| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Remainder condition then further work |
| Difficulty | Moderate -0.8 This is a straightforward application of the Remainder Theorem requiring direct substitution into f(-2) to find k, then substitution into f(2/3). Both parts are routine C2 exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires careful arithmetic with fractions. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(-2) = -35\) \(\therefore -24 - 8 - 2k + 9 = -35\), \(k = 6\) | M1 A1 | B1 |
| (b) \(= f(\frac{2}{3})\) \(= 3(\frac{2}{3})^3 - 2(\frac{2}{3}) + 6(\frac{2}{3}) + 9 = \frac{8}{9} - \frac{8}{9} + 4 + 9 = 13\) | M1 A1 | (5) |
(a) $f(-2) = -35$ $\therefore -24 - 8 - 2k + 9 = -35$, $k = 6$ | M1 A1 | B1
(b) $= f(\frac{2}{3})$ $= 3(\frac{2}{3})^3 - 2(\frac{2}{3}) + 6(\frac{2}{3}) + 9 = \frac{8}{9} - \frac{8}{9} + 4 + 9 = 13$ | M1 A1 | (5)\begin{enumerate}
\item $\quad \mathrm { f } ( x ) = 3 x ^ { 3 } - 2 x ^ { 2 } + k x + 9$.
\end{enumerate}
Given that when $\mathrm { f } ( x )$ is divided by ( $x + 2$ ) there is a remainder of - 35 ,\\
(a) find the value of the constant $k$,\\
(b) find the remainder when $\mathrm { f } ( x )$ is divided by ( $3 x - 2$ ).\\
\hfill \mbox{\textit{Edexcel C2 Q1 [5]}}