| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| 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 multi-part question testing standard C4 polynomial techniques: substitution, Factor Theorem application, polynomial division, and Remainder Theorem. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average, though the multiple steps and algebraic manipulation keep it close to typical difficulty. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks |
|---|---|
| \(f(-1) = 0\) | B1 (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) - 7\left(-\frac{1}{2}\right) - 3 = -\frac{1}{2} + \frac{7}{2} - 3 = 0 \Rightarrow\) factor | M1, A1 (2 marks) | Use of \(\pm\frac{1}{2}\); need to see simplification, \(\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}\), '=0' and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Third factor is \((2x-3)\) | B1 | PI |
| \(\frac{(x+1)(2x+1)(2x-3)}{(x+1)(2x+1)}\) simplifies to \(2x-3\) | M1, A1 (3 marks) | \(\frac{\text{3 linear factors}}{\text{2 linear factors}}\); simplified result stated |
| Answer | Marks | Guidance |
|---|---|---|
| Complete division to \(2x+b\); to \(2x-3\); simplifies to \(2x-3\) | (M1), (A1), (A1) (3 marks) | Simplified result stated |
| Answer | Marks |
|---|---|
| \(g\left(-\frac{1}{2}\right) = -\frac{1}{2} + \frac{7}{2} + d = 2\), \(d = -1\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Complete division leading to rem \(= 2\), \(d = -1\) | (M1), (A1) (2 marks) | Remainder \(= d + p = 2\) |
## Question 1:
### Part (a)(i)
| $f(-1) = 0$ | B1 (1 mark) | |
### Part (a)(ii)
| $f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) - 7\left(-\frac{1}{2}\right) - 3 = -\frac{1}{2} + \frac{7}{2} - 3 = 0 \Rightarrow$ factor | M1, A1 (2 marks) | Use of $\pm\frac{1}{2}$; need to see simplification, $\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$, '=0' and conclusion |
### Part (a)(iii)
| Third factor is $(2x-3)$ | B1 | PI |
| $\frac{(x+1)(2x+1)(2x-3)}{(x+1)(2x+1)}$ simplifies to $2x-3$ | M1, A1 (3 marks) | $\frac{\text{3 linear factors}}{\text{2 linear factors}}$; simplified result stated |
**Alternative:**
| Complete division to $2x+b$; to $2x-3$; simplifies to $2x-3$ | (M1), (A1), (A1) (3 marks) | Simplified result stated |
### Part (b)
| $g\left(-\frac{1}{2}\right) = -\frac{1}{2} + \frac{7}{2} + d = 2$, $d = -1$ | M1, A1 | |
**Alternative:**
| Complete division leading to rem $= 2$, $d = -1$ | (M1), (A1) (2 marks) | Remainder $= d + p = 2$ |
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\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 4 x ^ { 3 } - 7 x - 3$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { f } ( - 1 )$.
\item Use the Factor Theorem to show that $2 x + 1$ is a factor of $\mathrm { f } ( x )$.
\item Simplify the algebraic fraction $\frac { 4 x ^ { 3 } - 7 x - 3 } { 2 x ^ { 2 } + 3 x + 1 }$.
\end{enumerate}\item The polynomial $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 4 x ^ { 3 } - 7 x + d$. When $\mathrm { g } ( x )$ is divided by $2 x + 1$, the remainder is 2 . Find the value of $d$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q1 [8]}}