| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Division then Solve Polynomial Equation |
| Difficulty | Moderate -0.3 Part (a) is straightforward algebraic manipulation (equating two expressions and rearranging). Part (b) requires using factor theorem with a given root to factorize a cubic, then solving a quadratic - all standard C2 techniques with no novel insight required. The 8 marks reflect routine multi-step work rather than conceptual difficulty, making this slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (a) \(7 - 2x - 3x^2 = \frac{2}{x} \Rightarrow 7x - 2x^2 - 3x^3 = 2\) | M1 |
| \(3x^3 + 2x^2 - 7x + 2 = 0\) | A1 |
| (b) \(x = -2\) is a solution \(\therefore (x+2)\) is a factor | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[x + 2 \mid \begin{array}{c} 3x^2 - 4x + 1 \\ 3x^3 + 2x^2 - 7x + 2 \\ 3x^3 + 6x^2 \\ -4x^2 - 7x \\ -4x^2 - 8x \\ x + 2 \\ x + 2 \end{array}\] | M1 A1 | |
| \((x+2)(3x^2 - 4x + 1) = 0\) | M1 | |
| \((x+2)(3x-1)(x-1) = 0\) | M1 | |
| \(x = -2\) (at P), \(\frac{1}{3}\), \(1\) \(\therefore (\frac{1}{3}, 6), (1, 2)\) | A2 | (8 marks) |
**(a)** $7 - 2x - 3x^2 = \frac{2}{x} \Rightarrow 7x - 2x^2 - 3x^3 = 2$ | M1 |
$3x^3 + 2x^2 - 7x + 2 = 0$ | A1 |
**(b)** $x = -2$ is a solution $\therefore (x+2)$ is a factor | B1 |
Polynomial division shown:
$$x + 2 \mid \begin{array}{c} 3x^2 - 4x + 1 \\ 3x^3 + 2x^2 - 7x + 2 \\ 3x^3 + 6x^2 \\ -4x^2 - 7x \\ -4x^2 - 8x \\ x + 2 \\ x + 2 \end{array}$$ | M1 A1 |
$(x+2)(3x^2 - 4x + 1) = 0$ | M1 |
$(x+2)(3x-1)(x-1) = 0$ | M1 |
$x = -2$ (at P), $\frac{1}{3}$, $1$ $\therefore (\frac{1}{3}, 6), (1, 2)$ | A2 | (8 marks)\includegraphics{figure_2}
Figure 2 shows the curves with equations $y = 7 - 2x - 3x^2$ and $y = \frac{2}{x}$.
The two curves intersect at the points $P$, $Q$ and $R$.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinates of $P$, $Q$ and $R$ satisfy the equation
$$3x^3 + 2x^2 - 7x + 2 = 0.$$ [2]
Given that $P$ has coordinates $(-2, -1)$,
\item find the coordinates of $Q$ and $R$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q3 [8]}}