| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Simple Algebraic Fraction Simplification |
| Difficulty | Moderate -0.3 This is a straightforward C4 polynomial question requiring routine techniques: substitution to evaluate f(-1), factor theorem verification by showing f(2/5)=0, then polynomial division to factorise f(x) completely before simplifying the algebraic fraction. All steps are standard textbook procedures with no novel insight 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 |
| Answer | Marks |
|---|---|
| \(f(-1) = -15 + 19 - 4 = 0\) | B1 (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Evaluate \(f\left(\frac{2}{5}\right)\) | M1 | evaluate or complete division leading to a numerical remainder |
| \(\left(15\times\frac{8}{125}+19\times\frac{4}{25}-4\right)=0 \Rightarrow\) factor | A1 (2 marks) | Or decimal equivalent \((0.96+3.04-4)\) or zero remainder \(\Rightarrow\) factor |
| Answer | Marks | Guidance |
|---|---|---|
| \((x+1)\) is a factor | B1 | Stated or implied |
| Third factor is \((3x+2)\) | M1, A1 | Any appropriate method to find third factor |
| \(\frac{15x^2-6x}{f(x)}=\frac{3x(5x-2)}{(x+1)(5x-2)(3x+2)}\) | M1 | \(\left[(5x-2)(3x^2\pm5x\pm2)+\text{attempt to factorise}\right]\); Factorise numerator correctly and attempt to simplify |
| \(=\frac{3x}{(x+1)(3x+2)}\) | A1 (5 marks) | CSO, no ISW |
## Question 1:
### Part (a)(i)
| $f(-1) = -15 + 19 - 4 = 0$ | B1 (1 mark) | |
### Part (a)(ii)
| Evaluate $f\left(\frac{2}{5}\right)$ | M1 | evaluate **or** complete division leading to a numerical remainder |
| $\left(15\times\frac{8}{125}+19\times\frac{4}{25}-4\right)=0 \Rightarrow$ factor | A1 (2 marks) | Or decimal equivalent $(0.96+3.04-4)$ **or** zero remainder $\Rightarrow$ factor |
### Part (b)
| $(x+1)$ is a factor | B1 | Stated or implied |
| Third factor is $(3x+2)$ | M1, A1 | Any appropriate method to find third factor |
| $\frac{15x^2-6x}{f(x)}=\frac{3x(5x-2)}{(x+1)(5x-2)(3x+2)}$ | M1 | $\left[(5x-2)(3x^2\pm5x\pm2)+\text{attempt to factorise}\right]$; Factorise numerator correctly and attempt to simplify |
| $=\frac{3x}{(x+1)(3x+2)}$ | A1 (5 marks) | CSO, no ISW |
**Total: 8 marks**
---1 The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 15 x ^ { 3 } + 19 x ^ { 2 } - 4$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { f } ( - 1 )$.
\item Show that $( 5 x - 2 )$ is a factor of $\mathrm { f } ( x )$.
\end{enumerate}\item Simplify
$$\frac { 15 x ^ { 2 } - 6 x } { f ( x ) }$$
giving your answer in a fully factorised form.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2010 Q1 [8]}}