OCR MEI FP1 2013 January — Question 3 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJanuary
Marks6
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Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeComplex roots with real coefficients
DifficultyStandard +0.3 This is a standard Further Maths question testing the conjugate root theorem and polynomial division. Given one complex root with real coefficients, students know the conjugate must also be a root, then find the third root and parameter p through expansion or substitution. Straightforward application of well-practiced techniques with no novel insight required.
Spec4.02g Conjugate pairs: real coefficient polynomials4.02j Cubic/quartic equations: conjugate pairs and factor theorem
3 You are given that \(z = 2 + \mathrm { j }\) is a root of the cubic equation \(2 z ^ { 3 } + p z ^ { 2 } + 22 z - 15 = 0\), where \(p\) is real. Find the other roots and the value of \(p\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(z = 2 - j\) is also a rootB1 Stated, not just used
\(\alpha\beta\gamma = \frac{15}{2}\), or \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{22}{2}\), with \(\alpha\beta = (2+j)(2-j) = 5\) usedM1, A1 Attempt to use roots in a relationship; correct equation obtained for \(\gamma\). Allow incorrect signs
\(\mathbf{OR}\ (az+b)(z-2+j)(z-2-j) = 2z^3 + pz^2 + 22z - 15\)M1 Attempt use of complex factors. Allow incorrect signs \((z+\ldots)\)
\(\Rightarrow (az+b)(z^2-4z+5) = 2z^3 + pz^2 + 22z - 15\)A1 Correct complex factors; one pair of factors correctly multiplied
\(\mathbf{OR}\ 2(2+11j) + p(3+4j) + 22(2+j) - 15 = 0\)M1, A1 Substitution; correct equation. Allow an incorrect sign
Complete valid method for obtaining the other unknownM1 Root relation, obtaining linear factor, equating real and imaginary parts. Signs correct
real root \(= \frac{3}{2}\), \(p = -11\)A1 A1 FT one value
[6] 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = 2 - j$ is also a root | B1 | Stated, not just used |
| $\alpha\beta\gamma = \frac{15}{2}$, or $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{22}{2}$, with $\alpha\beta = (2+j)(2-j) = 5$ used | M1, A1 | Attempt to use roots in a relationship; correct equation obtained for $\gamma$. Allow incorrect signs |
| $\mathbf{OR}\ (az+b)(z-2+j)(z-2-j) = 2z^3 + pz^2 + 22z - 15$ | M1 | Attempt use of complex factors. Allow incorrect signs $(z+\ldots)$ |
| $\Rightarrow (az+b)(z^2-4z+5) = 2z^3 + pz^2 + 22z - 15$ | A1 | Correct complex factors; one pair of factors correctly multiplied |
| $\mathbf{OR}\ 2(2+11j) + p(3+4j) + 22(2+j) - 15 = 0$ | M1, A1 | Substitution; correct equation. Allow an incorrect sign |
| Complete valid method for obtaining the other unknown | M1 | Root relation, obtaining linear factor, equating real and imaginary parts. Signs correct |
| real root $= \frac{3}{2}$, $p = -11$ | A1 A1 | FT one value |
| | **[6]** | |

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3 You are given that $z = 2 + \mathrm { j }$ is a root of the cubic equation $2 z ^ { 3 } + p z ^ { 2 } + 22 z - 15 = 0$, where $p$ is real. Find the other roots and the value of $p$.

\hfill \mbox{\textit{OCR MEI FP1 2013 Q3 [6]}}
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