OCR GCSE Mathematics: Algebraic language, proof, simplifying and expanding (J560) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow OCR GCSE (9-1) Mathematics, specification Version 2.1 (August 2026), sections 6.01a to 6.01d. Questions 1 to 7 are for both tiers. Questions 8 to 11 are marked Higher tier only. Question 12 is for both tiers apart from part (c), which is marked Higher tier only. They are algebra skills, so work them by hand and show every line.

Revise first with the study guide or the revision notes. Tick off topics on the checklist, see the course hub, and find your weak spots with the free diagnostics.

Questions

1. Simplify. (a) 9m − 4n + 2m + 7n [2] (b) 5x² + 3x − 2x² − 8x [2]

2. Here are four items. (i) 3x + 7 (ii) 4t > 20 (iii) P = 2l + 2w (iv) 5y − 1 = 9 (a) For each item, write expression, equation, formula or inequality. Each word is used once. [4] (b) Write down a factor of 12ab that is a letter. [1]

3. (a) Say whether each of these is an equation or an identity. [2] (i) 3(x + 2) = 3x + 6 (ii) 5x − 1 = 14 (b) Show that (x + 4)(x − 4) ≡ x² − 16. [2] (c) Ben says that (x + 3)² = x² + 9. Use x = 1 to show that Ben is wrong. [1]

4. A triangle has sides of length (4x − 3) cm, (2x + 5) cm and (3x + 1) cm. (a) Show that the perimeter of the triangle is (9x + 3) cm. [2] (b) Jo says: “If x is a whole number, the perimeter is always odd.” Show that Jo is wrong. [2]

5. Expand. (a) 4(3a − 2b) [2] (b) Expand and simplify 5(2x + 3) − 2(3x − 4) [3]

6. Simplify. (a) 3a²b × 5ab³ [2] (b) 20x⁵y² ÷ 8x²y [2] (c) 15m⁴n / 10mn³ [2]

7. (a) Expand and simplify (x + 6)(x − 2). [2] (b) Expand and simplify (3x − 4)(2x + 5). [3] (c) Expand (2a − 5)². [2] (d) Show that (x + 3)² − (x − 3)² ≡ 12x. [3]

8. (Higher tier only) Simplify. (a) 2a⁶ × 4a⁻² [2] (b) 12y³ ÷ 3y⁵, giving your answer as a fraction [2] (c) (27p⁶)^(2/3) [2]

9. (Higher tier only) (a) Expand and simplify (x + 4)(x − 1)(x + 2). [4] (b) Expand and simplify (x + 2)³. [3]

10. (Higher tier only) Prove that the sum of four consecutive integers is never a multiple of 4. [4]

11. (Higher tier only) Prove that (2n + 3)² − (2n − 1)² is a multiple of 8 for every integer n. [5]

12. A cuboid has dimensions x cm, (x + 2) cm and (x + 5) cm. (a) Show that its volume is (x³ + 7x² + 10x) cm³. [2] (b) Find an expression for the total surface area of the cuboid. Give your answer as simplified as possible. [4] (c) (Higher tier only) A second cuboid has dimensions (x − 2) cm, (x + 1) cm and (x + 4) cm. Show that its volume is (x³ + 3x² − 6x − 8) cm³. [3]

Answers

1. (a) m terms: 9m + 2m = 11m [1]; n terms: −4n + 7n = 3n. Answer: 11m + 3n [1] (b) x² terms: 5x² − 2x² = 3x² [1]; x terms: 3x − 8x = −5x. Answer: 3x² − 5x [1] Examiner insight: Treat x² and x as different kinds of term; adding them into one term is a common slip that makes the final answer wrong.

2. (a) (i) expression [1]; (ii) inequality [1]; (iii) formula [1]; (iv) equation [1] (b) Any of a, b: for example a [1] Examiner insight: Look for the symbol. An equals sign with several letters linking real quantities is a formula, while an equals sign you could solve for one letter is an equation.

3. (a) (i) identity, because it is true for every x [1]; (ii) equation, because it is true only when x = 3 [1] (b) (x + 4)(x − 4) = x² − 4x + 4x − 16 [1] = x² − 16, which matches the right-hand side [1] (c) At x = 1 the left side is 4² = 16 and the right side is 1 + 9 = 10, so they differ and Ben is wrong [1] Examiner insight: In a “show that”, write the expanded line before the simplified line, and end by saying the result matches the target. A bare final answer gives the reader nothing to follow.

4. (a) Perimeter = (4x − 3) + (2x + 5) + (3x + 1) [1] = 9x + 3 [1] (b) Choose x = 3, which gives sides 9, 11 and 10 [1]. The perimeter is 9 × 3 + 3 = 30, which is even, so Jo is wrong [1] Examiner insight: A disproof needs one specific value that is allowed in the question and a clear statement of what it shows. Check that your chosen value gives valid side lengths.

5. (a) 4 × 3a = 12a [1] and 4 × (−2b) = −8b. Answer: 12a − 8b [1] (b) 5(2x + 3) = 10x + 15 [1]; −2(3x − 4) = −6x + 8 [1]; combined: 10x + 15 − 6x + 8 = 4x + 23 [1] Examiner insight: Write each bracket as its own line first. The second bracket has a negative multiplier, so check that both signs inside have changed.

