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WMA12/01 Jun 2022, Question 7: The Factor and Remainder Theorems

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 7 marks

The question, in short

Given f(x) = Ax³ + 6x² − 4x + B with (x + 2) a factor and the integral from 3 to 5 of f(x) dx equal to 176, find constants A and B.

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Solution · 7 marks

  1. 1(x + 2) is a factor, so f(−2) = 0: −8A + 24 + 8 + B = 0, so 8A − B = 32
  2. 2∫ f(x) dx = (A/4)x⁴ + 2x³ − 2x² + Bx
  3. 3Apply limits 5 and 3: (A/4)(625 − 81) + 2(125 − 27) − 2(25 − 9) + B(5 − 3) = 176
  4. 4136A + 196 − 32 + 2B = 176, so 136A + 2B = 12
  5. 5Substitute B = 8A − 32: 136A + 16A − 64 = 12, so 152A = 76
  6. 6A = 1/2 and B = 8(1/2) − 32 = −28

AnswerA = 1/2, B = −28

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