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4PM1/01R Jun 2023, Question 5: The Factor and Remainder Theorems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 16 marks

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The question, in short

A cubic f(x) = 2x³ + ax² − 14x + b has remainder 39 when divided by (x − 4) and (x − 1) as a factor; determine a and b, factorise fully, read intercepts p, q, r, and find the shaded area.

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

Part (a) · 5 marks

Using the remainder 39 on division by (x − 4) and the factor (x − 1), show a = −3 and work out b.

  1. 1f(1) = 0: 2 + a − 14 + b = 0, so a + b = 12
  2. 2f(4) = 39: 128 + 16a − 56 + b = 39, so 16a + b = −33
  3. 3Subtract: 15a = −45, a = −3
  4. 4b = 12 − (−3) = 15

Answera = −3 shown, b = 15

Part (b) · 4 marks

With a = −3 and b found, express the cubic as a product of three linear factors.

  1. 1f(x) = 2x³ − 3x² − 14x + 15
  2. 2Divide by (x − 1): quotient 2x² − x − 15
  3. 32x² − x − 15 = (2x + 5)(x − 3)
  4. 4f(x) = (x − 1)(2x + 5)(x − 3)

Answerf(x) = (x − 1)(2x + 5)(x − 3)

Part (c) · 3 marks

From the graph, state the three x-axis crossing values p, q and r.

  1. 1Roots of f(x) = 0: x = −5/2, 1, 3
  2. 2p < q < r, so p = −5/2, q = 1, r = 3

Answerp = −5/2, q = 1, r = 3

Part (d) · 4 marks

Integrate between the roots to calculate the exact area of the shaded region.

  1. 1Area = ∫ from −5/2 to 1 of f(x) dx + |∫ from 1 to 3 of f(x) dx|
  2. 2∫ f(x) dx = x⁴/2 − x³ − 7x² + 15x
  3. 3From −5/2 to 1: 15/2 − (−1475/32) = 1715/32
  4. 4From 1 to 3: (−9/2) − 15/2 = −12, so that area is 12
  5. 5Total = 1715/32 + 12 = 2099/32

Answer2099/32

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