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4PM1/02R Jun 2024, Question 4: The Factor and Remainder Theorems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 11 marks

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The question, in short

A cubic f(x) = px³ + qx² − 37x − 12q has f′(x) leaving remainder −33 on division by (x + 2), and (x + 5) is a factor of f(x). Use these to find p and q, factorise fully, and solve f(x) = 0.

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

Part (a) · 6 marks

(i) Use the remainder condition on f′(x) to prove p = 2. (ii) Use the factor condition to work out q.

  1. 1f′(x) = 3px² + 2qx − 37
  2. 2f′(−2) = 12p − 4q − 37 = −33, so 12p − 4q = 4
  3. 3f(−5) = −125p + 25q + 185 − 12q = 0, so −125p + 13q = −185
  4. 4Solve: from the first, q = 3p − 1
  5. 5−125p + 39p − 13 = −185, so p = 2
  6. 6q = 5

Answer(i) p = 2 shown (ii) q = 5

Part (b) · 3 marks

With your p and q, factorise the cubic completely using algebra.

  1. 1f(x) = 2x³ + 5x² − 37x − 60
  2. 2Divide by (x + 5): quotient 2x² − 5x − 12
  3. 32x² − 5x − 12 = (2x + 3)(x − 4)
  4. 4f(x) = (x + 5)(2x + 3)(x − 4)

Answerf(x) = (x + 5)(2x + 3)(x − 4)

Part (c) · 2 marks

Use your factorised form to state all solutions of f(x) = 0.

  1. 1(x + 5)(2x + 3)(x − 4) = 0
  2. 2x = −5, x = −3/2, x = 4

Answerx = −5, x = 4, x = −3/2

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