Edexcel International GCSE Mathematics A 4MA1: Use of symbols and algebraic manipulation – 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 cover sections 2.1 Use of symbols and 2.2 Algebraic manipulation of Topic 2, Equations, formulae and identities, in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Questions 1–4 use content on both tiers. Questions 5–10 and 12, and part 11(c), are Higher tier only and are marked (Higher); the rest of question 11 uses content on both tiers. A calculator may be used on every 4MA1 paper, but no calculator will do the algebra for you, so show every line: most marks here are for method.

Learn the content first in the study guide and the revision notes. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement. For more questions on the same content, the earlier topic-level practice set uses different expressions.

Questions

1. (Both tiers) Simplify

(a) t⁴ × t⁵ [1] (b) w¹⁰ ÷ w⁻³ [1] (c) (h³)⁴ [1] (d) Write 1/y⁶ as a single power of y. [1]

2. (Both tiers) p = 3, q = −2 and r = 0.5.

(a) Work out the value of 4p − 5q. [2] (b) Work out the value of p²q − 6r. [2]

3. (Both tiers)

(a) Simplify 7a + 3b − 2a − 8b. [1] (b) Expand and simplify 3(2x − 5) − 4(x − 2). [2] (c) Factorise fully 15m²n − 20mn². [2]

4. (Both tiers)

(a) Expand and simplify (y + 6)(y − 9). [2] (b) Factorise x² + 3x − 40. [2] (c) Factorise x² − 13x + 36. [2]

5. (Higher)

(a) Work out 81^(3/4). [1] (b) Work out 125^(−1/3). [1] (c) Work out (4/9)^(−3/2), giving your answer as a fraction. [2] (d) Simplify fully (27x⁶y⁻³)^(2/3), giving your answer with positive powers only. [2]

6. (Higher) Expand and simplify (3x + 2)(x − 1)(2x + 5). [3]

7. (Higher) Factorise fully

(a) 8x² + 14x − 15 [2] (b) 3x² − 75 [2] (c) 25a² − 16b² [1]

8. (Higher)

(a) Simplify fully (2x² − 7x − 4)/(x² − 16). [3] (b) Write 5/(x − 2) − 3/(x + 1) as a single fraction in its simplest form. [3]

9. (Higher)

(a) Write x² + 10x + 31 in the form (x + p)² + q. [2] (b) Hence explain why x² + 10x + 31 is positive for all values of x. [1] (c) Write 2x² − 12x + 7 in the form a(x + b)² + c. [3]

10. (Higher) Prove that the sum of the squares of any two consecutive integers is always odd. [3]

11. A rectangle has length (2x + 3) cm and width (x − 1) cm. A square has sides of length (x + 2) cm.

(a) Show that the area of the rectangle is (2x² + x − 3) cm². [2] (b) Show that the area of the rectangle minus the area of the square is (x² − 3x − 7) cm². [2] (c) (Higher) Write x² − 3x − 7 in the form (x + p)² + q. [2] (d) Explain why x must be greater than 1. [1]

12. (Higher)

(a) Express (x + 3)/4 − (2x − 1)/6 as a single fraction in its simplest form. [3] (b) Simplify fully (x² + 5x)/(x² + 2x − 15) ÷ 2x/(x + 1). [4]

Answers

1. (a) t⁴⁺⁵ = t⁹ [1] (b) w¹⁰⁻⁽⁻³⁾ = w¹³ [1] (c) h³ˣ⁴ = h¹² [1] (d) y⁻⁶ [1] Examiner insight: Each part is a single accuracy mark, so w⁷ in (b), from subtracting 3 instead of −3, scores nothing; there is no method mark to fall back on.

2. (a) 4(3) − 5(−2) [1] = 12 + 10 = 22 [1] (b) (3)²(−2) − 6(0.5) [1] = −18 − 3 = −21 [1] Examiner insight: The method mark needs the correct substitution seen; writing 12 − 10 = 2 in (a) shows the sign of q was lost, so neither mark is earned.

3. (a) 5a − 5b [1] (b) 6x − 15 − 4x + 8 [1] = 2x − 7 [1] (c) 5mn(3m − 4n) [1] [1] Examiner insight: In (c) a correct but partial factorisation, such as 5m(3mn − 4n²) or mn(15m − 20n), usually earns 1 of the 2 marks; “fully” needs both the 5 and mn outside.

4. (a) y² − 9y + 6y − 54 (at least three of the four terms correct) [1] = y² − 3y − 54 [1] (b) Two numbers with product −40 and sum +3: 8 and −5 [1]; (x + 8)(x − 5) [1] (c) Product +36, sum −13: −4 and −9 [1]; (x − 4)(x − 9) [1] Examiner insight: For factorising, the method mark is typically given for brackets that expand to give two of the three terms correctly, such as (x − 8)(x + 5); expand your answer to catch this.

