| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Given factor, find all roots |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing basic factor theorem application. Part (a) is simple substitution, part (b) requires polynomial division and factoring a quadratic (which factors neatly), and part (c) is direct remainder theorem application. All steps are routine with no problem-solving insight required, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2+1-5+c=0\) or \(-2+c=0\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(f(x) = (x-1)(2x^2+3x-2)\) | B1 | |
| \((x-1)\) division | M1 | |
| \(= \ldots (2x-1)(x+2)\) | M1 A1 | (4 marks) |
| (c) \(f\left(\frac{3}{2}\right) = 2 \times \frac{27}{8} + \frac{9}{4} + \frac{15}{2} + c\) | M1 | |
| Remainder \(= c + 1.5 = 3.5\) | A1 ft their \(c\) | (2 marks) |
(a) $2+1-5+c=0$ or $-2+c=0$ | M1 A1 | (2 marks)
$c=2$
(b) $f(x) = (x-1)(2x^2+3x-2)$ | B1 |
$(x-1)$ division | M1
$= \ldots (2x-1)(x+2)$ | M1 A1 | (4 marks)
(c) $f\left(\frac{3}{2}\right) = 2 \times \frac{27}{8} + \frac{9}{4} + \frac{15}{2} + c$ | M1
Remainder $= c + 1.5 = 3.5$ | A1 ft their $c$ | (2 marks)
**Total: 8 marks**
---\begin{enumerate}
\item $\mathrm { f } ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 5 x + c$, where $c$ is a constant.
\end{enumerate}
Given that $\mathrm { f } ( 1 ) = 0$,\\
(a) find the value of $c$,\\
(b) factorise $\mathrm { f } ( x )$ completely,\\
(c) find the remainder when $\mathrm { f } ( x )$ is divided by ( $2 x - 3$ ).\\
\hfill \mbox{\textit{Edexcel C2 2006 Q1 [8]}}