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 available free from your board.

Related: Algebraic Manipulation revision notes

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Section A

1. Expand and simplify fully: (a) 3(2x − 5) + 4(x + 2) (b) (x + 6)(x − 3) [4]

2. Factorise fully: (a) 12x² − 18x (b) x² − 49 (c) x² + 9x + 20 [5]

Section B

3. Expand and simplify (2x − 3)². [3]

4. Factorise 6x² − x − 12. [4]

5. Simplify fully, factorising first where it helps:

(a) (x² − 4) ÷ (x² + 5x + 6) [4] (b) 3/(x + 1) + 2/(x − 2) [4]

6. Make x the subject, showing every rearrangement step:

(a) y = (3x + 2)/(x − 1) [4] (b) T = 2π√(x/g) [3]

7. Show that (n + 1)² − n² is always odd for any positive integer n. [3]

8. Write x² + 8x + 3 in completed square form, and use it to state the minimum value of the expression. [4]

9. Simplify: (a) 5a/6 × 3a/10 (b) 5a/6 ÷ 3a/10 [4]

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Answers

1. (a) 6x − 15 + 4x + 8 [1] = 10x − 7 [1]. (b) x² − 3x + 6x − 18 [1] = x² + 3x − 18 [1].

2. (a) 6x(2x − 3) [1] [1]. (b) (x + 7)(x − 7) [1]. (c) (x + 4)(x + 5) [1] [1].

3. (2x − 3)(2x − 3) = 4x² − 6x − 6x + 9 [1] [1] = 4x² − 12x + 9 [1].

4. Find two numbers multiplying to 6 × (−12) = −72 and adding to −1: 8 and −9 [1]. 6x² + 8x − 9x − 12 [1]; 2x(3x + 4) − 3(3x + 4) [1]; = (2x − 3)(3x + 4) [1].

5. (a) Numerator = (x + 2)(x − 2) [1]; denominator = (x + 2)(x + 3) [1]; cancelling (x + 2) [1] gives (x − 2)/(x + 3), valid for x ≠ −2 and x ≠ −3 [1]. (b) Common denominator (x + 1)(x − 2) [1]; numerator = 3(x − 2) + 2(x + 1) [1] = 3x − 6 + 2x + 2 = 5x − 4 [1]; = (5x − 4)/((x + 1)(x − 2)) [1].

6. (a) y(x − 1) = 3x + 2 [1]; xy − y = 3x + 2; xy − 3x = y + 2 [1]; x(y − 3) = y + 2 [1]; x = (y + 2)/(y − 3) [1]. (b) T/(2π) = √(x/g) [1]; T²/(4π²) = x/g [1]; x = gT²/(4π²) [1].

7. (n + 1)² − n² = n² + 2n + 1 − n² [1] = 2n + 1 [1]. Since 2n is always even, 2n + 1 is always odd for any integer n [1].

8. Half of 8 is 4, so start from (x + 4)² = x² + 8x + 16 [1]; that has 16 where the original only has 3, so subtract 13 [1]: x² + 8x + 3 = (x + 4)² − 13 [1]. Since (x + 4)² can never be negative, the expression is smallest when (x + 4)² = 0, giving a minimum value of −13 [1].

9. (a) 5a/6 × 3a/10 = 15a²/60 = a²/4 [1] [1]. (b) 5a/6 ÷ 3a/10 = 5a/6 × 10/3a = 50a/18a = 25/9 [1] [1].

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Where marks are usually lost

• Writing (2x − 3)² as 4x² + 9.
• Cancelling terms rather than factors in algebraic fractions.
• Failing to collect all the x terms on one side when rearranging.
• Giving examples instead of a general algebraic argument in a proof.
• Forgetting to correct the constant term when completing the square — halving the coefficient of x and squaring the bracket is only the first step; the difference between the bracket’s expansion and the original expression must be added or subtracted afterwards.
• Multiplying denominators together and simplifying too early in an algebraic fraction multiplication, before checking whether factors could have cancelled first — factorising before multiplying often avoids handling much larger numbers.
• Not stating the minimum (or maximum) value explicitly after completing the square, when a question asks for it — the completed-square form itself is not a full answer to that part of the question.
• Cancelling a term rather than a factor when simplifying an algebraic fraction — cancelling is only ever valid for a factor shared by the whole numerator and the whole denominator, never for a term merely added or subtracted somewhere inside an unfactorised expression.

Where completing the square fits in

Completing the square is the bridge between simplifying an expression and solving or sketching it: once x² + bx + c is written as (x + p)² + q, the minimum value of the expression is read off directly as q (occurring when x = −p), and the same rearranged form gives an alternative route to solving the quadratic equation x² + bx + c = 0 without factorising or using the formula.

For condensed recall notes on this topic, see the Algebraic Manipulation revision notes; for the full explanation with additional worked examples, see the Algebraic Manipulation study guide.