June 2024 Paper 1 Q13
13
(a) It is given that\[\mathrm{P}(x) = 4x^3 + 8x^2 + 11x + 4\]Use the factor theorem to show that \((2x + 1)\) is a factor of \(\mathrm{P}(x)\) [2 marks]
(b) Express \(\mathrm{P}(x)\) in the form\[\mathrm{P}(x) = (2x + 1)(ax^2 + bx + c)\]where \(a\), \(b\) and \(c\) are constants to be found. [2 marks]
(c) Given that \(n\) is a positive integer, use your answer to part (b) to explain why \(4n^3 + 8n^2 + 11n + 4\) is never prime. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(x = -\dfrac{1}{2}\) into \(\mathrm{P}(x)\) and obtains zero. Must see \(-\dfrac{1}{2}\) bracketed correctly. If bracket(s) missing must see a further step to indicate correct evaluation eg \(-\dfrac{4}{8} + \dfrac{8}{4} - \dfrac{11}{2} + 4 = 0\) or better. | M1 | 1.1a |
| Completes factor theorem argument by showing \(\mathrm{P}\left(-\dfrac{1}{2}\right) = 0\) and stating \(\therefore (2x + 1)\) is a factor of \(\mathrm{P}(x)\) OE | R1 | 2.1 |
| (2) |
Typical solution
\[\mathrm{P}\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 + 8\left(-\frac{1}{2}\right)^2 + 11\left(-\frac{1}{2}\right) + 4\]\[= 0\]\(\therefore (2x + 1)\) is a factor of \(\mathrm{P}(x)\)
| Scheme | Marks | AO |
|---|---|---|
| Obtains two correct coefficients of \(2x^2 + 3x + 4\) | M1 | 1.1a |
| Obtains \((2x + 1)(2x^2 + 3x + 4)\) | A1 | 1.1b |
| (2) |
Typical solution
\[\mathrm{P}(x) = (2x + 1)(2x^2 + 3x + 4)\]| Scheme | Marks | AO |
|---|---|---|
| Begins argument by explaining that either \((2n + 1) \neq 1\) Or \((an^2 + bn + c) \neq 1\) Or \((2n + 1) \neq\) the cubic expression Or \((an^2 + bn + c) \neq\) the cubic expression Condone \(x\) instead of \(n\) | M1 | 2.1 |
| States that Either both \((2n + 1) \neq 1\) and their \((an^2 + bn + c) \neq 1\) Or both \((2n + 1)\) is not equal to \(4n^3 + 8n^2 + 11n + 4\) and their \((an^2 + bn + c)\) is not equal to \(4n^3 + 8n^2 + 11n + 4\) Or \((2n + 1) \neq 1\) and \((2n + 1)\) is not equal to \(4n^3 + 8n^2 + 11n + 4\) Or their \((an^2 + bn + c) \neq 1\) and their \((an^2 + bn + c)\) is not equal to \(4n^3 + 8n^2 + 11n + 4\) | R1F | 2.2a |
| (2) | ||
| (6 marks) |
Typical solution
\[4n^3 + 8n^2 + 11n + 4 = (2n + 1)(2n^2 + 3n + 4)\]There are two factors. Both factors are integers not equal to 1 so \(4n^3 + 8n^2 + 11n + 4\) is never prime.