| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Partial Fraction Form via Division |
| Difficulty | Moderate -0.8 Part (a) is direct application of the remainder theorem (substitute x=1/2), and part (b) requires polynomial long division or comparing coefficients - both are standard C4 techniques with straightforward execution. This is routine practice requiring no problem-solving insight, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right) - 2 = -1\) | M1A1 | Allow M1 for \(f\left(-\frac{1}{2}\right)\); Long division: M1 for their complete attempt |
| (b) \(\frac{x^2(2x-1) + 2x - 2}{2x - 1}\) | M1M1 | Reasonable start, complete method |
| \(x^2 + \frac{2x - 1 - 1}{2x - 1} = x^2 + 1 - \frac{1}{2x-1}\) | A1Bf | ft (a) and (b)(i) |
| Total: 6 marks | A1Bf | \(a = 1\) CAO, \(b = -1\) ft part (a) |
**(a)** $f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right) - 2 = -1$ | M1A1 | Allow M1 for $f\left(-\frac{1}{2}\right)$; Long division: M1 for their complete attempt
**(b)** $\frac{x^2(2x-1) + 2x - 2}{2x - 1}$ | M1M1 | Reasonable start, complete method
$x^2 + \frac{2x - 1 - 1}{2x - 1} = x^2 + 1 - \frac{1}{2x-1}$ | A1Bf| ft (a) and (b)(i)
**Total: 6 marks** | A1Bf | $a = 1$ CAO, $b = -1$ ft part (a)
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\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $2 x ^ { 3 } - x ^ { 2 } + 2 x - 2$ is divided by $2 x - 1$.
\item Given that $\frac { 2 x ^ { 3 } - x ^ { 2 } + 2 x - 2 } { 2 x - 1 } = x ^ { 2 } + a + \frac { b } { 2 x - 1 }$, find the values of $a$ and $b$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2005 Q3 [6]}}