OCR GCSE Mathematics: Factorising, completing the square and algebraic fractions (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 (J560), Version 2.1 (August 2026), content 6.01e, 6.01f and 6.01g. Questions 1 to 4 are on both tiers. Questions 5 to 12 are Higher tier only. Work by hand and check each factorisation by expanding.

Links: study guide, revision notes, OCR GCSE Mathematics hub, checklist and free diagnostics.

Questions

1. Factorise fully.

(a) 7x − 21 [1] (b) 6y² + 15y [2] (c) 12ab² − 8a²b [2]

2. Factorise.

(a) x² + 8x + 15 [2] (b) x² − 3x − 28 [2] (c) x² − 12x + 35 [2]

3.

(a) Factorise x² − 121. [1] (b) Factorise 16a² − 9b². [2] (c) Factorise fully 5x² − 45. [2] (d) Write x² − 11 as a product of two brackets. [1]

4. A rectangular lawn has area (x² + 9x + 14) m² and width (x + 2) m.

(a) Factorise x² + 9x + 14 and write down the length of the lawn in terms of x. [2] (b) Show that the perimeter of the lawn is (4x + 18) m. [2]

5. (Higher tier only) Factorise.

(a) 2x² + 9x + 10 [2] (b) 5x² + 17x + 6 [2] (c) 8x² + 10x − 3 [2]

6. (Higher tier only) Factorise fully 18x² − 3x − 6. [3]

7. (Higher tier only) Write each expression in the form (x + p)² + q.

(a) x² + 12x + 5 [2] (b) x² − 9x + 3 [2]

8. (Higher tier only) Write each expression in the form a(x + p)² + q.

(a) 4x² − 16x + 9 [3] (b) 2x² + 3x − 4 [3]

9. (Higher tier only) Simplify.

(a) (x² − 25)/(x² + 2x − 15) [3] (b) (3x² + x − 10)/(x² − 4) [3]

10. (Higher tier only) Write as a single fraction, simplifying your numerator.

(a) 3/(x + 1) + 4/(x − 2) [3] (b) 5/(2x − 1) − 2/x [3]

11. (Higher tier only) Simplify fully [(x² + 4x)/(x² − x − 2)] × [(x + 1)/(x² − 16)]. [4]

12. (Higher tier only)

(a) Show that (x² + 3x − 10)/(x² − 4) + 3/(x + 2) simplifies to (x + 8)/(x + 2). [4] (b) Write (x + 8)/(x + 2) in the form 1 + k/(x + 2), where k is an integer. [2]

Answers

1. (a) 7(x − 3) [1] (b) Common factor 3y [1]; 3y(2y + 5) [1] (c) Common factor 4ab [1]; 4ab(3b − 2a) [1] Examiner insight: In part (c) the number and each letter must be taken out. 2ab(6b − 4a) is only partly factorised, so check for a common factor left in the bracket.

2. (a) Numbers with product 15 and sum 8: 3 and 5 [1]; (x + 3)(x + 5) [1] (b) Product −28 and sum −3: 4 and −7 [1]; (x − 7)(x + 4) [1] (c) Product 35 and sum −12, so both negative: −5 and −7 [1]; (x − 5)(x − 7) [1] Examiner insight: Finding the right pair is the main step, so write the pair down. Writing the pair down also makes it easy to spot where a slip happened.

3. (a) (x − 11)(x + 11) [1] (b) 16a² = (4a)² and 9b² = (3b)² [1]; (4a − 3b)(4a + 3b) [1] (c) Take out 5: 5(x² − 9) [1]; 5(x − 3)(x + 3) [1] (d) 11 = (√11)², so (x − √11)(x + √11) [1] Examiner insight: Part (c) is a common trap. Looking for a difference of two squares before taking out the common factor can leave you stuck, so check for a common factor first.

4. (a) x² + 9x + 14 = (x + 2)(x + 7) [1]; length = (x + 7) m [1] (b) Perimeter = 2(x + 2) + 2(x + 7) [1]; = 2x + 4 + 2x + 14 = 4x + 18 [1] Examiner insight: On a “show that” question write every line of the algebra, including the unsimplified perimeter, and finish on the given result.

5. (a) ac = 20, numbers 4 and 5 [1]; 2x² + 4x + 5x + 10 = 2x(x + 2) + 5(x + 2) = (2x + 5)(x + 2) [1] (b) ac = 30, numbers 15 and 2 [1]; 5x² + 15x + 2x + 6 = 5x(x + 3) + 2(x + 3) = (5x + 2)(x + 3) [1] (c) ac = −24, numbers 12 and −2 [1]; 8x² + 12x − 2x − 3 = 4x(2x + 3) − (2x + 3) = (4x − 1)(2x + 3) [1] Examiner insight: If the two brackets after splitting do not match, a sign or a number pair is wrong. Expanding your answer is a quick way to catch this.

