| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Apply remainder theorem only |
| Difficulty | Moderate -0.8 This is a straightforward application of the remainder theorem requiring substitution of x values and basic algebraic manipulation. Part (a) requires evaluating f(1/2), part (b)(i) requires f(-2), and part (b)(ii) uses the remainder to identify a factor, leading to a routine factorization. All steps are standard textbook exercises with no problem-solving insight required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks |
|---|---|
| (a) \(f\left(\frac{1}{2}\right) = \frac{1}{4} + \frac{3}{4} - 3 + 1 = -1\) | M1 A1 |
| (b)(i) \(f(-2) = -16 + 12 + 12 + 1 = 9\) | B1 |
| (b)(ii) \(x = -2\) is a solution to \(f(x) = 9\) i.e. \(2x^3 + 3x^2 - 6x - 8 = 0\) | M1 A1 |
| Answer | Marks |
|---|---|
| \[\begin{array}{c | c} |
| Answer | Marks |
|---|---|
| M1 A1 | |
| Therefore \((x + 2)(2x^2 - x - 4) = 0\) | |
| \(x = -2\) or \(x = \frac{1 \pm \sqrt{1 + 32}}{4} = -2, -1.19 \text{ (3sf)}, 1.69 \text{ (3sf)}\) | M1 A1 (9) |
**(a)** $f\left(\frac{1}{2}\right) = \frac{1}{4} + \frac{3}{4} - 3 + 1 = -1$ | M1 A1 |
**(b)(i)** $f(-2) = -16 + 12 + 12 + 1 = 9$ | B1 |
**(b)(ii)** $x = -2$ is a solution to $f(x) = 9$ i.e. $2x^3 + 3x^2 - 6x - 8 = 0$ | M1 A1 |
Polynomial division: $\frac{2x^3 + 3x^2 - 6x - 8}{x + 2}$
$$\begin{array}{c|c}
& 2x^2 - x - 4 \\
\hline
x + 2 & 2x^3 + 3x^2 - 6x - 8 \\
& 2x^3 + 4x^2 \\
\hline
& -x^2 - 6x \\
& -x^2 - 2x \\
\hline
& -4x - 8 \\
& -4x - 8 \\
\hline
& 0
\end{array}$$
| M1 A1 |
Therefore $(x + 2)(2x^2 - x - 4) = 0$ | |
$x = -2$ or $x = \frac{1 \pm \sqrt{1 + 32}}{4} = -2, -1.19 \text{ (3sf)}, 1.69 \text{ (3sf)}$ | M1 A1 (9) |6.
$$f ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } - 6 x + 1$$
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by ( $2 x - 1$ ).
\item \begin{enumerate}[label=(\roman*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by $( x + 2 )$.
\item Hence, or otherwise, solve the equation
$$2 x ^ { 3 } + 3 x ^ { 2 } - 6 x - 8 = 0$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q6 [9]}}