)

2 marks5
  1. List factor pairs of c, with signs.
  2. Choose the pair that adds to b.
  3. Write the two brackets and expand to check.

Example. Factorise \(x^2 - 3x - 28\)

Show the model answer
\(-7 \times 4 = -28\) and \(-7 + 4 = -3\)
\((x - 7)(x + 4)\) B2 (B1 for \((x \pm 7)(x \pm 4)\))

Difference of two squares, then 'hence'

3 marks5
  1. Write each term as a square: \(x^2 - 64 = x^2 - 8^2\).
  2. Use \((a + b)(a - b)\).
  3. For 'hence', use the same pattern with the numbers given.

Example. (a) Factorise \(x^2 - 64\).
(b) Hence, without a calculator, work out \(58^2 - 42^2\).

Show the model answer
(a) \((x + 8)(x - 8)\) B1
(b) \((58 + 42)(58 - 42)\) M1
\(= 100 \times 16 = 1600\) A1

Shortcuts and memory tricks

  • Signs: if c is positive, both numbers have the same sign as b; if c is negative, one is positive and one is negative, and the larger one takes the sign of b.
  • Learn the squares 1, 4, 9, …, 144 so you spot differences of two squares quickly.
  • Expanding your answer to check takes a few seconds and catches sign errors.
  • If a question says 'fully', expect more than one step.

Where marks are lost

  • Not factorising fully, e.g. leaving a common factor of 2 inside the bracket.
  • Getting the signs the wrong way round, e.g. \((x + 7)(x - 4)\) for \(x^2 - 3x - 28\) (this expands to \(x^2 + 3x - 28\)).
  • Trying to factorise a sum of two squares such as \(x^2 + 25\).
  • Solving instead of factorising: the answer to 'factorise' is brackets, not values of x.

Exam technique

  • 'Factorise' wants brackets; 'solve' wants values of x. Read the command word.
  • If part (a) says 'factorise' and part (b) says 'hence solve', use your brackets from part (a).
  • Write down the pair of numbers you found as working.

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks
(a) Factorise \(28y + 21\)[1]
(b) Factorise fully \(7y^{2} - 21y\)[2]
Show the answer and mark scheme
(a) Answer: \(7(4y + 3)\)
  • B1 for \(7(4y + 3)\)
(b) Answer: \(7y(y - 3)\)
  • B2 for \(7y(y - 3)\) (B1 for a correct partial factorisation, e.g. \(y(7y - 21)\) or \(7(y^{2} - 3y)\))

Worked solution: The highest common factor of the terms is \(7y\), so \(7y^{2} - 21y = 7y(y - 3)\)

Question 2Medium3 marks
(a) Factorise fully \(30pq^{3} + 18p^{2}q^{3}\)[2]
(b) Factorise \(a(a - 1) - 6(a - 1)\)[1]
Show the answer and mark scheme
(a) Answer: \(6pq^{3}(5 + 3p)\)
  • B2 for \(6pq^{3}(5 + 3p)\) (B1 for a correct factorisation that is not complete, e.g. \(6(5pq^{3} + 3p^{2}q^{3})\) or \(pq^{3}(30 + 18p)\))

Worked solution: The highest common factor is \(6pq^{3}\), so \(30pq^{3} + 18p^{2}q^{3} = 6pq^{3}(5 + 3p)\)

(b) Answer: \((a - 1)(a - 6)\)
  • B1 for \((a - 1)(a - 6)\)

Worked solution: \((a - 1)\) is a common factor: \(a(a - 1) - 6(a - 1) = (a - 1)(a - 6)\)

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