| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Simplify algebraic fractions by addition or subtraction |
| Difficulty | Moderate -0.5 This question requires factoring the second denominator, finding a common denominator, and simplifying—standard algebraic manipulation for C3 level. While it involves multiple steps, the techniques are routine and no novel insight is required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 - x - 2 = (x-2)(x+1)\) | B1 | At any stage |
| \(\frac{2x^2+3x}{(2x+3)(x-2)} = \frac{x(2x+3)}{(2x+3)(x-2)} = \frac{x}{x-2}\) | B1 | |
| \(\frac{2x^2+3x}{(2x+3)(x-2)} - \frac{6}{x^2-x-2} = \frac{x(x+1)-6}{(x-2)(x+1)}\) | M1 | |
| \(= \frac{x^2+x-6}{(x-2)(x+1)}\) | A1 | |
| \(= \frac{(x+3)(x-2)}{(x-2)(x+1)}\) | M1 A1 | |
| \(= \frac{x+3}{x+1}\) | A1 | (7) [7] |
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 - x - 2 = (x-2)(x+1)$ | B1 | At any stage |
| $\frac{2x^2+3x}{(2x+3)(x-2)} = \frac{x(2x+3)}{(2x+3)(x-2)} = \frac{x}{x-2}$ | B1 | |
| $\frac{2x^2+3x}{(2x+3)(x-2)} - \frac{6}{x^2-x-2} = \frac{x(x+1)-6}{(x-2)(x+1)}$ | M1 | |
| $= \frac{x^2+x-6}{(x-2)(x+1)}$ | A1 | |
| $= \frac{(x+3)(x-2)}{(x-2)(x+1)}$ | M1 A1 | |
| $= \frac{x+3}{x+1}$ | A1 | **(7) [7]** |
---\begin{enumerate}
\item Express
\end{enumerate}
$$\frac { 2 x ^ { 2 } + 3 x } { ( 2 x + 3 ) ( x - 2 ) } - \frac { 6 } { x ^ { 2 } - x - 2 }$$
as a single fraction in its simplest form.\\
\hfill \mbox{\textit{Edexcel C3 2006 Q2 [7]}}