Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Algebra – 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 section 2.1 Algebra of topic 2, Pure Mathematics 2, in the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). This content is examined on Paper 2, which is offered only as part of AS Level on the Pure Mathematics only route. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Give exact answers where the question asks for them, and non-exact answers to 3 significant figures.

Related: the study guide, the revision notes, the A Level Mathematics hub, the printable 9709 checklist, the free AS Level diagnostic and the 9709 self-check bank. For mixed Paper 2 questions across all six sections, use the Pure Mathematics 2 practice set.

Questions

1. Sketch the graph of y = |3x + 6|, stating the coordinates of the points where the graph meets the axes. [2]

2. Solve the equation |x + 5| = |3x − 1|. [3]

3. Solve the inequality |2x − 7| < 3. [2]

4. Find the remainder when 4x³ − 6x² + 5x − 7 is divided by (2x − 1). [2]

5. Find the quotient and the remainder when x⁴ + 2x³ − 3x + 1 is divided by x² + x − 1. [4]

6. Solve the inequality |3x + 2| ≥ x + 6. [4]

7. The polynomial p(x) is defined by p(x) = x³ + ax² + bx + 12, where a and b are constants. It is given that (x + 3) is a factor of p(x), and that the remainder is 4 when p(x) is divided by (x − 1).

(a) Find the values of a and b. [4] (b) Hence factorise p(x) completely and state the roots of p(x) = 0. [2]

8. The polynomial q(x) is defined by q(x) = 3x³ − 11x² − 6x + 8.

(a) Use the factor theorem to show that (3x − 2) is a factor of q(x). [2] (b) Factorise q(x) completely. [2] (c) Hence solve the equation 3y⁶ − 11y⁴ − 6y² + 8 = 0, giving your answers in exact form. [3]

9. The polynomial p(x) is defined by p(x) = x⁴ + ax³ + bx² − 7x + 6, where a and b are constants. When p(x) is divided by x² − 3x + 2, the remainder is 3x − 2.

(a) Explain why p(1) = 1 and p(2) = 4. [2] (b) Find the values of a and b. [3] (c) Find the quotient when p(x) is divided by x² − 3x + 2. [2]

10. (a) Sketch, on the same diagram, the graphs of y = |3x − 4| and y = 6 − x, showing where each graph meets the axes. [2] (b) Solve the equation |3x − 4| = 6 − x. [3] (c) Hence solve the inequality |3x − 4| < 6 − x. [1] (d) Hence solve the inequality |3√t − 4| < 6 − √t. [2]

11. The polynomial p(x) is defined by p(x) = 6x³ + ax² + bx − 4, where a and b are constants. It is given that (2x − 1) is a factor of p(x), and that the remainder is −15 when p(x) is divided by (x + 1).

(a) Find the values of a and b. [4] (b) Show that the equation p(x) = 0 has exactly one real root. [3] (c) Find the quotient and the remainder when p(x) is divided by x² + 1. [3]

Answers

1. V-shaped graph with vertex on the x-axis, both arms straight [1]. Meets the axes at (−2, 0) and (0, 6) [1] Examiner insight: both intercepts must be stated or labelled; a correct V with no coordinates earns only the shape mark.

2. (x + 5)² = (3x − 1)², so x² + 10x + 25 = 9x² − 6x + 1 [1]. 8x² − 16x − 24 = 0, so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0 [1]. x = 3 or x = −1 [1] Examiner insight: squaring is safe here because both sides are moduli; solving only the case x + 5 = 3x − 1 gives x = 3 alone and loses the final mark.

3. −3 < 2x − 7 < 3, so 4 < 2x < 10 [1]. 2 < x < 5 [1] Examiner insight: the answer is a single interval; “x > 2, x < 5” is usually accepted, but “x > 2 or x < 5” describes every number and is marked wrong.

4. Remainder = p(1/2) = 4(1/8) − 6(1/4) + 5(1/2) − 7 [1] = 1/2 − 3/2 + 5/2 − 7 = −11/2 [1] Examiner insight: the method mark needs substitution of x = 1/2; using x = 2 or x = −1/2 scores nothing, even with correct arithmetic after it.

5. Write the dividend as x⁴ + 2x³ + 0x² − 3x + 1 [1]. First term x⁴ ÷ x² = x²; x⁴ + 2x³ + 0x² − 3x + 1 − x²(x² + x − 1) = x³ + x² − 3x + 1 [1]. Next term x³ ÷ x² = x; x³ + x² − 3x + 1 − x(x² + x − 1) = −2x + 1 [1]. Quotient x² + x, remainder 1 − 2x [1] Examiner insight: the remainder after dividing by a quadratic may be linear; stopping with a constant, or dividing further into −2x, loses the accuracy marks.

6. Case 3x + 2 = x + 6 gives x = 2 [1]. Case −(3x + 2) = x + 6 gives −4x = 8, so x = −2 [1]. Both values make x + 6 positive, so both are valid meeting points [1]. From the sketch the V is on or above the line outside these values: x ≤ −2 or x ≥ 2 [1] Examiner insight: the final mark needs the non-strict signs (≤, ≥) copied from the question; strict signs lose accuracy even when the critical values are right.

