Edexcel CP AS 2021 June — Question 7 9 marks

Exam BoardEdexcel
ModuleCP AS (Core Pure AS)
Year2021
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeComplex roots with real coefficients
DifficultyStandard +0.3 This is a straightforward application of the complex conjugate root theorem and Vieta's formulas. Part (a) requires knowing that complex roots come in conjugate pairs for polynomials with real coefficients. Parts (b) and (c) involve routine algebraic manipulation using sum/product of roots. Part (d) is a simple transformation. While multi-part, each step follows standard procedures with no novel insight required, making it slightly easier than average.
Spec4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem

7. $$f ( z ) = z ^ { 4 } - 6 z ^ { 3 } + p z ^ { 2 } + q z + r$$ where \(p , q\) and \(r\) are real constants.
The roots of the equation \(\mathrm { f } ( \mathrm { z } ) = 0\) are \(\alpha , \beta , \gamma\) and \(\delta\) where \(\alpha = 3\) and \(\beta = 2 + \mathrm { i }\) Given that \(\gamma\) is a complex root of \(\mathrm { f } ( \mathrm { z } ) = 0\)
    1. write down the root \(\gamma\),
    2. explain why \(\delta\) must be real.
  1. Determine the value of \(\delta\).
  2. Hence determine the values of \(p , q\) and \(r\).
  3. Write down the roots of the equation \(\mathrm { f } ( - 2 \mathrm { z } ) = 0\)

Question 7:
Part (a)(i):
AnswerMarks Guidance
AnswerMark Guidance
\(2 - i\)B1 Correct complex number
Part (a)(ii):
AnswerMarks Guidance
AnswerMark Guidance
Roots of polynomials with real coefficients occur in conjugate pairs, \(\beta\) and \(\gamma\) form a conjugate pair, \(\alpha\) is real so \(\delta\) must also be real. OR Quartics have either 4 real roots, 2 real roots and 2 complex roots or 4 complex roots. As 2 complex roots and 1 real root therefore \(\delta\) must also be real.B1 Correct explanation
Part (b):
AnswerMarks Guidance
AnswerMark Guidance
\(\alpha + \beta + \gamma + \delta = 6 \Rightarrow 3 + 2+i + 2-i + \delta = 6\)M1 Uses \(2 \pm i\) and 1 together with sum of roots \(= \pm 6\) to find \(\delta\)
\(\delta = -1\)A1 Correct value
Part (c):
AnswerMarks Guidance
AnswerMark Guidance
\(f(z) = (z-3)(z+1)(z-(2+i))(z-(2-i)) = \ldots\)M1 Uses \((z-3)\) and \((z - \delta)\) and conjugate pair correctly as factors and attempts expansion. Alternatively attempts pair sum, triple sum and product
\(= (z^2 - 2z - 3)(z^2 - 4z + 5)\)A1 Establishes at least 2 of the required coefficients correctly
\(= z^4 - 6z^3 + 10z^2 + 2z - 15\), \(p = 10, q = 2, r = -15\)A1 Correct quartic or correct constants
Part (d):
AnswerMarks Guidance
AnswerMark Guidance
\(z = \frac{1}{2}, -\frac{3}{2}\)B1ft For \(-\frac{3}{2}\) and \(-\frac{\delta}{2}\) as the real roots
\(z = -1 \pm \frac{i}{2}\)B1ft For \(-1 - \frac{i}{2}\) and \(-\frac{\gamma}{2}\) as the complex roots
# Question 7:

## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 - i$ | B1 | Correct complex number |

## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Roots of polynomials with real coefficients occur in conjugate pairs, $\beta$ and $\gamma$ form a conjugate pair, $\alpha$ is real so $\delta$ must also be real. OR Quartics have either 4 real roots, 2 real roots and 2 complex roots or 4 complex roots. As 2 complex roots and 1 real root therefore $\delta$ must also be real. | B1 | Correct explanation |

## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\alpha + \beta + \gamma + \delta = 6 \Rightarrow 3 + 2+i + 2-i + \delta = 6$ | M1 | Uses $2 \pm i$ and 1 together with sum of roots $= \pm 6$ to find $\delta$ |
| $\delta = -1$ | A1 | Correct value |

## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(z) = (z-3)(z+1)(z-(2+i))(z-(2-i)) = \ldots$ | M1 | Uses $(z-3)$ and $(z - \delta)$ and conjugate pair correctly as factors and attempts expansion. Alternatively attempts pair sum, triple sum and product |
| $= (z^2 - 2z - 3)(z^2 - 4z + 5)$ | A1 | Establishes at least 2 of the required coefficients correctly |
| $= z^4 - 6z^3 + 10z^2 + 2z - 15$, $p = 10, q = 2, r = -15$ | A1 | Correct quartic or correct constants |

## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = \frac{1}{2}, -\frac{3}{2}$ | B1ft | For $-\frac{3}{2}$ and $-\frac{\delta}{2}$ as the real roots |
| $z = -1 \pm \frac{i}{2}$ | B1ft | For $-1 - \frac{i}{2}$ and $-\frac{\gamma}{2}$ as the complex roots |

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7.

$$f ( z ) = z ^ { 4 } - 6 z ^ { 3 } + p z ^ { 2 } + q z + r$$

where $p , q$ and $r$ are real constants.\\
The roots of the equation $\mathrm { f } ( \mathrm { z } ) = 0$ are $\alpha , \beta , \gamma$ and $\delta$ where $\alpha = 3$ and $\beta = 2 + \mathrm { i }$\\
Given that $\gamma$ is a complex root of $\mathrm { f } ( \mathrm { z } ) = 0$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item write down the root $\gamma$,
\item explain why $\delta$ must be real.
\end{enumerate}\item Determine the value of $\delta$.
\item Hence determine the values of $p , q$ and $r$.
\item Write down the roots of the equation $\mathrm { f } ( - 2 \mathrm { z } ) = 0$
\end{enumerate}

\hfill \mbox{\textit{Edexcel CP AS 2021 Q7 [9]}}