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0606/11 May/Jun 2026, Question 5: The Factor and Remainder Theorems

Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 8 marks

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The question, in short

Given p(x) = ax³ + 4x² − ax − 28 with positive constant a and p(a) = 0, find a, factorise p(x), and show p(x) = 0 has exactly one root.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 4 marks

Use the condition p(a) = 0 to form an equation in a and solve for the positive value of a.

  1. 1p(a) = a⁴ + 4a² − a² − 28 = 0
  2. 2a⁴ + 3a² − 28 = 0
  3. 3(a² + 7)(a² − 4) = 0
  4. 4a² = 4 (a² = −7 impossible)
  5. 5a = 2 (a is positive)

Answera = 2

Part (b) · 2 marks

With the found value of a, express p(x) as a linear factor times a quadratic factor.

  1. 1p(x) = 2x³ + 4x² − 2x − 28
  2. 2p(2) = 0, so (x − 2) is a factor
  3. 3p(x) = (x − 2)(2x² + 8x + 14)

Answerp(x) = (x − 2)(2x² + 8x + 14)

Part (c) · 2 marks

Using the factorisation, show that p(x) = 0 has exactly one real root.

  1. 1Discriminant of 2x² + 8x + 14 = 8² − 4 × 2 × 14 = −48
  2. 2−48 < 0, so 2x² + 8x + 14 = 0 has no real roots
  3. 3So p(x) = 0 has exactly one root, x = 2

AnswerDiscriminant of 2x² + 8x + 14 is −48 < 0, so the quadratic has no real roots and x = 2 is the only root

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Also searched as: 0606/11/M/J/26 · 0606_s26_qp_11 · 0606 s26 11 · question 5.
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