| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Known polynomial, verify then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem followed by routine polynomial division or comparison of coefficients. Part (a) requires only substituting x=2 and verifying the result is zero (simple arithmetic), while part (b) involves standard algebraic manipulation to find the remaining quadratic factor and its roots. This is easier than average as it's a direct textbook exercise with no problem-solving insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(2^3+6\times2^2-2-30=0\) so \((x-2)\) is a factor | B1 | Factor theorem must be used with concluding statement. Synthetic/Long Division is B0. Must see evidence of substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-2)(x^2+8x+15)\) | M1, A1 | By inspection or long division, allow sign errors only |
| \((x-2)(x+5)(x+3)\) | A1 | Fully correct and fully factorised |
| *Alternatively:* \(f(k)\) evaluated where \(k\) is \(-3\) and \(-5\) | M1 | Allow a slip with either but not both |
| \((x+3)\) or \((x+5)\) identified as a factor | A1 | |
| \((x-2)(x+5)(x+3)\) | A1 |
# Question 7:
## Part (a)
$2^3+6\times2^2-2-30=0$ so $(x-2)$ is a factor | B1 | Factor theorem must be used with concluding statement. Synthetic/Long Division is B0. Must see evidence of substitution
## Part (b)
$(x-2)(x^2+8x+15)$ | M1, A1 | By inspection or long division, allow sign errors only
$(x-2)(x+5)(x+3)$ | A1 | Fully correct and fully factorised
*Alternatively:* $f(k)$ evaluated where $k$ is $-3$ and $-5$ | M1 | Allow a slip with either but not both
$(x+3)$ or $(x+5)$ identified as a factor | A1 |
$(x-2)(x+5)(x+3)$ | A1 |
---7
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $( x - 2 )$ is a factor of $x ^ { 3 } + 6 x ^ { 2 } - x - 30$.
\item Factorise $x ^ { 3 } + 6 x ^ { 2 } - x - 30$ completely.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2023 Q7 [4]}}