6. Common factor 3: 3(6x² − x − 2) [1]; ac = −12, numbers −4 and 3: 6x² − 4x + 3x − 2 = 2x(3x − 2) + (3x − 2) [1]; 3(3x − 2)(2x + 1) [1] Examiner insight: “Fully” signals a common factor. Keep the 3 in front of your final answer, since dropping it changes the expression.

7. (a) (x + 6)² = x² + 12x + 36 [1]; (x + 6)² − 31 [1] (b) (x − 9/2)² = x² − 9x + 81/4 [1]; 3 − 81/4 = (x − 9/2)² − 69/4 [1] Examiner insight: The sign inside the bracket matches the sign of the x coefficient halved, and the bracket’s constant is subtracted. Expand to check.

8. (a) 4(x² − 4x) + 9 [1]; = 4[(x − 2)² − 4] + 9 [1]; 4(x − 2)² − 7 [1] (b) 2(x² + 3x/2) − 4 [1]; = 2[(x + 3/4)² − 9/16] − 4 [1]; 2(x + 3/4)² − 41/8 [1] Examiner insight: Multiplying the subtracted square by the outside factor is an easy step to miss. Writing the bracket in square brackets before multiplying out helps.

9. (a) Numerator (x − 5)(x + 5) [1]; denominator (x + 5)(x − 3) [1]; (x − 5)/(x − 3) [1] (b) Numerator (3x − 5)(x + 2) [1]; denominator (x − 2)(x + 2) [1]; (3x − 5)/(x − 2) [1] Examiner insight: Factorise top and bottom on separate lines before cancelling. Cancelling the x or the 5 directly is not allowed, so factorise first.

10. (a) Common denominator (x + 1)(x − 2) [1]; numerator 3(x − 2) + 4(x + 1) [1]; (7x − 2)/[(x + 1)(x − 2)] [1] (b) Common denominator x(2x − 1) [1]; numerator 5x − 2(2x − 1) = 5x − 4x + 2 [1]; (x + 2)/[x(2x − 1)] [1] Examiner insight: In part (b) the minus sign applies to the whole of 2(2x − 1). Writing the bracket stops the sign error.

11. (x² + 4x) = x(x + 4); x² − x − 2 = (x − 2)(x + 1); x² − 16 = (x − 4)(x + 4) [2]; product = x(x + 4)(x + 1) / [(x − 2)(x + 1)(x − 4)(x + 4)] [1]; x/[(x − 2)(x − 4)] [1] Examiner insight: Factorising all three quadratics first is the efficient route. Multiplying out the numerators and denominators makes cancelling much harder.

12. (a) Numerator x² + 3x − 10 = (x + 5)(x − 2) [1]; denominator x² − 4 = (x − 2)(x + 2) [1]; fraction simplifies to (x + 5)/(x + 2) [1]; (x + 5)/(x + 2) + 3/(x + 2) = (x + 8)/(x + 2) [1] (b) x + 8 = (x + 2) + 6 [1]; 1 + 6/(x + 2), so k = 6 [1] Examiner insight: A fraction with the same denominator is added by adding numerators only. Keep the denominator as (x + 2) and do not add it twice.

How to check your answers

• Factorising: expand your brackets. You should get the original expression back, with every term and sign.
• Completing the square: expand the answer. For (x + p)² + q the x term is 2p and the constant is p² + q. With a factor a outside, multiply both by a.
• Algebraic fractions: substitute a simple value such as x = 1 or x = 2 into the question and your answer. Both should give the same number, as long as that value does not make a denominator zero.

Where marks are usually lost

• Leaving a common factor outside the final brackets.
• Using a number pair that adds correctly but does not multiply correctly.
• Sign errors when c is negative.
• Not expanding to check a factorisation.
• Cancelling terms in an algebraic fraction instead of factors.
• Losing a sign when subtracting one algebraic fraction from another.
• Forgetting to multiply the subtracted square by a when completing the square.
• Giving q with the wrong sign in (x + p)² + q.
• Not showing the intermediate line in a “show that” question.
• Stopping at a single bracket pair when a numerical factor, such as the 5 in question 3(c) or the 3 in question 6, is still outside.

Next steps

• Re-read the study guide for any question you got wrong.
• Use the revision notes for the self-test.
• Visit the OCR GCSE Mathematics hub and tick items off 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. Content statements 6.01e, 6.01f and 6.01g.