| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Apply remainder theorem only |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard remainder theorem application and algebraic manipulation. Part (a) requires simple substitution of x=-1/2, part (b)(i) is trivial given the remainder from (a), and subsequent parts involve routine factorization and algebraic fraction simplification. All techniques are standard C4 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
1 The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 8 x - 7$.
\begin{enumerate}[label=(\alph*)]
\item Use the Remainder Theorem to find the remainder when $\mathrm { f } ( x )$ is divided by $( 2 x + 1 )$.\\
(2 marks)
\item The polynomial $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = \mathrm { f } ( x ) + d$, where $d$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Given that $( 2 x + 1 )$ is a factor of $\mathrm { g } ( x )$, show that $\mathrm { g } ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 8 x - 4$.\\
(1 mark)
\item Given that $\mathrm { g } ( x )$ can be written as $\mathrm { g } ( x ) = ( 2 x + 1 ) \left( x ^ { 2 } + a \right)$, where $a$ is an integer, express $\mathrm { g } ( x )$ as a product of three linear factors.
\item Hence, or otherwise, show that $\frac { \mathrm { g } ( x ) } { 2 x ^ { 3 } - 3 x ^ { 2 } - 2 x } = p + \frac { q } { x }$, where $p$ and $q$ are integers.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2013 Q1 [7]}}