Identities and algebraic proof - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Identities and algebraic proof for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Here are two statements.

\(4(x + 2) = 20\)
\(4(x + 2) \equiv 4x + 8\)

Explain the difference between them.

Model answer

\(4(x + 2) = 20\) is an equation: it is only true for one value of x. Solving it, \(x + 2 = 5\), so x = 3.
\(4(x + 2) \equiv 4x + 8\) is an identity: the two sides are equal for every value of x, because expanding \(4(x + 2)\) always gives \(4x + 8\).

Explanation

An equation can be solved to find the particular value(s) that make it true. An identity is true whatever value the letter takes, and the symbol ≡ ("is identical to") shows this. Substituting different values of x into each statement shows the difference.

Mark scheme

1 mark
B1 for the first is an equation, which is true only for one value of x (x = 3)
oe; must say it is only true for a particular value
1 mark
B1 for the second is an identity, which is true for all values of x
oe

Question 2

3 marks

Show that \((2x - 1)(x + 3) - (x + 2)^2 \equiv x^2 + x - 7\)

Model answer

\((2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3\)
\((x + 2)^2 = x^2 + 4x + 4\)
\((2x - 1)(x + 3) - (x + 2)^2 = 2x^2 + 5x - 3 - (x^2 + 4x + 4)\)
\(= 2x^2 + 5x - 3 - x^2 - 4x - 4\)
\(= x^2 + x - 7\)

Explanation

Expand each product separately and keep the second one in a bracket, because the minus sign in front applies to all three of its terms. A common mistake is to write \(-x^2 + 4x + 4\), changing only the first sign.

Mark scheme

1 mark
M1 for \((2x - 1)(x + 3) = 2x^2 + 5x - 3\)
at least 3 of the 4 terms of \(2x^2 + 6x - x - 3\) correct
1 mark
M1 for \((x + 2)^2 = x^2 + 4x + 4\)
at least 3 of the 4 terms of \(x^2 + 2x + 2x + 4\) correct
1 mark
C1 for \(2x^2 + 5x - 3 - x^2 - 4x - 4 = x^2 + x - 7\), with all steps shown
answer only scores 0

Question 3

4 marks

n is an integer.

Prove that \(n^3 - n\) is always a multiple of 6.

Model answer

\(n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1) = (n - 1)n(n + 1)\)
\(n - 1\), n and \(n + 1\) are three consecutive integers.
In any two consecutive integers one is even, so the product is a multiple of 2.
In any three consecutive integers one is a multiple of 3, so the product is a multiple of 3.
A number that is a multiple of both 2 and 3 is a multiple of 6, so \(n^3 - n\) is always a multiple of 6.

Explanation

Factorise fully: take out n, then use the difference of two squares. This shows \(n^3 - n\) is the product of three consecutive integers. One of them must be even and one must be a multiple of 3, and because 2 and 3 have no common factor, the product is a multiple of 6.

Mark scheme

1 mark
M1 for \(n(n^2 - 1)\)
1 mark
M1 for \((n - 1)n(n + 1)\)
oe, in any order
1 mark
C1 for stating these are three consecutive integers, so at least one is a multiple of 2 and one is a multiple of 3
oe
1 mark
C1 for a complete conclusion, e.g. so the product is a multiple of 2 × 3 = 6
all steps needed

Question 4

5 marks

The diagram shows triangle ABC.
AB = \(n^2 - 1\), BC = \(2n\) and AC = \(n^2 + 1\), where n is an integer greater than 1.

Prove that triangle ABC is right-angled for every value of n.

Hence find the lengths of AB and BC for the triangle of this type in which AC = 145

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle ABC. Side AB is labelled n² − 1. Side BC is labelled 2n. Side AC is labelled n² + 1.

Model answer

\(AB^2 + BC^2 = (n^2 - 1)^2 + (2n)^2 = n^4 - 2n^2 + 1 + 4n^2 = n^4 + 2n^2 + 1\)
\(AC^2 = (n^2 + 1)^2 = n^4 + 2n^2 + 1\)
So \(AB^2 + BC^2 = AC^2\) for every value of n, and by Pythagoras' theorem the triangle is right-angled, with the right angle at B.
AC = 145: \(n^2 + 1 = 145\), \(n^2 = 144\), n = 12
AB = \(12^2 - 1 = 143\), BC = 2 × 12 = 24
(Check: \(143^2 + 24^2 = 20\,449 + 576 = 21\,025 = 145^2\))

Explanation

A triangle is right-angled exactly when the square of the longest side equals the sum of the squares of the other two (the converse of Pythagoras' theorem). Expand both sides in terms of n and show they are identical. Then use AC = 145 to find n and substitute into the other two expressions.

Mark scheme

1 mark
M1 for \((n^2 - 1)^2 + (2n)^2\)
oe; attempting Pythagoras with the two shorter sides
1 mark
M1 for \(n^4 - 2n^2 + 1 + 4n^2\)
1 mark
A1 for \(n^4 + 2n^2 + 1\)
1 mark
C1 for showing \((n^2 + 1)^2 = n^4 + 2n^2 + 1\), so \(AB^2 + BC^2 = AC^2\) and the triangle is right-angled (at B)
must have a conclusion
1 mark
B1 for AB = 143 and BC = 24
from \(n^2 + 1 = 145\), n = 12

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