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4PM1/01 Nov 2021, Question 11: The Factor and Remainder Theorems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 12 marks

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The question, in short

Given f′(x) = ax² − 14x − 10 with a an integer, and conditions on factors and remainder of f(x), show a = 6 then solve f(x) = 0 algebraically.

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Part (a) · 6 marks

Use the factor (x − 4) and the remainder 25 on division by (x + 1) to show that a = 6.

  1. 1f(x) = ax³/3 − 7x² − 10x + c
  2. 2f(4) = 0: 64a/3 − 152 + c = 0
  3. 3f(−1) = 25: −a/3 − 7 + 10 + c = 25, so −a/3 − 22 + c = 0
  4. 4Subtract: 65a/3 = 130, so 65a = 390
  5. 5a = 6

Answera = 6 shown

Part (b) · 6 marks

With a = 6, solve f(x) = 0 using algebra.

  1. 1c = 22 + 6/3 = 24
  2. 2f(x) = 2x³ − 7x² − 10x + 24
  3. 3Divide by (x − 4): f(x) = (x − 4)(2x² + x − 6)
  4. 4f(x) = (x − 4)(2x − 3)(x + 2)
  5. 5x = 4, 3/2, −2

Answerx = 4, x = 3/2, x = −2

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