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A4aExpanding brackets

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

Practise Expanding brackets. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

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

Question 1Easy3 marks
Jake says, '\((x - 5)^2 = x^2 - 25\) for all values of \(x\).'
(a) By substituting a value of \(x\), show that Jake is wrong.[1]
(b) Expand and simplify \((x - 5)^2\).[2]
Show the answer and mark scheme
(a) Answer: E.g. \(x = 1\): \((1 - 5)^2 = 16\) but \(1^2 - 25 = -24\).
  • C1 for a correct substitution showing the two sides are different, e.g. \(x = 1\) gives 16 and −24

Worked solution: When \(x = 1\): \((x - 5)^2 = (-4)^2 = 16\) and \(x^2 - 25 = -24\). These are different, so Jake is wrong.

(b) Answer: \(x^2 - 10x + 25\)
  • M1 for \(x^2 - 5x - 5x + 25\) (four terms, at least three correct)
  • A1 for \(x^2 - 10x + 25\)

Worked solution: \((x - 5)(x - 5) = x^2 - 5x - 5x + 25 = x^2 - 10x + 25\)

Question 2Medium4 marks
(a) Expand and simplify \(3s(2s - 5t) - 2t(5s + 2t)\)[2]
(b) Expand \(6b^{3}(5b^{2} + 3)\)[2]
Show the answer and mark scheme
(a) Answer: \(6s^2 - 25st - 4t^2\)
  • M1 for at least three of the four terms in \(6s^2 - 15st - 10st - 4t^2\) correct
  • A1 for \(6s^2 - 25st - 4t^2\) cao

Worked solution: \(3s(2s - 5t) - 2t(5s + 2t) = 6s^2 - 15st - 10st - 4t^2 = 6s^2 - 25st - 4t^2\)

(b) Answer: \(30b^{5} + 18b^{3}\)
  • B2 for \(30b^{5} + 18b^{3}\) (B1 for one correct term)

Worked solution: Multiply each term in the bracket by \(6b^{3}\) and add the powers of \(b\): \(6b^{3}(5b^{2} + 3) = 30b^{5} + 18b^{3}\)

Question 3Hard3 marks
\((3x + p)^2 - (3x - 1)(3x + 1) \equiv 42x + q\)
where \(p\) and \(q\) are integers.
Work out the value of \(p\) and the value of \(q\).[3]
Show the answer and mark scheme
Answer: \(p = 7, \ q = 50\)
  • M1 for expanding both brackets correctly, \(9x^2 + 6px + p^2\) and \(9x^{2} - 1\)
  • M1 for comparing the coefficients of \(x\): \(6p = 42\)
  • A1 for \(p = 7\) and \(q = 50\)

Worked solution: \((3x + p)^2 - (3x - 1)(3x + 1) = 9x^2 + 6px + p^2 - (9x^{2} - 1) = 6px + p^2 + 1\)
Coefficients of \(x\): \(6p = 42\), so \(p = 7\).
Constant terms: \(q = 7^2 + 1 = 50\).

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