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WMA12/01 Jun 2019, Question 6: Dividing Polynomials

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 8 marks

The question, in short

Given cubic f(x) = kx³ − 15x² − 32x − 12 with (x − 3) a factor, find k, fully factorise f(x), then solve a related equation in cos θ for 0 ≤ θ < 360°.

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

Part (a) · 2 marks

Use the factor (x − 3) to show that k = 9.

  1. 1(x − 3) is a factor, so f(3) = 0
  2. 227k − 15(9) − 32(3) − 12 = 0
  3. 327k − 243 = 0
  4. 427k = 243, so k = 9

AnswerShown: f(3) = 27k − 135 − 96 − 12 = 0, so 27k = 243 and k = 9

Part (b) · 4 marks

Fully factorise f(x) using algebra, showing each step of the working.

  1. 1f(x) = 9x³ − 15x² − 32x − 12
  2. 2Divide by (x − 3): f(x) = (x − 3)(9x² + 12x + 4)
  3. 39x² + 12x + 4 = (3x + 2)²
  4. 4f(x) = (x − 3)(3x + 2)²

Answerf(x) = (x − 3)(3x + 2)²

Part (c) · 2 marks

Solve 9cos³θ − 15cos²θ − 32cosθ − 12 = 0 for 0 ≤ θ < 360°, answers to one decimal place.

  1. 1(cosθ − 3)(3cosθ + 2)² = 0
  2. 2cosθ = 3 has no solutions, so cosθ = −2/3
  3. 3θ = cos⁻¹(−2/3) = 131.8°
  4. 4Second solution: 360° − 131.8° = 228.2°

Answerθ = 131.8°, 228.2°

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