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4PM1/01 Jan 2022, Question 5: The Factor and Remainder Theorems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 9 marks

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The question, in short

For f(x) = ax³ + 5bx² + 8ax − 4b with integers a and b, use the factor (x + 2) and remainder 21 on division by (x + 3) to find a and b, then solve f(x) = 0.

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Part (a) · 5 marks

Show that a = 2 and determine the integer b using the factor and remainder conditions.

  1. 1f(−2) = 0: −8a + 20b − 16a − 4b = 0, so 16b = 24a, i.e. 2b = 3a
  2. 2f(−3) = 21: −27a + 45b − 24a − 4b = 21, so 41b = 51a + 21
  3. 3Substitute b = 3a/2: 123a/2 = 51a + 21
  4. 421a/2 = 21, so a = 2
  5. 5b = 3

Answera = 2 shown; b = 3

Part (b) · 4 marks

Solve f(x) = 0 algebraically using the values of a and b found.

  1. 1f(x) = 2x³ + 15x² + 16x − 12
  2. 2Divide by (x + 2): 2x² + 11x − 6
  3. 32x² + 11x − 6 = (2x − 1)(x + 6)
  4. 4x = −6, −2, 1/2

Answerx = −6, x = −2, x = 1/2

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