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0606/12 Feb/Mar 2025, Question 4: The Factor and Remainder Theorems

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 6 marks

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The question, in short

Given p(x) = a²x³ + 2ax² + ax + 2 with positive integer a and factor 2x + 1, find a, factorise p, and show p(x) = 0 has one real root.

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

Part (a) · 3 marks

Use the factor 2x + 1 to find the positive integer value of a.

  1. 1p(−1/2) = 0: a²(−1/2)³ + 2a(−1/2)² + a(−1/2) + 2 = 0
  2. 2−a²/8 + a/2 − a/2 + 2 = 0
  3. 3a² = 16
  4. 4a = 4 (a is positive)

Answera = 4

Part (b) · 2 marks

Using your value of a, factorise p(x) fully.

  1. 1p(x) = 16x³ + 8x² + 4x + 2
  2. 2Divide by (2x + 1): quotient 8x² + 2
  3. 3p(x) = (2x + 1)(8x² + 2) = 2(2x + 1)(4x² + 1)

Answer2(2x + 1)(4x² + 1)

Part (c) · 1 mark

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

  1. 1p(x) = 0 means 2x + 1 = 0 or 4x² + 1 = 0
  2. 22x + 1 = 0 gives x = −1/2
  3. 34x² + 1 = 0: discriminant = 0² − 4(4)(1) = −16 < 0, no real roots
  4. 4So x = −1/2 is the only real root

Answer4x² + 1 = 0 has no real roots (discriminant 0 − 16 < 0, or 4x² = −1), so the only real root is x = −1/2

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