AQA FP2 2016 June — Question 2 8 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2016
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeParametric polynomials with root conditions
DifficultyStandard +0.3 This is a standard Further Pure complex roots question testing well-rehearsed techniques: complex conjugate roots theorem, Vieta's formulas, and basic complex arithmetic. The multi-part structure guides students through the solution methodically with no novel insight required—slightly easier than average A-level difficulty.
Spec4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots

The cubic equation \(3z^3 + pz^2 + 17z + q = 0\), where \(p\) and \(q\) are real, has a root \(\alpha = 1 + 2\mathrm{i}\).
    1. Write down the value of another non-real root, \(\beta\), of this equation. [1 mark]
    2. Hence find the value of \(\alpha\beta\). [1 mark]
  1. Find the value of the third root, \(\gamma\), of this equation. [3 marks]
  2. Find the values of \(p\) and \(q\). [3 marks]

AnswerMarks Guidance
(a)(i) \(1 - 2i\)B1 1 mark
(a)(ii) \((\alpha\beta = 1 + 4 =) 5\)B1 1 mark
(b) \(\sum\alpha\beta = \frac{17}{3}\)B1 PI by next line
\(\alpha\gamma + \beta\gamma + \text{"their"} 5 = \text{"their"} \frac{17}{3}\)M1 FT "their" \(\alpha\beta\) and \(\sum\alpha\beta\) values
\(\Rightarrow \gamma = \frac{1}{3}\)A1 3 marks
Alternative: \(z^3 + \frac{p}{3}z^2 + \frac{17}{3}z + \frac{q}{3}\); quadratic factor \(z^2 - 2z + 5\) B1; \((z^2 - 2z + 5)(z - \gamma)\) comparing coefficient of \(z\): \(5 + 2\gamma = \frac{17}{3}\) M1; \(\Rightarrow \gamma = \frac{1}{3}\) A1 (3)
AnswerMarks Guidance
(c) \(\alpha + \beta + \gamma = \frac{-p}{3}\), \(\alpha\beta\gamma = \frac{-q}{3}\)M1 Either of these expressions correct
\(p = -7\)A1 PI by correct \(p\) or \(q\)
\(q = -5\)A1 3 marks
Alternative: comparing coefficients either \(-5y = \frac{q}{3}\) or \(-\gamma - 2 = \frac{p}{3}\) M1; \(p = -7\) A1; \(q = -5\) A1 (3)
Total: 8 marks
Notes:
- Allow M1 for \(5 + 2\gamma = \frac{17}{3}\) if \(\sum\alpha\beta\) not seen
- (c) Example: \(\alpha + \beta + \gamma = -p\); \(\alpha + \beta + \gamma = 2 + \frac{1}{3} - \frac{7}{3}\) \(\Rightarrow p = -7\) Award M1 A1 assuming first statement was meant as candidate's "reminder" for signs but "wiggly underline" incorrect statement
- (c) Example: \(\gamma = \frac{4}{3}\); \(\alpha + \beta + \gamma = \frac{10}{3}\) \(\Rightarrow p = -10\) Award M1 (implied) A0
- Alternative: substituting \(z = 1 + 2i\) or \(1 - 2i\) leading to correct simultaneous equations \(3p - q + 16 = 0\) and \(4p + 28 = 0\) M1 then \(p = -7\) A1; \(q = -5\) A1
**(a)(i)** $1 - 2i$ | B1 | **1 mark**

**(a)(ii)** $(\alpha\beta = 1 + 4 =) 5$ | B1 | **1 mark**

**(b)** $\sum\alpha\beta = \frac{17}{3}$ | B1 | PI by next line

$\alpha\gamma + \beta\gamma + \text{"their"} 5 = \text{"their"} \frac{17}{3}$ | M1 | FT "their" $\alpha\beta$ and $\sum\alpha\beta$ values

$\Rightarrow \gamma = \frac{1}{3}$ | A1 | **3 marks**

**Alternative:** $z^3 + \frac{p}{3}z^2 + \frac{17}{3}z + \frac{q}{3}$; quadratic factor $z^2 - 2z + 5$ B1; $(z^2 - 2z + 5)(z - \gamma)$ comparing coefficient of $z$: $5 + 2\gamma = \frac{17}{3}$ M1; $\Rightarrow \gamma = \frac{1}{3}$ A1 (3)

**(c)** $\alpha + \beta + \gamma = \frac{-p}{3}$, $\alpha\beta\gamma = \frac{-q}{3}$ | M1 | Either of these expressions correct

$p = -7$ | A1 | PI by correct $p$ or $q$

$q = -5$ | A1 | **3 marks**

**Alternative:** comparing coefficients either $-5y = \frac{q}{3}$ or $-\gamma - 2 = \frac{p}{3}$ M1; $p = -7$ A1; $q = -5$ A1 (3)

**Total: 8 marks**

**Notes:**
- Allow M1 for $5 + 2\gamma = \frac{17}{3}$ if $\sum\alpha\beta$ not seen
- **(c) Example:** $\alpha + \beta + \gamma = -p$; $\alpha + \beta + \gamma = 2 + \frac{1}{3} - \frac{7}{3}$ $\Rightarrow p = -7$ Award M1 A1 assuming first statement was meant as candidate's "reminder" for signs but "wiggly underline" incorrect statement
- **(c) Example:** $\gamma = \frac{4}{3}$; $\alpha + \beta + \gamma = \frac{10}{3}$ $\Rightarrow p = -10$ Award M1 (implied) A0
- **Alternative:** substituting $z = 1 + 2i$ or $1 - 2i$ leading to correct simultaneous equations $3p - q + 16 = 0$ and $4p + 28 = 0$ M1 then $p = -7$ A1; $q = -5$ A1

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The cubic equation $3z^3 + pz^2 + 17z + q = 0$, where $p$ and $q$ are real, has a root $\alpha = 1 + 2\mathrm{i}$.

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Write down the value of another non-real root, $\beta$, of this equation. [1 mark]
\item Hence find the value of $\alpha\beta$. [1 mark]
\end{enumerate}

\item Find the value of the third root, $\gamma$, of this equation. [3 marks]

\item Find the values of $p$ and $q$. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2016 Q2 [8]}}