| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Factor theorem and finding roots |
| Difficulty | Moderate -0.5 This is a straightforward Further Maths question requiring polynomial division (given one factor) and then solving a quadratic. While it's FP1, the techniques are routine: expand and compare coefficients for part (a), then use the quadratic formula for part (b). No novel insight required, just methodical application of standard procedures. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = -2, b = 13\) | B1 | At least one of either \(a = -2\) or \(b = 13\) or seen as their coefficients. |
| B1 | Both \(a = -2\) and \(b = 13\) or seen as their coefficients. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{z =\} 1\) is a root | B1 | 1 is a root, seen anywhere. |
| \(2z^2 - 2z + 13 = 0 \Rightarrow z^2 - z + \frac{13}{2} = 0\) | M1 | Correct method for solving a 3-term quadratic equation. Do not allow M1 here for an attempt at factorising. |
| Either \(z = \frac{2 \pm \sqrt{4-4(2)(13)}}{2(2)}\) | ||
| or \(\left(z - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{13}{2} = 0\) and \(z = ...\) | ||
| or \((2z-1)^2 - 1 + 13 = 0\) and \(z = ...\) | ||
| \(\text{So, } \{z =\} \frac{1}{2} + \frac{5}{2}i, \frac{1}{2} - \frac{5}{2}i\) | A1 | At least one of either \(\frac{1}{2} + \frac{5}{2}i\) or \(\frac{1}{2} - \frac{5}{2}i\) or any equivalent form. |
| A1ft | For conjugate of first complex root |
**Part (a)**
| $a = -2, b = 13$ | B1 | At least one of either $a = -2$ or $b = 13$ or seen as their coefficients. |
| | B1 | Both $a = -2$ and $b = 13$ or seen as their coefficients. |
**Part (b)**
| $\{z =\} 1$ is a root | B1 | 1 is a root, seen anywhere. |
| $2z^2 - 2z + 13 = 0 \Rightarrow z^2 - z + \frac{13}{2} = 0$ | M1 | Correct method for solving a 3-term quadratic equation. Do not allow M1 here for an attempt at factorising. |
| Either $z = \frac{2 \pm \sqrt{4-4(2)(13)}}{2(2)}$ | | |
| or $\left(z - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{13}{2} = 0$ and $z = ...$ | | |
| or $(2z-1)^2 - 1 + 13 = 0$ and $z = ...$ | | |
| $\text{So, } \{z =\} \frac{1}{2} + \frac{5}{2}i, \frac{1}{2} - \frac{5}{2}i$ | A1 | At least one of either $\frac{1}{2} + \frac{5}{2}i$ or $\frac{1}{2} - \frac{5}{2}i$ or any equivalent form. |
| | A1ft | For conjugate of first complex root |
**Total: 6 marks**
---1.
$$f ( z ) = 2 z ^ { 3 } - 4 z ^ { 2 } + 15 z - 13$$
Given that $\mathrm { f } ( z ) \equiv ( z - 1 ) \left( 2 z ^ { 2 } + a z + b \right)$, where $a$ and $b$ are real constants,
\begin{enumerate}[label=(\alph*)]
\item find the value of $a$ and the value of $b$.
\item Hence use algebra to find the three roots of the equation $\mathrm { f } ( \mathrm { z } ) = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2018 Q1 [6]}}