Identities
Algebra
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Show that $(x + 2)^{2} \equiv x^{2} + 4x + 4$.
2 marks
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Is $x^{2} + 4 \equiv (x + 2)^{2}$ an identity? Give a reason.
2 marks
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Show that $(x + 3)(x - 3) \equiv x^{2} - 9$.
3 marks
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Find $a$ and $b$ such that $x^{2} + 8x + 7 \equiv (x + a)^{2} + b$.
3 marks
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Show that $(n + 1)^{2} - (n - 1)^{2} \equiv 4n$.
4 marks
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Prove that $(2n + 1)^{2} - (2n - 1)^{2}$ is a multiple of $8$ for every integer $n$.
4 marks
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Find $p$ and $q$ such that $2(x + 3)^{2} + 5 \equiv 2x^{2} + px + q$.
5 marks
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Show that $(3n + 2)^{2} - (3n)^{2} \equiv 12n + 4$. Explain why this value is never a multiple of $3$. Find its value when $n = 5$.
6 marks
Mark scheme
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Answer. Both sides expand to the same expression.
- $(x + 2)^{2} = x^{2} + 4x + 4$.
- The two sides are identical for every $x$.
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Answer. No. $(x + 2)^{2} = x^{2} + 4x + 4$.
- Expand the right side.
- The extra $4x$ means the sides are not always equal. For example, when $x = 1$ they are $5$ and $9$.
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Answer. The brackets expand to $x^{2} - 9$.
- $(x + 3)(x - 3) = x^{2} - 3x + 3x - 9$
- $x^{2} - 9$
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Answer. $a = 4$ and $b = -9$
- $(x + 4)^{2} = x^{2} + 8x + 16$
- $(x + 4)^{2} - 9$
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Answer. The left side simplifies to $4n$.
- $(n^{2} + 2n + 1) - (n^{2} - 2n + 1)$
- $4n$
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Answer. The difference is $8n$.
- $(4n^{2} + 4n + 1) - (4n^{2} - 4n + 1) = 8n$
- $8n$ is a multiple of $8$.
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Answer. $p = 12$ and $q = 23$
- $(x + 3)^{2} = x^{2} + 6x + 9$
- $2(x^{2} + 6x + 9) + 5 = 2x^{2} + 12x + 23$
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Answer. The difference is $12n + 4$, which leaves remainder $1$ when divided by $3$. When $n = 5$ it is $64$.
- $(9n^{2} + 12n + 4) - 9n^{2} = 12n + 4$.
- $12n$ is a multiple of $3$, and $4$ is not, so $12n + 4$ is never a multiple of $3$.
- When $n = 5$, $12 \times 5 + 4 = 64$.