| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find remainder(s) then factorise |
| Difficulty | Moderate -0.3 This is a standard C2 polynomial question testing the remainder theorem and factorization. Part (a) is direct application of the remainder theorem. Part (b) uses the remainder theorem to identify a factor, then requires polynomial division and solving a quadratic—all routine techniques for this level with no novel insight required. Slightly easier than average due to the structured guidance ('hence') and straightforward execution. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (a) \(= f(\frac{1}{2}) = \frac{1}{4} + \frac{3}{4} - 3 + 1 = -1\) | M1 A1 |
| Answer | Marks |
|---|---|
| (i) \(= f(-2) = -16 + 12 + 12 + 1 = 9\) | B1 |
| (ii) \(x = -2\) is a solution to \(f(x) = 9\) i.e. \(2x^3 + 3x^2 - 6x - 8 = 0\) | M1 A1 |
| \[\begin{array}{c | cc cc} |
| Answer | Marks | Guidance |
|---|---|---|
| \end{array}\] | M1 A1 | |
| \(\therefore (x+2)(2x^2 - x - 4) = 0\) | ||
| \(x = -2\) or \(\frac{1 \pm \sqrt{1+32}}{4} = \frac{1 \pm \sqrt{33}}{4}\) | M1 | |
| \(x = -2, -1.19 (2\text{dp}), 1.69 (2\text{dp})\) | A1 | (9) |
**(a)** $= f(\frac{1}{2}) = \frac{1}{4} + \frac{3}{4} - 3 + 1 = -1$ | M1 A1 |
**(b)**
**(i)** $= f(-2) = -16 + 12 + 12 + 1 = 9$ | B1 |
**(ii)** $x = -2$ is a solution to $f(x) = 9$ i.e. $2x^3 + 3x^2 - 6x - 8 = 0$ | M1 A1 |
$$\begin{array}{c|cc cc}
& 2x^3 & +3x^2 & -6x & -8 \\
x+2 & 2x^3 & +4x^2 & & \\
& & x^2 & -6x & \\
& & x^2 & -2x & \\
& & & -4x & -8 \\
& & & -4x & -8
\end{array}$$ | M1 A1 |
$\therefore (x+2)(2x^2 - x - 4) = 0$ | |
$x = -2$ or $\frac{1 \pm \sqrt{1+32}}{4} = \frac{1 \pm \sqrt{33}}{4}$ | M1 |
$x = -2, -1.19 (2\text{dp}), 1.69 (2\text{dp})$ | A1 | (9)$f(x) = 2x^3 + 3x^2 - 6x + 1$.
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $f(x)$ is divided by $(2x - 1)$. [2]
\item \begin{enumerate}[label=(\roman*)]
\item Find the remainder when $f(x)$ is divided by $(x + 2)$.
\item Hence, or otherwise, solve the equation
$$2x^3 + 3x^2 - 6x - 8 = 0,$$
giving your answers to 2 decimal places where appropriate. [7]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [9]}}