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0606/12 May/Jun 2025, Question 5: The Factor and Remainder Theorems

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 10 marks

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The question, in short

Given cubic p(x) = 3x³ − 7x² + ax + b with p′(−1) = 21 and x − 2 a factor, determine a and b, factorise fully, then solve a related equation in e^(2y).

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

Part (a) · 4 marks

Use the derivative condition at x = −1 and the factor condition at x = 2 to form equations and work out the integers a and b.

  1. 1p′(x) = 9x² − 14x + a
  2. 2p′(−1) = 9 + 14 + a = 21, so a = −2
  3. 3p(2) = 24 − 28 + 2a + b = 0
  4. 4−4 − 4 + b = 0, so b = 8

Answera = −2, b = 8

Part (b) · 3 marks

With your a and b, express the cubic as a product of linear factors having integer coefficients.

  1. 1p(x) = 3x³ − 7x² − 2x + 8
  2. 2Divide by (x − 2): p(x) = (x − 2)(3x² − x − 4)
  3. 33x² − x − 4 = (x + 1)(3x − 4)
  4. 4p(x) = (x − 2)(x + 1)(3x − 4)

Answerp(x) = (x − 2)(x + 1)(3x − 4)

Part (c) · 3 marks

Substitute your a and b into 3e^(6y) − 7e^(4y) + ae^(2y) + b = 0 and determine the values of y.

  1. 1Let u = e^(2y): 3u³ − 7u² − 2u + 8 = 0, i.e. p(u) = 0
  2. 2(u − 2)(u + 1)(3u − 4) = 0
  3. 3e^(2y) = 2 or e^(2y) = 4/3 (e^(2y) = −1 has no solution)
  4. 4y = (1/2)ln 2 or y = (1/2)ln(4/3)

Answery = (1/2)ln(4/3), y = (1/2)ln 2

Stuck on a step? Read the free lesson: The Factor and Remainder Theorems

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Also searched as: 0606/12/M/J/25 · 0606_s25_qp_12 · 0606 s25 12 · question 5.
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