| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor and Remainder Theorems with routine algebraic manipulation. Part (a) involves simple substitution and polynomial division to find constants, while part (b) applies the Remainder Theorem directly. All steps are standard textbook exercises requiring no problem-solving insight, making it easier than average for A-level. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(1) = 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(-2) = -24 + 8 + 14 + 2 = 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(x-1)(x+2)}{3x^3+2x^2-7x+2} = \frac{(x-1)(x+2)}{(x-1)(x+2)(ax+b)}\) | B1 | Recognising \((x-1),(x+2)\) as factors; PI |
| \(ax^3 = 3x^3 \Rightarrow a=3\) | B1 | \(a\) |
| \(-2b = 2 \Rightarrow b=-1\) | B1 | \(b\); Or by division M1 attempt started, M1 complete division, A1 correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(\frac{1}{3}\) | B1 | |
| \(3\left(\frac{1}{3}\right)^3 + 2\left(\frac{1}{3}\right)^2 - 7 \times \frac{1}{3} + d = 2\) | M1 | Remainder Th\(^M\) with \(\pm\frac{1}{3}, \pm 3\) |
| \(d = 4\) | A1F | Ft on \(-\frac{1}{3}\) (answer \(-\frac{4}{9}\)); Or by division M1 M1 A1 as above |
## Question 1:
### Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(1) = 0$ | B1 | |
### Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(-2) = -24 + 8 + 14 + 2 = 0$ | B1 | |
### Part (a)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(x-1)(x+2)}{3x^3+2x^2-7x+2} = \frac{(x-1)(x+2)}{(x-1)(x+2)(ax+b)}$ | B1 | Recognising $(x-1),(x+2)$ as factors; PI |
| $ax^3 = 3x^3 \Rightarrow a=3$ | B1 | $a$ |
| $-2b = 2 \Rightarrow b=-1$ | B1 | $b$; Or by division M1 attempt started, M1 complete division, A1 correct answers |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $\frac{1}{3}$ | B1 | |
| $3\left(\frac{1}{3}\right)^3 + 2\left(\frac{1}{3}\right)^2 - 7 \times \frac{1}{3} + d = 2$ | M1 | Remainder Th$^M$ with $\pm\frac{1}{3}, \pm 3$ |
| $d = 4$ | A1F | Ft on $-\frac{1}{3}$ (answer $-\frac{4}{9}$); Or by division M1 M1 A1 as above |
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\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x + 2$.
\begin{enumerate}[label=(\roman*)]
\item Find f(1).
\item Show that $\mathrm { f } ( - 2 ) = 0$.
\item Hence, or otherwise, show that
$$\frac { ( x - 1 ) ( x + 2 ) } { 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x + 2 } = \frac { 1 } { a x + b }$$
where $a$ and $b$ are integers.
\end{enumerate}\item The polynomial $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x + d$.
When $\mathrm { g } ( x )$ is divided by $( 3 x - 1 )$, the remainder is 2 . Find the value of $d$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2006 Q1 [8]}}