| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor and Remainder Theorems requiring routine substitution into p(x) and solving simultaneous equations. The multi-part structure guides students through each step with no conceptual challenges beyond standard C1 techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(-2) = (-2)^3 + (-2)^2c + (-2)d - 12\) | M1 | \(p(-2)\) attempted or long division by \(x+2\) as far as remainder |
| 'their' \(-8 + 4c - 2d - 12 = -150\) | m1 | putting expression for remainder \(= -150\) |
| \(\Rightarrow 2c - d + 65 = 0\) | A1cso (3 marks) | AG terms all on one side in any order (check no errors in working) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(3) = 3^3 + 3^2c + 3d - 12\) | M1 | \(p(3)\) attempted or long division by \(x-3\) as far as remainder |
| \(9c + 3d + 15 = 0\) | A1 (2 marks) | any correct equation with terms collected eg \(3c + d = -5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left.\begin{array}{l}2c - d + 65 = 0\\ 3c + d + 5 = 0\end{array}\right\} \Rightarrow 5c = -70\) | M1 | Elimination of \(c\) or \(d\) |
| \(\Rightarrow c = -14\), \(d = 37\) OE | A1, A1 (3 marks) | value of \(c\) or \(d\) correct unsimplified; both \(c\) and \(d\) correct unsimplified |
# Question 5:
## Part 5(a):
$p(-2) = (-2)^3 + (-2)^2c + (-2)d - 12$ | M1 | $p(-2)$ attempted or long division by $x+2$ as far as remainder
'their' $-8 + 4c - 2d - 12 = -150$ | m1 | putting expression for remainder $= -150$
$\Rightarrow 2c - d + 65 = 0$ | A1cso (3 marks) | AG terms all on one side in any order (check no errors in working)
## Part 5(b):
$p(3) = 3^3 + 3^2c + 3d - 12$ | M1 | $p(3)$ attempted or long division by $x-3$ as far as remainder
$9c + 3d + 15 = 0$ | A1 (2 marks) | any correct equation with terms collected eg $3c + d = -5$
## Part 5(c):
$\left.\begin{array}{l}2c - d + 65 = 0\\ 3c + d + 5 = 0\end{array}\right\} \Rightarrow 5c = -70$ | M1 | Elimination of $c$ or $d$
$\Rightarrow c = -14$, $d = 37$ OE | A1, A1 (3 marks) | value of $c$ or $d$ correct unsimplified; both $c$ and $d$ correct unsimplified
---5 The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } + c x ^ { 2 } + d x - 12$, where $c$ and $d$ are constants.
\begin{enumerate}[label=(\alph*)]
\item When $\mathrm { p } ( x )$ is divided by $x + 2$, the remainder is - 150 .
Show that $2 c - d + 65 = 0$.
\item Given that $x - 3$ is a factor of $\mathrm { p } ( x )$, find another equation involving $c$ and $d$.
\item By solving these two equations, find the value of $c$ and the value of $d$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2012 Q5 [8]}}