AQA C1 2005 January — Question 4 18 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2005
SessionJanuary
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFactorise polynomial completely
DifficultyModerate -0.8 This is a highly structured, multi-part question that guides students through standard C1 techniques: remainder theorem (given), writing down factors (given information), completing factorisation by inspection/division, finding intercepts, and basic integration. Each part scaffolds the next with minimal problem-solving required, making it easier than average despite covering multiple topics.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals

4
  1. The function f is defined for all values of \(x\) by \(\mathrm { f } ( x ) = x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8\).
    1. Find the remainder when \(\mathrm { f } ( x )\) is divided by \(x + 1\).
    2. Given that \(\mathrm { f } ( 1 ) = 0\) and \(\mathrm { f } ( - 2 ) = 0\), write down two linear factors of \(\mathrm { f } ( x )\).
    3. Hence express \(x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8\) as the product of three linear factors.
  2. The curve with equation \(y = x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{10bca9b4-5327-4b35-8b75-612b396e8a76-3_543_796_897_623}
    1. The curve intersects the \(y\)-axis at the point \(A\). Find the \(y\)-coordinate of \(A\).
    2. The curve crosses the \(x\)-axis when \(x = - 2\), when \(x = 1\) and also at the point \(B\). Use the results from part (a) to find the \(x\)-coordinate of \(B\).
    1. Find \(\int \left( x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8 \right) d x\).
    2. Hence find the area of the shaded region bounded by the curve and the \(x\)-axis.

4
\begin{enumerate}[label=(\alph*)]
\item The function f is defined for all values of $x$ by $\mathrm { f } ( x ) = x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8$.
\begin{enumerate}[label=(\roman*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by $x + 1$.
\item Given that $\mathrm { f } ( 1 ) = 0$ and $\mathrm { f } ( - 2 ) = 0$, write down two linear factors of $\mathrm { f } ( x )$.
\item Hence express $x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8$ as the product of three linear factors.
\end{enumerate}\item The curve with equation $y = x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{10bca9b4-5327-4b35-8b75-612b396e8a76-3_543_796_897_623}
\begin{enumerate}[label=(\roman*)]
\item The curve intersects the $y$-axis at the point $A$. Find the $y$-coordinate of $A$.
\item The curve crosses the $x$-axis when $x = - 2$, when $x = 1$ and also at the point $B$. Use the results from part (a) to find the $x$-coordinate of $B$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8 \right) d x$.
\item Hence find the area of the shaded region bounded by the curve and the $x$-axis.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2005 Q4 [18]}}