Lemmara logoLemmaraWorked solution

0606/12 Oct/Nov 2025, Question 2: The Factor and Remainder Theorems

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 7 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Given cubic p(x) = 2x³ + ax² + 13x + b with x + 2 a factor and remainder 6 on division by x + 1, find a and b, then show p(x) = 0 has one real root.

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

Part (a) · 4 marks

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

  1. 1p(−2) = 0: −16 + 4a − 26 + b = 0 ⇒ 4a + b = 42
  2. 2p(−1) = 6: −2 + a − 13 + b = 6 ⇒ a + b = 21
  3. 3Subtract: 3a = 21 ⇒ a = 7
  4. 4b = 14

Answera = 7, b = 14

Part (b) · 3 marks

Show that the cubic equation p(x) = 0 has exactly one real root.

  1. 1p(x) = 2x³ + 7x² + 13x + 14
  2. 2Divide by x + 2: p(x) = (x + 2)(2x² + 3x + 7)
  3. 3Discriminant of 2x² + 3x + 7: 3² − 4 × 2 × 7 = −47 < 0
  4. 4So the quadratic has no real roots and x = −2 is the only real root

Answerp(x) = (x + 2)(2x² + 3x + 7) and 2x² + 3x + 7 has discriminant 9 − 56 = −47 < 0, so x = −2 is the only real root — shown

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

Practise more the factor and remainder theorems questions

Real 0606 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on the factor and remainder theorems

Also searched as: 0606/12/O/N/25 · 0606_w25_qp_12 · 0606 w25 12 · question 2.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.