CAIE FP1 2015 November — Question 5 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with given sum conditions
DifficultyStandard +0.8 This is a multi-step Further Maths question requiring systematic application of Vieta's formulas and algebraic manipulation of symmetric functions of roots. While the techniques are standard for FP1 (relating sums of roots to coefficients), it requires careful bookkeeping across multiple equations and the insight to use the identity α²+β²+γ² = (α+β+γ)² - 2(αβ+αγ+βγ). The final part requires solving for individual roots given partial information, which adds complexity beyond routine exercises.
Spec1.04e Sequences: nth term and recurrence relations4.05a 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
\(\frac{36}{\alpha} = 15 - \alpha\ \ (= [\beta + \gamma])\)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
\(\Rightarrow r = -\alpha\beta\gamma = -3 \times 35 = -105\) (extra answer penalised)A1
Total: 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 | |
| $\frac{36}{\alpha} = 15 - \alpha\ \ (= [\beta + \gamma])$ | 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 | |
| $\Rightarrow r = -\alpha\beta\gamma = -3 \times 35 = -105$ (extra answer penalised) | A1 | |
| **Total: 8** | | |

---
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 [8]}}
This paper (2 questions)
View full paper