| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Factor theorem and finding roots |
| Difficulty | Moderate -0.3 This is a straightforward application of the factor theorem and polynomial division. Part (a) requires simple substitution to find k, and part (b) involves routine factorization and solving a quadratic. While it's a multi-step problem, each step uses standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02h Square roots: of complex numbers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f\!\left(\tfrac{1}{2}\right) = 2\!\left(\tfrac{1}{2}\right)^3 - 9\!\left(\tfrac{1}{2}\right)^2 + k\!\left(\tfrac{1}{2}\right) - 13\) | M1 | Attempts \(f(0.5)\) |
| \(\left(\tfrac{1}{4}\right) - \left(\tfrac{9}{4}\right) + \left(\tfrac{k}{2}\right) - 13 = 0 \Rightarrow k =\) ... | dM1 | Sets \(f(0.5) = 0\) and leading to \(k=\) |
| \(k = 30\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (2x-1)(x^2 - 4x + 13)\) | M1 | \(x^2 + bx \pm 13\) or \(2x^2 + bx \pm 26\); uses inspection, long division or compares coefficients with \((2x-1)\) |
| \(x^2 - 4x + 13\) or \(2x^2 - 8x + 26\) | A1 | |
| \(x = \dfrac{4 \pm \sqrt{4^2 - 4\times 13}}{2}\) or equivalent | M1 | Use of correct quadratic formula or completes the square |
| \(x = \dfrac{4 \pm 6i}{2} = 2 \pm 3i\) | A1 | oe |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f\!\left(\tfrac{1}{2}\right) = 2\!\left(\tfrac{1}{2}\right)^3 - 9\!\left(\tfrac{1}{2}\right)^2 + k\!\left(\tfrac{1}{2}\right) - 13$ | M1 | Attempts $f(0.5)$ |
| $\left(\tfrac{1}{4}\right) - \left(\tfrac{9}{4}\right) + \left(\tfrac{k}{2}\right) - 13 = 0 \Rightarrow k =$ ... | dM1 | Sets $f(0.5) = 0$ **and** leading to $k=$ |
| $k = 30$ | A1 | cao |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (2x-1)(x^2 - 4x + 13)$ | M1 | $x^2 + bx \pm 13$ or $2x^2 + bx \pm 26$; uses inspection, long division or compares coefficients with $(2x-1)$ |
| $x^2 - 4x + 13$ or $2x^2 - 8x + 26$ | A1 | |
| $x = \dfrac{4 \pm \sqrt{4^2 - 4\times 13}}{2}$ or equivalent | M1 | Use of correct quadratic formula or completes the square |
| $x = \dfrac{4 \pm 6i}{2} = 2 \pm 3i$ | A1 | oe |
---3. Given that $x = \frac { 1 } { 2 }$ is a root of the equation
$$2 x ^ { 3 } - 9 x ^ { 2 } + k x - 13 = 0 , \quad k \in \mathbb { R }$$
find
\begin{enumerate}[label=(\alph*)]
\item the value of $k$,
\item the other 2 roots of the equation.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q3 [7]}}