| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 3 |
| Topic | Factor & Remainder Theorem |
| Type | Find remainder(s) then factorise |
| Difficulty | Easy -1.3 This is a straightforward application of the remainder theorem requiring simple substitution (part i) and immediate deduction (part ii). The calculations are trivial: f(-2) = -8 - 4 = -12, then k = 12. No problem-solving or conceptual depth required, just direct recall of a basic theorem. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f(-2) = -12\) | M1, A1 [2] | Substitute \(x = -2\), or any other complete method – must get as far as attempting the remainder but allow no more than 2 errors. If using inspection then allow M1 for \((x+2)(x^2-2x+k)-2k\). Obtain \(-12\) (no iw if then given as 12 or if given as \(\frac{-12}{(x+2)}\)). Must be identified as remainder so A0 if just left at bottom of division attempt. |
| (ii) \(12\) | B1FT [1] | FT on their (i) |
**(i)** $f(-2) = -12$ | M1, A1 [2] | Substitute $x = -2$, or any other complete method – must get as far as attempting the remainder but allow no more than 2 errors. If using inspection then allow M1 for $(x+2)(x^2-2x+k)-2k$. Obtain $-12$ (no iw if then given as 12 or if given as $\frac{-12}{(x+2)}$). Must be identified as remainder so A0 if just left at bottom of division attempt.
**(ii)** $12$ | B1FT [1] | FT on their (i)\begin{enumerate}[label=(\roman*)]
\item Find the remainder when $x^3 + 2x$ is divided by $x + 2$. [2]
\item Write down the value of $k$ for which $x + 2$ is a factor of $x^3 + 2x + k$. [1]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2016 Q1 [3]}}