| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Simple Algebraic Fraction Simplification |
| Difficulty | Moderate -0.8 Part (a) is straightforward application of the remainder theorem (substitute x = -1/2) or simple polynomial division for 2 marks. Part (b) requires factorizing both numerator and denominator then canceling common factors—a routine algebraic manipulation tested at C4 level with no conceptual challenges or problem-solving insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 3 = -3\) | M1A1 | 2 marks |
| Alt: algebraic division: | ||
| \(2x+1\enclose{longdiv}{2x^2+x-3}\) | M1 | |
| \(2x^2 + x\) | ||
| \(-3\) | A1 | 2 marks |
| Alt: | ||
| \(\frac{x(2x+1)-3}{2x+1}\) | M1 | |
| remainder = –3 stated or –3 highlighted | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{(2x+3)(x-1)}{(x+1)(x-1)} = \frac{2x+3}{x+1}\) | B1, B1 | |
| CAO in this form. Not \(\frac{2x+3}{x+1} \cancel{x-1}\) | B1 | 3 marks |
| Alternative: | ||
| \(\frac{2x^2-2x-1}{x^2-1} = 2 + \frac{x-1}{x^2-1}\) | M1 | |
| \(= 2 + \frac{x-1}{(x-1)(x+1)}\) | B1 | |
| \(= 2 + \frac{1}{x+1}\) | A1 | 3 marks |
**1(a)**
| $2\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 3 = -3$ | M1A1 | 2 marks | use of $\pm\frac{1}{2}$; SC NMS –3 1/2, No ISW, subsequent answer "3" AO |
| **Alt: algebraic division:** | | | |
| $2x+1\enclose{longdiv}{2x^2+x-3}$ | M1 | | complete division with integer remainder |
| $2x^2 + x$ | | | |
| $-3$ | A1 | 2 marks | remainder = –3 stated or –3 highlighted |
| **Alt:** | | | |
| $\frac{x(2x+1)-3}{2x+1}$ | M1 | | attempt to rearrange numerator with $(2x+1)$ as a factor |
| remainder = –3 stated or –3 highlighted | A1 | 2 marks | |
**1(b)**
| $\frac{(2x+3)(x-1)}{(x+1)(x-1)} = \frac{2x+3}{x+1}$ | B1, B1 | | numerator, denominator – not necessarily in fraction |
| CAO in this form. Not $\frac{2x+3}{x+1} \cancel{x-1}$ | B1 | 3 marks | |
| **Alternative:** | | | |
| $\frac{2x^2-2x-1}{x^2-1} = 2 + \frac{x-1}{x^2-1}$ | M1 | | |
| $= 2 + \frac{x-1}{(x-1)(x+1)}$ | B1 | | |
| $= 2 + \frac{1}{x+1}$ | A1 | 3 marks | |1
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $2 x ^ { 2 } + x - 3$ is divided by $2 x + 1$.\\
(2 marks)
\item Simplify the algebraic fraction $\frac { 2 x ^ { 2 } + x - 3 } { x ^ { 2 } - 1 }$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q1 [5]}}