5. (a) Fourth root of 81 is 3, 3³ = 27 [1] (b) 1/125^(1/3) = 1/5 [1] (c) (9/4)^(3/2) [1] = (3/2)³ = 27/8 [1] (d) 27^(2/3) = 9 and x⁶ˣ⁽²ᐟ³⁾ = x⁴ (at least two of the three parts correct) [1]; y⁻³ˣ⁽²ᐟ³⁾ = y⁻², so 9x⁴/y² [1] Examiner insight: In (d) the answer 9x⁴y⁻² gets the method mark but not the accuracy mark, because the question asks for positive powers only.

6. (3x + 2)(x − 1) = 3x² − x − 2 [1]; (3x² − x − 2)(2x + 5) = 6x³ + 15x² − 2x² − 5x − 4x − 10 (all six terms, at most one error) [1] = 6x³ + 13x² − 9x − 10 [1] Examiner insight: Marks are usually given for the first pair of brackets expanded correctly, then for the six-term expansion; an answer with no working that is wrong gets nothing.

7. (a) ac = −120, numbers 20 and −6: 8x² + 20x − 6x − 15 = 4x(2x + 5) − 3(2x + 5) [1]; (4x − 3)(2x + 5) [1] (b) 3(x² − 25) [1] = 3(x − 5)(x + 5) [1] (c) (5a − 4b)(5a + 4b) [1] Examiner insight: In (b), 3(x² − 25) or (3x − 15)(x + 5) gets only the method mark; “fully” means taking out the 3 and then using the difference of two squares.

8. (a) Numerator (2x + 1)(x − 4) [1]; denominator (x − 4)(x + 4) [1]; (2x + 1)/(x + 4) [1] (b) [5(x + 1) − 3(x − 2)] / [(x − 2)(x + 1)] [1]; numerator 5x + 5 − 3x + 6 = 2x + 11 [1]; (2x + 11)/[(x − 2)(x + 1)] [1] Examiner insight: In (b) the common-denominator mark can still be earned with a sign error, but 5x + 5 − 3x − 6 loses both the numerator mark and the final answer.

9. (a) (x + 5)² − 25 + 31 [1] = (x + 5)² + 6 [1] (b) (x + 5)² ≥ 0 for all x, so (x + 5)² + 6 ≥ 6 > 0 [1] (c) 2(x² − 6x) + 7 [1]; 2[(x − 3)² − 9] + 7 [1] = 2(x − 3)² − 11 [1] Examiner insight: “Hence” in (b) means the explanation must use your completed-square form; testing a few values of x scores 0.

10. Let the integers be n and n + 1 [1]. n² + (n + 1)² = n² + n² + 2n + 1 = 2n² + 2n + 1 [1]. 2n² + 2n + 1 = 2(n² + n) + 1, which is one more than a multiple of 2, so the sum is always odd [1] Examiner insight: The last mark needs the expression written as 2(…) + 1 with a statement; stopping at 2n² + 2n + 1 does not show it is odd.

11. (a) (2x + 3)(x − 1) = 2x² − 2x + 3x − 3 [1] = 2x² + x − 3 [1] (b) Square: (x + 2)² = x² + 4x + 4 [1]; 2x² + x − 3 − (x² + 4x + 4) = x² − 3x − 7 [1] (c) (x − 1.5)² − 2.25 − 7 [1] = (x − 1.5)² − 9.25 [1] (d) The width x − 1 must be positive, so x > 1 [1] Examiner insight: On a “show that”, every line must be written; jumping from (2x + 3)(x − 1) straight to the given answer earns no marks because the result is printed in the question.

12. (a) Common denominator 12: [3(x + 3) − 2(2x − 1)] / 12 [1]; 3x + 9 − 4x + 2 [1] = (11 − x)/12 [1] (b) x² + 5x = x(x + 5) and x² + 2x − 15 = (x + 5)(x − 3) [1]; multiply by the reciprocal (x + 1)/(2x) [1]; cancel x and (x + 5) [1] = (x + 1)/[2(x − 3)] [1] Examiner insight: A common denominator of 24 in (a) is acceptable, but the answer must then be cancelled to (11 − x)/12 for the final mark, as the question asks for simplest form.

Where marks are usually lost

• Multiplying indices when they should be added: t⁴ × t⁵ written as t²⁰.
• Substituting negatives without brackets, so (−2) becomes −2 and signs go wrong.
• Leaving out the HCF of the letters when factorising fully, or leaving out the number.
• Missing a common factor before factorising a quadratic, as in 3x² − 75.
• Losing a term when expanding three brackets: there must be six products before collecting.
• Putting a sign wrong in the second numerator when subtracting algebraic fractions.
• Cancelling terms that are not whole factors, such as the x² in (2x² − 7x − 4)/(x² − 16).
• Not multiplying the subtracted square by a when completing the square with a ≠ 1.
• On “show that” questions, jumping to the printed answer instead of showing each line.
• In proofs, giving numerical examples instead of general algebra, or omitting the final statement.

Next steps

• Refresh the key results: revision notes
• Re-learn anything you missed: study guide
• Course hub: Edexcel IGCSE Mathematics
• Track coverage: printable checklist
• Check other topics: all free 10-minute diagnostics
• Want help from a teacher? Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities: sections 2.1 Use of symbols and 2.2 Algebraic manipulation (Foundation and Higher Tier content).