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WMA12/01 Oct 2019, Question 4: Dividing Polynomials

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 8 marks

The question, in short

For f(x) = (x − 3)(3x² + x + a) − 35 with constant a, find a remainder, determine a given a linear factor, then fully factorise the cubic.

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

Part (a) · 1 mark

Give the remainder on dividing f(x) by (x − 3).

  1. 1By the remainder theorem the remainder is f(3)
  2. 2f(3) = (0)(27 + 3 + a) − 35 = −35

Answer−35

Part (b) · 2 marks

Given that (3x − 2) divides f(x), prove that a = −17.

  1. 1(3x − 2) is a factor, so f(2/3) = 0
  2. 2(2/3 − 3)(3(2/3)² + 2/3 + a) − 35 = 0
  3. 3(−7/3)(4/3 + 2/3 + a) = 35
  4. 4(−7/3)(2 + a) = 35, so 2 + a = −15
  5. 5a = −17

AnswerShown: f(2/3) = (−7/3)(2 + a) − 35 = 0, so 2 + a = −15 and a = −17

Part (c) · 5 marks

Fully factorise f(x), showing every algebraic step of your method.

  1. 1f(x) = (x − 3)(3x² + x − 17) − 35
  2. 2= 3x³ + x² − 17x − 9x² − 3x + 51 − 35 = 3x³ − 8x² − 20x + 16
  3. 3Divide by (3x − 2): 3x³ − 8x² − 20x + 16 = (3x − 2)(x² − 2x − 8)
  4. 4x² − 2x − 8 = (x + 2)(x − 4)
  5. 5f(x) = (3x − 2)(x + 2)(x − 4)

Answerf(x) = (3x − 2)(x + 2)(x − 4)

Stuck on a step? Read the free lesson: Dividing Polynomials

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Also searched as: WMA12 October 2019 · WMA12/01 · wma12 oct19 · IAS Maths Pure Mathematics 2 October 2019 · question 4.
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