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4PM1/02R Jun 2022, Question 4: The Factor and Remainder Theorems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 8 marks

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The question, in short

Given cubic f(x) = x³ + px² + qx + 7 with (x + 1) a factor and remainder −5 on division by (x + 2), find p and q, then show the equation has just one real root.

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Part (a) · 5 marks

Use the factor and remainder conditions to form equations in p and q, then solve for both integers.

  1. 1f(−1) = 0: −1 + p − q + 7 = 0, so p − q + 6 = 0
  2. 2f(−2) = −5: −8 + 4p − 2q + 7 = −5, so 4p − 2q = −4
  3. 3From the first, p = q − 6: 4(q − 6) − 2q = −4
  4. 42q = 20, so q = 10 and p = 4

Answerp = 4, q = 10

Part (b) · 3 marks

Using your p and q, demonstrate that f(x) = 0 has exactly one real root.

  1. 1f(x) = x³ + 4x² + 10x + 7
  2. 2Divide by (x + 1): f(x) = (x + 1)(x² + 3x + 7)
  3. 3Discriminant of x² + 3x + 7: 3² − 4 × 1 × 7 = −19
  4. 4−19 < 0, so the quadratic has no real roots
  5. 5So the only real root of f(x) = 0 is x = −1

Answerf(x) = (x + 1)(x² + 3x + 7) and x² + 3x + 7 has discriminant −19 < 0, so f(x) = 0 has only one real root (x = −1) — shown

Stuck on a step? Read the free lesson: The Factor and Remainder Theorems

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