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WMA12/01 Oct 2023, Question 4: Dividing Polynomials

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 9 marks

The question, in short

Given cubic f(x) = 4x³ + ax² − 29x + b with (2x + 1) a factor, and remainder −25 on division by (x − 2), find a and b and factorise fully, showing all working without calculator.

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

Part (a) · 2 marks

Use the factor (2x + 1) to show that a + 4b = −56.

  1. 1(2x + 1) is a factor, so f(−1/2) = 0
  2. 24(−1/2)³ + a(−1/2)² − 29(−1/2) + b = 0
  3. 3−1/2 + a/4 + 29/2 + b = 0, so a/4 + b + 14 = 0
  4. 4Multiply by 4: a + 4b = −56

AnswerShown: f(−1/2) = −1/2 + a/4 + 29/2 + b = 0 ⇒ a/4 + b + 14 = 0 ⇒ a + 4b = −56

Part (b) · 2 marks

Use the remainder −25 when dividing by (x − 2) to derive a second equation in a and b.

  1. 1Remainder theorem: f(2) = −25
  2. 24(8) + a(4) − 29(2) + b = −25
  3. 332 + 4a − 58 + b = −25, so 4a + b = 1

Answer4a + b = 1

Part (ci) · 2 marks

Solve the two equations simultaneously to find the values of a and b.

  1. 1From (b): b = 1 − 4a
  2. 2Substitute into a + 4b = −56: a + 4 − 16a = −56
  3. 3−15a = −60, so a = 4
  4. 4b = 1 − 16 = −15

Answera = 4, b = −15

Part (cii) · 3 marks

Using your a and b, fully factorise f(x) into linear factors.

  1. 1f(x) = 4x³ + 4x² − 29x − 15
  2. 2Divide by the factor (2x + 1): f(x) = (2x + 1)(2x² + x − 15)
  3. 3Factorise the quadratic: 2x² + x − 15 = (2x − 5)(x + 3)
  4. 4f(x) = (2x + 1)(2x − 5)(x + 3)

Answerf(x) = (2x + 1)(2x − 5)(x + 3)

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Also searched as: WMA12 October 2023 · WMA12/01 · wma12 oct23 · IAS Maths Pure Mathematics 2 October 2023 · question 4.
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