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4PM1/02R Jun 2025, Question 5: 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

For the cubic x³ + x² + x + c, use the two given remainder conditions to determine c, then use the factor (x + 2) to establish how many real solutions f(x) = 0 has.

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

Part (a) · 4 marks

Apply the remainder theorem at x = 2 and x = −1, set the first remainder equal to 4 times the second, and solve for c to obtain 6.

  1. 1f(2) = 8 + 4 + 2 + c = 14 + c
  2. 2f(−1) = −1 + 1 − 1 + c = c − 1
  3. 314 + c = 4(c − 1)
  4. 43c = 18, so c = 6

Answerc = 6 shown

Part (b) · 4 marks

With c = 6 and (x + 2) known to be a factor, factorise the cubic and examine the remaining quadratic to prove exactly one real root exists.

  1. 1f(x) = x³ + x² + x + 6 = (x + 2)(x² − x + 3)
  2. 2For x² − x + 3: b² − 4ac = 1 − 12 = −11
  3. 3−11 < 0 so the quadratic has no real roots
  4. 4So f(x) = 0 has only one real root, x = −2

Answerf(x) = (x + 2)(x² − x + 3); discriminant of x² − x + 3 is −11 < 0, so x = −2 is the only real root

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