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0606/13 May/Jun 2026, Question 4: Dividing Polynomials

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 6 marks

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The question, in short

For p(x) = 4x³ − 5x² − 7x + 2, prove x − 2 divides it, factorise fully, then use that to solve 4u⁶ − 5u⁴ − 7u² + 2 = 0.

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

Part (a) · 1 mark

Verify that substituting x = 2 into p(x) gives zero, confirming x − 2 is a factor.

  1. 1p(2) = 4(8) − 5(4) − 7(2) + 2
  2. 2= 32 − 20 − 14 + 2 = 0
  3. 3So x − 2 is a factor

Answerp(2) = 32 − 20 − 14 + 2 = 0, so x − 2 is a factor shown

Part (b) · 3 marks

Express p(x) as a product of three linear factors.

  1. 1Divide by (x − 2): p(x) = (x − 2)(4x² + 3x − 1)
  2. 24x² + 3x − 1 = (4x − 1)(x + 1)
  3. 3p(x) = (x − 2)(4x − 1)(x + 1)

Answer(x − 2)(4x − 1)(x + 1)

Part (c) · 2 marks

Using your factorisation, find all values of u satisfying 4u⁶ − 5u⁴ − 7u² + 2 = 0.

  1. 1The equation is p(u²) = 0
  2. 2(u² − 2)(4u² − 1)(u² + 1) = 0
  3. 3u² = 2 or u² = 1/4 (u² = −1 has no real solution)
  4. 4u = ±√2, ±1/2

Answeru = ±√2, ±1/2

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Also searched as: 0606/13/M/J/26 · 0606_s26_qp_13 · 0606 s26 13 · question 4.
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