AQA C1 2006 June — Question 6 10 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2006
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeKnown polynomial, verify then factorise
DifficultyEasy -1.2 This is a straightforward C1 question testing routine application of Factor and Remainder Theorems with standard algebraic manipulation. Part (a) requires simple substitution, (b) involves factoring out x then factorising a quadratic, and (c) applies textbook methods with no problem-solving insight needed. All steps are mechanical and well-practiced at this level.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

6 The polynomial \(\mathrm { p } ( x )\) is given by \(\mathrm { p } ( x ) = x ^ { 3 } - 4 x ^ { 2 } + 3 x\).
  1. Use the Factor Theorem to show that \(x - 3\) is a factor of \(\mathrm { p } ( x )\).
  2. Express \(\mathrm { p } ( x )\) as the product of three linear factors.
    1. Use the Remainder Theorem to find the remainder, \(r\), when \(\mathrm { p } ( x )\) is divided by \(x - 2\).
    2. Using algebraic division, or otherwise, express \(\mathrm { p } ( x )\) in the form $$( x - 2 ) \left( x ^ { 2 } + a x + b \right) + r$$ where \(a , b\) and \(r\) are constants.

6 The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } - 4 x ^ { 2 } + 3 x$.
\begin{enumerate}[label=(\alph*)]
\item Use the Factor Theorem to show that $x - 3$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x )$ as the product of three linear factors.
\item \begin{enumerate}[label=(\roman*)]
\item Use the Remainder Theorem to find the remainder, $r$, when $\mathrm { p } ( x )$ is divided by $x - 2$.
\item Using algebraic division, or otherwise, express $\mathrm { p } ( x )$ in the form

$$( x - 2 ) \left( x ^ { 2 } + a x + b \right) + r$$

where $a , b$ and $r$ are constants.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C1 2006 Q6 [10]}}