Question 4 - A-Level Maths

AQA FP2 2010 June — Question 4 13 marks

Exam Board	AQA
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Roots of polynomials
Type	Roots with given sum conditions
Difficulty	Standard +0.8 This is a Further Maths FP2 question requiring systematic use of Vieta's formulas, Newton's identities for power sums, and algebraic manipulation across multiple connected parts. While the individual steps are guided, students must understand why roots satisfy the polynomial equation, manipulate symmetric functions (sum of squares, sum of cubes), and solve the resulting cubic. The multi-stage reasoning and FP2 content place it moderately above average difficulty.
Spec	4.05a Roots and coefficients: symmetric functions

4 The roots of the cubic equation $$z ^ { 3 } - 2 z ^ { 2 } + p z + 10 = 0$$ are $\alpha , \beta$ and $\gamma$.
It is given that $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } = - 4$.

Write down the value of $\alpha + \beta + \gamma$.
1. Explain why $\alpha ^ { 3 } - 2 \alpha ^ { 2 } + p \alpha + 10 = 0$.
2. Hence show that $$\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = p + 13$$
3. Deduce that $p = - 3$.
1. Find the real root $\alpha$ of the cubic equation $z ^ { 3 } - 2 z ^ { 2 } - 3 z + 10 = 0$.
2. Find the values of $\beta$ and $\gamma$.

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
$\alpha + \beta + \gamma = 2$	B1	From Vieta's formulas

Part (b)(i):

Answer	Marks	Guidance
$\alpha$ is a root so satisfies the equation $z^3 - 2z^2 + pz + 10 = 0$	B1	Correct explanation

Part (b)(ii):

Answer	Marks	Guidance
$(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha)$	M1	Correct identity used
$\alpha\beta+\beta\gamma+\gamma\alpha = p$	B1	From Vieta's formulas
$\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$	M1	Using sum of cubes identity
$-4 - 3(-10) = 2(\alpha^2+\beta^2+\gamma^2 - p)$	A1	Correct substitution
$\alpha^2+\beta^2+\gamma^2 = p + 13$	A1	Correct result

Part (b)(iii):

Answer	Marks	Guidance
$4 = \alpha^2+\beta^2+\gamma^2 - 2p \Rightarrow 4 = p+13-2p$	M1	Using $(α+β+γ)^2$
$p = -3$	A1	Correct deduction

Part (c)(i):

Answer	Marks	Guidance
Trial of factors of $z^3 - 2z^2 - 3z + 10 = 0$; $z = -2$: $-8-8+6+10=0$ ✓	M1	Correct method
$\alpha = -2$	A1	Correct real root

Part (c)(ii):

Answer	Marks	Guidance
$(z+2)(z^2 - 4z + 5) = 0$	M1	Correct factorisation
$z = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm 2i}{2}$	A1	Correct method
$\beta = 2+i,\ \gamma = 2-i$	A1	Correct values

# Question 4:

## Part (a):
| $\alpha + \beta + \gamma = 2$ | B1 | From Vieta's formulas |

## Part (b)(i):
| $\alpha$ is a root so satisfies the equation $z^3 - 2z^2 + pz + 10 = 0$ | B1 | Correct explanation |

## Part (b)(ii):
| $(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha)$ | M1 | Correct identity used |
|---|---|---|
| $\alpha\beta+\beta\gamma+\gamma\alpha = p$ | B1 | From Vieta's formulas |
| $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$ | M1 | Using sum of cubes identity |
| $-4 - 3(-10) = 2(\alpha^2+\beta^2+\gamma^2 - p)$ | A1 | Correct substitution |
| $\alpha^2+\beta^2+\gamma^2 = p + 13$ | A1 | Correct result |

## Part (b)(iii):
| $4 = \alpha^2+\beta^2+\gamma^2 - 2p \Rightarrow 4 = p+13-2p$ | M1 | Using $(α+β+γ)^2$ |
|---|---|---|
| $p = -3$ | A1 | Correct deduction |

## Part (c)(i):
| Trial of factors of $z^3 - 2z^2 - 3z + 10 = 0$; $z = -2$: $-8-8+6+10=0$ ✓ | M1 | Correct method |
|---|---|---|
| $\alpha = -2$ | A1 | Correct real root |

## Part (c)(ii):
| $(z+2)(z^2 - 4z + 5) = 0$ | M1 | Correct factorisation |
|---|---|---|
| $z = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm 2i}{2}$ | A1 | Correct method |
| $\beta = 2+i,\ \gamma = 2-i$ | A1 | Correct values |

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Show LaTeX source

4 The roots of the cubic equation

$$z ^ { 3 } - 2 z ^ { 2 } + p z + 10 = 0$$

are $\alpha , \beta$ and $\gamma$.\\
It is given that $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } = - 4$.\\
(a) Write down the value of $\alpha + \beta + \gamma$.\\
(b) (i) Explain why $\alpha ^ { 3 } - 2 \alpha ^ { 2 } + p \alpha + 10 = 0$.\\
(ii) Hence show that

$$\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = p + 13$$

(iii) Deduce that $p = - 3$.\\
(c) (i) Find the real root $\alpha$ of the cubic equation $z ^ { 3 } - 2 z ^ { 2 } - 3 z + 10 = 0$.\\
(ii) Find the values of $\beta$ and $\gamma$.

\hfill \mbox{\textit{AQA FP2 2010 Q4 [13]}}

This paper (7 questions)

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Q1 9 Q2 8 Q3 9 Q4 13 Q5 18 Q6 8 Q7 10

Answer	Marks	Guidance
\((\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha)\)	M1	Correct identity used
\(\alpha\beta+\beta\gamma+\gamma\alpha = p\)	B1	From Vieta's formulas
\(\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)\)	M1	Using sum of cubes identity
\(-4 - 3(-10) = 2(\alpha^2+\beta^2+\gamma^2 - p)\)	A1	Correct substitution
\(\alpha^2+\beta^2+\gamma^2 = p + 13\)	A1	Correct result

Answer	Marks	Guidance
\(4 = \alpha^2+\beta^2+\gamma^2 - 2p \Rightarrow 4 = p+13-2p\)	M1	Using \((α+β+γ)^2\)
\(p = -3\)	A1	Correct deduction

Answer	Marks	Guidance
Trial of factors of \(z^3 - 2z^2 - 3z + 10 = 0\); \(z = -2\): \(-8-8+6+10=0\) ✓	M1	Correct method
\(\alpha = -2\)	A1	Correct real root

Answer	Marks	Guidance
\((z+2)(z^2 - 4z + 5) = 0\)	M1	Correct factorisation
\(z = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm 2i}{2}\)	A1	Correct method
\(\beta = 2+i,\ \gamma = 2-i\)	A1	Correct values