| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.3 This is a standard C2 Factor/Remainder Theorem question requiring systematic application of f(1/2) = -5 and f(-2) = 0 to find two unknowns, then factorization. It's slightly easier than average because it's a routine textbook exercise with clear steps: substitute values, solve simultaneous equations, factor. No novel insight required, just methodical application of the theorem. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f\!\left(\frac{1}{2}\right) = 2 \times \frac{1}{8} + a \times \frac{1}{4} + b \times \frac{1}{2} - 6\) | M1 | Attempting \(f\!\left(\pm\frac{1}{2}\right)\); omission of \(-5\) treated as slip |
| \(f\!\left(\frac{1}{2}\right) = -5 \Rightarrow \frac{1}{4}a + \frac{1}{2}b = \frac{3}{4}\) or \(a + 2b = 3\) | A1 | First correct equation in \(a\) and \(b\) simplified to three non-zero terms |
| \(f(-2) = -16 + 4a - 2b - 6\) | M1 | Attempting \(f(\mp 2)\) |
| \(f(-2) = 0 \Rightarrow 4a - 2b = 22\) | A1 | Second correct equation in \(a\) and \(b\) simplified to three terms |
| Eliminating one variable; \(a = 5\) and \(b = -1\) | M1, A1 | M1 for attempt to eliminate one variable; A1 for both correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x^3 + 5x^2 - x - 6 = (x+2)(2x^2 + x - 3)\) | M1 | Attempt to divide by \((x+2)\) leading to 3TQ beginning with correct term \(2x^2\) |
| \(= (x+2)(2x+3)(x-1)\) | M1A1 | M1 for attempt to factorise quadratic (no remainder); A1 c.a.o. all three factors |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f\!\left(\frac{1}{2}\right) = 2 \times \frac{1}{8} + a \times \frac{1}{4} + b \times \frac{1}{2} - 6$ | M1 | Attempting $f\!\left(\pm\frac{1}{2}\right)$; omission of $-5$ treated as slip |
| $f\!\left(\frac{1}{2}\right) = -5 \Rightarrow \frac{1}{4}a + \frac{1}{2}b = \frac{3}{4}$ or $a + 2b = 3$ | A1 | First correct equation in $a$ and $b$ simplified to three non-zero terms |
| $f(-2) = -16 + 4a - 2b - 6$ | M1 | Attempting $f(\mp 2)$ |
| $f(-2) = 0 \Rightarrow 4a - 2b = 22$ | A1 | Second correct equation in $a$ and $b$ simplified to three terms |
| Eliminating one variable; $a = 5$ and $b = -1$ | M1, A1 | M1 for attempt to eliminate one variable; A1 for both correct answers |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x^3 + 5x^2 - x - 6 = (x+2)(2x^2 + x - 3)$ | M1 | Attempt to divide by $(x+2)$ leading to 3TQ beginning with correct term $2x^2$ |
| $= (x+2)(2x+3)(x-1)$ | M1A1 | M1 for attempt to factorise quadratic (no remainder); A1 c.a.o. all three factors |
---3.
$$f ( x ) = 2 x ^ { 3 } + a x ^ { 2 } + b x - 6$$
where $a$ and $b$ are constants.\\
When $\mathrm { f } ( x )$ is divided by $( 2 x - 1 )$ the remainder is - 5 .\\
When $\mathrm { f } ( x )$ is divided by $( x + 2 )$ there is no remainder.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$.
\item Factorise $\mathrm { f } ( x )$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2010 Q3 [9]}}