CAIE FP1 2015 November — Question 5

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeFinding polynomial from root properties
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring systematic application of Vieta's formulas and manipulation of symmetric functions of roots. While the techniques are standard for FP1 (sum of roots, sum of products, converting α²+β²+γ² to elementary symmetric functions), it requires careful algebraic manipulation across multiple steps and the additional constraint creates a system requiring problem-solving beyond routine exercises. The further maths context and multi-step nature place it moderately above average difficulty.
Spec4.05a Roots and coefficients: symmetric functions

5 The cubic equation \(x ^ { 3 } + p x ^ { 2 } + q x + r = 0\), where \(p , q\) and \(r\) are integers, has roots \(\alpha , \beta\) and \(\gamma\), such that $$\begin{aligned} \alpha + \beta + \gamma & = 15 , \\ \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } & = 83 . \end{aligned}$$ Write down the value of \(p\) and find the value of \(q\). Given that \(\alpha , \beta\) and \(\gamma\) are all real and that \(\alpha \beta + \alpha \gamma = 36\), find \(\alpha\) and hence find the value of \(r\).

Question 5:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\alpha + \beta + \gamma = -p = 15 \Rightarrow p = -15\)B1
\(2(\alpha\beta + \beta\gamma + \gamma\alpha) = (\alpha+\beta+\gamma)^2 - (\alpha^2+\beta^2+\gamma^2) = 2q\)M1
\(\Rightarrow q = \frac{1}{2}(225 - 83) = 71\)A1
\(\frac{36}{\alpha} = 15 - \alpha \quad (= [\beta + \gamma])\)M1
\(\Rightarrow \alpha^2 - 15\alpha + 36 = 0 \Rightarrow \alpha = 3\), \(\alpha \neq 12\), *e.g.* since \(12^2 > 83\) or other reasonM1A1
\(\beta\gamma = 71 - 36 = 35\)B1
\(\Rightarrow r = -\alpha\beta\gamma = -3 \times 35 = -105\)A1 extra answer penalised
Total: 8 marks[8]
## Question 5:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\alpha + \beta + \gamma = -p = 15 \Rightarrow p = -15$ | B1 | |
| $2(\alpha\beta + \beta\gamma + \gamma\alpha) = (\alpha+\beta+\gamma)^2 - (\alpha^2+\beta^2+\gamma^2) = 2q$ | M1 | |
| $\Rightarrow q = \frac{1}{2}(225 - 83) = 71$ | A1 | |
| $\frac{36}{\alpha} = 15 - \alpha \quad (= [\beta + \gamma])$ | M1 | |
| $\Rightarrow \alpha^2 - 15\alpha + 36 = 0 \Rightarrow \alpha = 3$, $\alpha \neq 12$, *e.g.* since $12^2 > 83$ or other reason | M1A1 | |
| $\beta\gamma = 71 - 36 = 35$ | B1 | |
| $\Rightarrow r = -\alpha\beta\gamma = -3 \times 35 = -105$ | A1 | extra answer penalised |
| **Total: 8 marks** | [8] | |

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5 The cubic equation $x ^ { 3 } + p x ^ { 2 } + q x + r = 0$, where $p , q$ and $r$ are integers, has roots $\alpha , \beta$ and $\gamma$, such that

$$\begin{aligned}
\alpha + \beta + \gamma & = 15 , \\
\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } & = 83 .
\end{aligned}$$

Write down the value of $p$ and find the value of $q$.

Given that $\alpha , \beta$ and $\gamma$ are all real and that $\alpha \beta + \alpha \gamma = 36$, find $\alpha$ and hence find the value of $r$.

\hfill \mbox{\textit{CAIE FP1 2015 Q5}}