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

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 12 marks

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The question, in short

For f(x) = 2ax³ + x² − bx + 3a with integer a, b and factors (x + 2) and (x − 1), find a and b, factorise fully, then solve h(y) = 0 for h(y) = 2^(3y + 2) + 2^(2y) − 11(2^y) + 6.

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

Part (a) · 5 marks

Use the two given factors of f(x) to show that a = 2 and determine the integer b.

  1. 1f(−2) = −16a + 4 + 2b + 3a = 0 → −13a + 4 + 2b = 0
  2. 2f(1) = 2a + 1 − b + 3a = 0 → 5a + 1 − b = 0
  3. 3From the second: b = 5a + 1; substitute: −13a + 4 + 10a + 2 = 0
  4. 4a = 2
  5. 5b = 11

Answera = 2 shown, b = 11

Part (b) · 2 marks

Using the values found, factorise f(x) completely into linear factors.

  1. 1f(x) = 4x³ + x² − 11x + 6
  2. 2Divide by (x + 2)(x − 1) = x² + x − 2: quotient 4x − 3
  3. 3f(x) = (4x − 3)(x + 2)(x − 1)

Answerf(x) = (4x − 3)(x + 2)(x − 1)

Part (c) · 5 marks

Solve h(y) = 0, giving answers to 3 decimal places where appropriate.

  1. 1h(y) = 4(2^y)³ + (2^y)² − 11(2^y) + 6 = f(2^y)
  2. 2(2^y + 2)(2^y − 1)(4(2^y) − 3) = 0
  3. 32^y = 1 → y = 0
  4. 42^y = 3/4 → y = log₂(3/4) = −0.415
  5. 52^y = −2 has no solution

Answery = 0, y = −0.415

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