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0606/12 May/Jun 2026, Question 3: Dividing Polynomials

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 7 marks

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The question, in short

For p(x) = 6x³ − x² − 5x + 2, work out the remainder on dividing by x + 2, factorise p fully into linear factors, and use this to solve 6u⁶ − u⁴ − 5u² + 2 = 0.

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

Part (a) · 1 mark

Work out the remainder left when p(x) = 6x³ − x² − 5x + 2 is divided by x + 2.

  1. 1Remainder = p(−2) = 6(−8) − 4 + 10 + 2
  2. 2= −40

Answer−40

Part (b) · 4 marks

Express p(x) = 6x³ − x² − 5x + 2 as a product of linear factors.

  1. 1p(−1) = −6 − 1 + 5 + 2 = 0, so (x + 1) is a factor
  2. 2p(x) = (x + 1)(6x² − 7x + 2)
  3. 36x² − 7x + 2 = (3x − 2)(2x − 1)
  4. 4p(x) = (x + 1)(3x − 2)(2x − 1)

Answerp(x) = (x + 1)(3x − 2)(2x − 1)

Part (c) · 2 marks

Using your factorisation, solve 6u⁶ − u⁴ − 5u² + 2 = 0 for u.

  1. 16u⁶ − u⁴ − 5u² + 2 = p(u²) = (u² + 1)(3u² − 2)(2u² − 1) = 0
  2. 2u² + 1 = 0 has no real solutions
  3. 3u² = 2/3, so u = ±√(2/3)
  4. 4u² = 1/2, so u = ±√(1/2)

Answeru = ±√(2/3), u = ±√(1/2)

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Also searched as: 0606/12/M/J/26 · 0606_s26_qp_12 · 0606 s26 12 · question 3.
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