7. (a) p(−3) = 0: −27 + 9a − 3b + 12 = 0, so 3a − b = 5 [1]. p(1) = 4: 1 + a + b + 12 = 4, so a + b = −9 [1]. Adding: 4a = −4 [1]. a = −1, b = −8 [1] (b) p(x) = x³ − x² − 8x + 12 = (x + 3)(x² − 4x + 4) [1] = (x + 3)(x − 2)², roots x = −3 and x = 2 [1] Examiner insight: in (a), the first two marks are method marks for setting p(−3) = 0 and p(1) = 4, so they can be earned even after a later slip; accuracy needs both values correct.

8. (a) q(2/3) = 3(8/27) − 11(4/9) − 6(2/3) + 8 = 8/9 − 44/9 − 4 + 8 [1] = 0, so (3x − 2) is a factor [1] (b) q(x) = (3x − 2)(x² − 3x − 4) [1] = (3x − 2)(x + 1)(x − 4) [1] (c) Let x = y², so (3y² − 2)(y² + 1)(y² − 4) = 0 [1]. y² = 2/3 or y² = 4; y² = −1 has no real solutions [1]. y = ±√6/3 or y = ±2 [1] Examiner insight: (a) is a “show that”: the substituted terms must be shown and the conclusion stated in words; “q(2/3) = 0” alone scores only the first mark.

9. (a) p(x) ≡ (x² − 3x + 2)Q(x) + 3x − 2, and x² − 3x + 2 = (x − 1)(x − 2), which is zero at x = 1 and x = 2 [1]. So p(1) = 3(1) − 2 = 1 and p(2) = 3(2) − 2 = 4 [1] (b) p(1) = 1 + a + b − 7 + 6 = 1, so a + b = 1 [1]. p(2) = 16 + 8a + 4b − 14 + 6 = 4, so 2a + b = −1 [1]. a = −2, b = 3 [1] (c) p(x) = x⁴ − 2x³ + 3x² − 7x + 6. Dividing: x² gives remainder x³ + x² − 7x + 6, then x gives 4x² − 9x + 6, then 4 gives 3x − 2 [1]. Quotient x² + x + 4 [1] Examiner insight: in (c) the remainder 3x − 2 appearing at the end is a check on a and b; if it does not appear, go back to (b) before writing the quotient.

10. (a) V-shape with vertex (4/3, 0) and y-intercept (0, 4); straight line through (0, 6) and (6, 0) [1]. Graphs drawn crossing twice, once on each arm of the V [1] (b) 3x − 4 = 6 − x gives x = 5/2 [1]. −(3x − 4) = 6 − x gives −2x = 2, so x = −1 [1]. Both make 6 − x positive: x = −1 or x = 5/2 [1] (c) −1 < x < 5/2 [1] (d) Replace x by √t: −1 < √t < 5/2, and √t ≥ 0, so 0 ≤ √t < 5/2 [1]. 0 ≤ t < 25/4 [1] Examiner insight: “hence” in (d) expects use of (c); the lower bound comes from √t ≥ 0; squaring −1 < √t to get t > 1 is wrong and loses the final mark.

11. (a) p(1/2) = 0: 6/8 + a/4 + b/2 − 4 = 0, so a + 2b = 13 [1]. p(−1) = −15: −6 + a − b − 4 = −15, so a − b = −5 [1]. Subtracting: 3b = 18 [1]. a = 1, b = 6 [1] (b) p(x) = 6x³ + x² + 6x − 4 = (2x − 1)(3x² + 2x + 4) [1]. Discriminant of 3x² + 2x + 4 is 2² − 4(3)(4) = −44 [1]. −44 < 0, so the quadratic has no real roots and the only real root is x = 1/2 [1] (c) 6x³ ÷ x² = 6x; 6x³ + x² + 6x − 4 − 6x(x² + 1) = x² − 4 [1]. x² ÷ x² = 1; x² − 4 − (x² + 1) = −5 [1]. Quotient 6x + 1, remainder −5 [1] Examiner insight: in (b), the discriminant value alone is not enough: the final mark needs “< 0, so no real roots” and a statement that x = 1/2 is the one real root.

Where marks are usually lost

• Squaring a modulus equation with an ordinary expression on one side, then not checking both roots.
• Reversing the inequality logic: “> b” needs two outer pieces, “< b” needs one interval.
• Dropping the “or equal to” part of ≤ and ≥ in the final answer.
• Using the wrong value in the remainder theorem for a divisor like (2x − 1) or (3x − 2).
• Leaving out a zero coefficient for a missing power in long division.
• Stopping the division before the remainder has lower degree than the divisor.
• A “show that” factor proof with no substituted working or no concluding sentence.
• In a “hence” part, starting again from scratch instead of using the previous answer, which can lose method credit.
• Forgetting the restriction √t ≥ 0 or y² ≥ 0 after a substitution.

Next steps

• Pure Mathematics 2 Algebra revision notes
• Pure Mathematics 2 Algebra study guide
• Pure Mathematics 2 mixed practice and Pure 1 mixed practice
• A Level Mathematics hub
• Printable 9709 checklist
• AS Level diagnostic, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.1 Algebra.