Pearson Edexcel IGCSE Mathematics: Use of Symbols and Algebraic Manipulation — Practice Questions (4MA1)

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 available free from your board.

Related: Use of symbols and algebraic manipulation study guide

Tiers: questions and parts marked (Higher tier only) use content that the 4MA1 specification (Issue 2) lists for the Higher Tier only: fractional indices (2.1A), expanding more than two brackets, factorising quadratics of the form ax² + bx + c, algebraic fractions and completing the square (2.2A–D). The Foundation Tier covers integer and zero indices and factorising x² + bx + c only. Questions 1(a) and 1(b) are for both tiers.

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Section A

1. Simplify: (a) x⁻³ (b) x⁰ (c) (Higher tier only) x^(1/2). [3]

2. (Higher tier only) Factorise 6x² + x − 12. [2]

Section B

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

4. (Higher tier only) Write 2x² − 10x + 3 in the form a(x + b)² + c. [4]

5. (Higher tier only) Express (2x + 3)/(x + 1) − (x − 4)/(x + 3) as a single fraction in its simplest form. [5]

6. (Higher tier only) Simplify (3x² − 6x)/(x² − 4). [4]

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Answers

1. (a) x⁻³ = 1/x³ [1]. (b) x⁰ = 1 [1]. (c) x^(1/2) = √x [1].

2. 6x² + x − 12 = (3x − 4)(2x + 3) [2] (1 mark for a partially correct factorisation with correct signs, full marks for the fully correct pair of brackets, verifiable by expanding back out).

3.

(x + 1)(x - 2) = x^2 - x - 2                        [1]
(x^2 - x - 2)(x + 4) = x^3 + 4x^2 - x^2 - 4x - 2x - 8
                     = x^3 + 3x^2 - 6x - 8           [1] [1] [1]

4.

2x^2 - 10x + 3
= 2(x^2 - 5x) + 3              [1]
= 2[(x - 2.5)^2 - 6.25] + 3    [1]
= 2(x - 2.5)^2 - 12.5 + 3
= 2(x - 2.5)^2 - 9.5           [1]

So a = 2, b = -2.5, c = -9.5   [1]

5.

Common denominator: (x + 1)(x + 3)                              [1]

(2x + 3)(x + 3) - (x - 4)(x + 1)
= (2x^2 + 9x + 9) - (x^2 - 3x - 4)                               [1] [1]
= x^2 + 12x + 13                                                 [1]

Result: (x^2 + 12x + 13) / [(x + 1)(x + 3)]                      [1]

6.

Numerator:   3x^2 - 6x = 3x(x - 2)                [1]
Denominator: x^2 - 4 = (x - 2)(x + 2)              [1]

Cancel shared (x - 2) factor:                      [1]
Result: 3x / (x + 2)                               [1]

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Exam technique for this topic

Higher-tier questions on this specification frequently combine two or three of 2.1-2.2’s skills within a single longer question — Q5 above, for instance, requires factorising-level fluency with finding a common denominator and then careful expansion of two brackets before subtracting, so a weakness in any one of those underlying skills surfaces immediately in a combined question like this. Practise checking completed-the-square answers by expanding a(x + b)² + c back out and confirming it matches the original expression — this single habit catches the most common error in this sub-topic, forgetting to multiply the adjustment term by the factored-out coefficient. For algebraic fraction questions, always factorise the numerator and denominator fully before attempting to cancel anything, since cancelling before fully factorising is a frequent source of an incorrect or incompletely simplified answer.

Worked example: an index-law calculation combining three rules

Simplify (x³ × x⁻⁵) / x⁰.

Step 1: combine the numerator using the multiplication law (add powers)
        x^3 x x^-5 = x^(3 + (-5)) = x^-2

Step 2: divide by x^0, and recall that x^0 = 1 for any non-zero base
        x^-2 / 1 = x^-2

Step 3: express with a positive index using the negative-power rule
        x^-2 = 1/x^2

Final answer: 1/x^2

This question deliberately combines all three index rules covered in 2.1 within a single calculation — practising short chained questions like this, rather than only single-rule questions in isolation, more accurately reflects how Higher-tier papers actually test index notation.

Worked example: choosing between factorising and completing the square

Given the quadratic x² + 4x − 5, a candidate must decide whether factorising or completing the square is the more efficient method. Because the roots are clean integers (x² + 4x − 5 = (x + 5)(x − 1)), factorising is the faster route here. By contrast, for 2x² − 10x + 3 (as in Q4), the roots are not clean integers, so completing the square is the more reliable method — attempting to factorise this expression by trial and error would likely fail or take considerably longer. Practising this judgement — trying factorising first, and switching to completing the square only when factors don’t appear cleanly — is a transferable exam-efficiency skill worth building deliberately rather than defaulting to one method for every quadratic regardless of its form.

Checking answers by substitution

A quick way to catch errors in expansion or simplification questions like Q3 and Q5 is to substitute a simple value, such as x = 2, into both the original expression and your simplified answer, and confirm the two numerical results match. This will not catch every possible error, but it reliably flags sign mistakes and dropped terms — the two most common sources of lost marks in longer algebraic manipulation questions — and takes only a few seconds once the simplified answer has been reached, making it a habit worth building into every longer algebra question attempted under exam conditions.

Where marks are usually lost

• Applying an index law to terms with different bases, when the rules for combining powers only apply to the same base.
• Sign errors when expanding three linear expressions, particularly with a negative bracket like (x − 1).
• Cancelling additive terms in an algebraic fraction as though they were multiplicative common factors.
• Forgetting to multiply the adjustment term by the factored-out coefficient when completing the square.