6. (a) Numbers: 3 × 5 = 15 [1]; a: a² × a = a³, b: b × b³ = b⁴. Answer: 15a³b⁴ [1] (b) Numbers: 20 ÷ 8 = 5/2 [1]; x⁵ ÷ x² = x³ and y² ÷ y = y. Answer: 5x³y / 2 [1] (c) Numbers: 15/10 = 3/2 [1]; m⁴ ÷ m = m³, and n ÷ n³ leaves n² on the bottom. Answer: 3m³ / 2n² [1] Examiner insight: Deal with the number, then each letter in turn. Keep a letter that ends up on the bottom of a fraction there; do not turn it into a negative power unless the question asks.

7. (a) x² − 2x + 6x − 12 [1] = x² + 4x − 12 [1] (b) 6x² + 15x − 8x − 20 [2] = 6x² + 7x − 20 [1] (c) (2a − 5)(2a − 5) = 4a² − 10a − 10a + 25 [1] = 4a² − 20a + 25 [1] (d) (x + 3)² = x² + 6x + 9 [1]; (x − 3)² = x² − 6x + 9 [1]; subtracting: 12x [1] Examiner insight: Write the squared bracket out twice. The error (a − b)² = a² − b² removes the middle term and leaves the answer wrong.

8. (a) Numbers: 2 × 4 = 8 [1]; a^(6 + (−2)). Answer: 8a⁴ [1] (b) 12 ÷ 3 = 4 [1]; y^(3 − 5) = y⁻² = 1/y². Answer: 4 / y² [1] (c) 27^(2/3) = 3² = 9 [1]; p^(6 × 2/3) = p⁴. Answer: 9p⁴ [1] Examiner insight: For a fractional power, take the root first, then the power. Treating every part of the bracket separately, number and letter, shows the method clearly.

9. (a) (x + 4)(x − 1) = x² − x + 4x − 4 [1] = x² + 3x − 4 [1]. Multiply by (x + 2): x³ + 2x² + 3x² + 6x − 4x − 8 [1]. Answer: x³ + 5x² + 2x − 8 [1] (b) (x + 2)² = x² + 4x + 4 [1]; (x² + 4x + 4)(x + 2) = x³ + 2x² + 4x² + 8x + 4x + 8 [1]. Answer: x³ + 6x² + 12x + 8 [1] Examiner insight: Tidy the first product before bringing in the third bracket. Substituting x = 1 in the original and your answer is a quick check: here (a) gives 5 × 0 × 3 = 0 and 1 + 5 + 2 − 8 = 0.

10. Let the integers be n, n + 1, n + 2 and n + 3 [1]. Sum = 4n + 6 [1] = 4(n + 1) + 2 [1]. This is a multiple of 4 plus 2, so it leaves remainder 2 and is never a multiple of 4 [1] Examiner insight: A proof needs general letters, a simplified form and a closing sentence. Four worked numerical examples would not prove a statement about every integer.

11. (2n + 3)² = 4n² + 12n + 9 [1]; (2n − 1)² = 4n² − 4n + 1 [1]. Difference = 16n + 8 [1] = 8(2n + 1) [1]. As 2n + 1 is an integer, the expression is 8 times an integer, so it is a multiple of 8 [1] Examiner insight: Put brackets around the second square before subtracting, so the sign of every term is changed. The factorised form and the final sentence are what complete the proof.

12. (a) (x + 2)(x + 5) = x² + 7x + 10 [1]; multiply by x: x³ + 7x² + 10x [1] (b) Face areas: x(x + 2) = x² + 2x, x(x + 5) = x² + 5x and (x + 2)(x + 5) = x² + 7x + 10 [2]. Sum = 3x² + 14x + 10 [1]. Each face appears twice, so the total is 6x² + 28x + 20 [1] (c) (x − 2)(x + 1) = x² − x − 2 [1]; (x² − x − 2)(x + 4) = x³ + 4x² − x² − 4x − 2x − 8 [1] = x³ + 3x² − 6x − 8, which matches the target [1] Examiner insight: In extended questions, define each face before adding. Remember the factor of 2 for opposite faces. In part (c), expand two brackets and tidy before using the third.

Where marks are usually lost

• Expanding (x + a)² as x² + a², with the middle term missing.
• Not changing every sign inside a bracket that follows a minus sign or a negative multiplier.
• Collecting terms that are not like terms, such as x² and x.
• Adding indices when raising a power to a power, or multiplying when adding is needed.
• Reading a negative power as a negative number.
• Using one letter for two different numbers in a proof.
• Giving numerical examples as a proof instead of a general argument.
• Leaving a proof without a final sentence connecting the form to the claim.
• Working on both sides at once in a “show that” question.

Next steps

• Review the revision notes and the study guide
• Visit the OCR GCSE Mathematics hub and the checklist
• Take one of the free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, published by OCR. These questions cover sections 6.01a to 6.01d.