| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find remainder(s) then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the remainder theorem followed by routine factorisation. Part (a) requires direct substitution into f(x) at x = -1/2 and x = 3, which is pure recall. Part (b) uses the zero remainder from (a)(ii) to find one factor, then polynomial division or inspection to complete the factorisation. All steps are standard textbook procedures with no problem-solving insight required, making it easier than average but not trivial due to the arithmetic involved. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(f(\pm\frac{1}{2})\) or long division as far as remainder | M1 | Achieves a numerical remainder \(R\) |
| Remainder \(= 49\) | A1 | cao; accept \(f(-\frac{1}{2}) = 49\) or just 49 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(f(\pm3)\) or long division as far as remainder | M1 | Quotient must start \(6x^2\) if long division used |
| Remainder \(= 0\) | A1 | cao; accept \(f(3) = 0\) or just 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6x^3 - 7x^2 - 43x + 30 = (x-3)(6x^2+11x-10)\) | M1 A1 | Recognises \((x-3)\) is factor; obtains quadratic with two correct terms |
| \((6x^2+11x-10) = (ax+b)(cx+d)\) where \(ac=6\) and \(bd=-10\) | M1 | Attempt to factorise quadratic |
| \(= (x-3)(2x+5)(3x-2)\) | A1 | cao — need all three factors; do not penalise if roots also stated |
## Question 4:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $f(\pm\frac{1}{2})$ or long division as far as remainder | M1 | Achieves a numerical remainder $R$ |
| Remainder $= 49$ | A1 | cao; accept $f(-\frac{1}{2}) = 49$ or just 49 |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $f(\pm3)$ or long division as far as remainder | M1 | Quotient must start $6x^2$ if long division used |
| Remainder $= 0$ | A1 | cao; accept $f(3) = 0$ or just 0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6x^3 - 7x^2 - 43x + 30 = (x-3)(6x^2+11x-10)$ | M1 A1 | Recognises $(x-3)$ is factor; obtains quadratic with two correct terms |
| $(6x^2+11x-10) = (ax+b)(cx+d)$ where $ac=6$ and $bd=-10$ | M1 | Attempt to factorise quadratic |
| $= (x-3)(2x+5)(3x-2)$ | A1 | cao — need all three factors; do not penalise if roots also stated |4.
$$f ( x ) = 6 x ^ { 3 } - 7 x ^ { 2 } - 43 x + 30$$
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by
\begin{enumerate}[label=(\roman*)]
\item $2 x + 1$
\item $x - 3$
\end{enumerate}\item Hence factorise $\mathrm { f } ( x )$ completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q4 [8]}}