| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| 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 with routine algebraic manipulation. Part (a) uses p(-3)=0, part (b) uses p(2)=65, and part (c) solves two simultaneous linear equations. All steps are standard C1 techniques with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p(-3) = 0\): \(-27 + 9c - 3d + 3 = 0\) | M1 | Substituting \(x=-3\) |
| \(9c - 3d = 24 \Rightarrow 3c - d = 8\) | A1 | Must show working clearly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p(2) = 65\): \(8 + 4c + 2d + 3 = 65\) | M1 | Substituting \(x=2\) into \(p(x)\) |
| \(4c + 2d = 54 \Rightarrow 2c + d = 27\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Adding equations: \(5c = 35 \Rightarrow c = 7\) | M1 | Correct method to solve simultaneous equations |
| \(d = 27 - 14 = 13\) | A1 | |
| Both \(c=7\) and \(d=13\) correct | A1 |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $p(-3) = 0$: $-27 + 9c - 3d + 3 = 0$ | M1 | Substituting $x=-3$ |
| $9c - 3d = 24 \Rightarrow 3c - d = 8$ | A1 | Must show working clearly |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $p(2) = 65$: $8 + 4c + 2d + 3 = 65$ | M1 | Substituting $x=2$ into $p(x)$ |
| $4c + 2d = 54 \Rightarrow 2c + d = 27$ | A1 | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Adding equations: $5c = 35 \Rightarrow c = 7$ | M1 | Correct method to solve simultaneous equations |
| $d = 27 - 14 = 13$ | A1 | |
| Both $c=7$ and $d=13$ correct | A1 | |5 The polynomial $\mathrm { p } ( x )$ is given by
$$\mathrm { p } ( x ) = x ^ { 3 } + c x ^ { 2 } + d x + 3$$
where $c$ and $d$ are integers.
\begin{enumerate}[label=(\alph*)]
\item Given that $x + 3$ is a factor of $\mathrm { p } ( x )$, show that
$$3 c - d = 8$$
\item The remainder when $\mathrm { p } ( x )$ is divided by $x - 2$ is 65 .
Obtain a further equation in $c$ and $d$.
\item Use the equations from parts (a) and (b) to find the value of $c$ and the value of $d$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2014 Q5 [7]}}