4PM1/02 Jun 2021, Question 6: Simultaneous Equations with a Curve
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 13 marks
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The question, in short
Cubic f(x) = x³ + (p + 1)x² − 10x + q with integer p, q; use two given factors to find p and q, then factorise f(x) fully.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 3 marks
Given (x − 3) is a factor, show that 9p + q + 6 = 0.
- 1f(3) = 0
- 227 + 9(p + 1) − 30 + q = 0
- 327 + 9p + 9 − 30 + q = 0 ⇒ 9p + q + 6 = 0
Answer9p + q + 6 = 0 shown
Part (b) · 3 marks
Given (x + p), p > 0, is also a factor, show that p² + 10p + q = 0.
- 1f(−p) = 0
- 2(−p)³ + (p + 1)p² + 10p + q = 0
- 3−p³ + p³ + p² + 10p + q = 0 ⇒ p² + 10p + q = 0
Answerp² + 10p + q = 0 shown
Part (c) · 5 marks
Hence find the values of p and q.
- 1q = −9p − 6
- 2p² + 10p − 9p − 6 = 0 ⇒ p² + p − 6 = 0
- 3(p + 3)(p − 2) = 0
- 4p > 0 so p = 2
- 5q = −18 − 6 = −24
Answerp = 2, q = −24
Part (d) · 2 marks
Using your p and q, factorise f(x) completely.
- 1f(x) = x³ + 3x² − 10x − 24
- 2(x − 3) and (x + 2) are factors, so f(x) = (x + a)(x − 3)(x + 2)
- 3Constant term: −6a = −24 ⇒ a = 4
- 4f(x) = (x + 4)(x − 3)(x + 2)
Answerf(x) = (x + 4)(x − 3)(x + 2)
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