| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Simplify algebraic fractions by addition or subtraction |
| Difficulty | Moderate -0.8 This is a straightforward algebraic manipulation question testing partial fractions in reverse and factorization of cubics/quadratics. Part (a) requires finding a common denominator and simplifying (routine), while part (b) involves recognizing difference of cubes and factoring a quadratic—both standard techniques with no problem-solving insight required. Easier than average for C3. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{4x}{(x+3)(x-3)} - \frac{2}{x+3}\) | M1 | |
| \(= \frac{4x - 2(x-3)}{(x+3)(x-3)}\) | M1 | |
| \(= \frac{2x + 6}{(x+3)(x-3)} = \frac{2(x+3)}{(x+3)(x-3)}\) | M1 | |
| \(= \frac{2}{x-3}\) | A1 | |
| (b) \(2^3 - 8 = 0\) ∴ \((x-2)\) is a factor of \((x^3-8)\) | B1 | |
| \(x - 2 \mid \begin{array}{c | cc} & x^2 + 2x + 4 \\ \hline x^3 + 0x^2 + 0x - 8 \\ & x^3 - 2x^2 \\ \hline & 2x^2 + 0x \\ & 2x^2 - 4x \\ \hline & 4x - 8 \\ & 4x - 8 \end{array}\) | M1 A1 |
| \(\therefore x^3 - 8 = (x-2)(x^2 + 2x + 4)\) | M1 A1 | |
| \(\therefore \frac{x^3 - 8}{3x^2 - 8x + 4} = \frac{(x-2)(x^2+2x+4)}{(3x-2)(x-2)} = \frac{x^2+2x+4}{3x-2}\) | (9) |
(a) $\frac{4x}{(x+3)(x-3)} - \frac{2}{x+3}$ | M1 |
$= \frac{4x - 2(x-3)}{(x+3)(x-3)}$ | M1 |
$= \frac{2x + 6}{(x+3)(x-3)} = \frac{2(x+3)}{(x+3)(x-3)}$ | M1 |
$= \frac{2}{x-3}$ | A1 |
(b) $2^3 - 8 = 0$ ∴ $(x-2)$ is a factor of $(x^3-8)$ | B1 |
$x - 2 \mid \begin{array}{c|cc} & x^2 + 2x + 4 \\ \hline x^3 + 0x^2 + 0x - 8 \\ & x^3 - 2x^2 \\ \hline & 2x^2 + 0x \\ & 2x^2 - 4x \\ \hline & 4x - 8 \\ & 4x - 8 \end{array}$ | M1 A1 |
$\therefore x^3 - 8 = (x-2)(x^2 + 2x + 4)$ | M1 A1 |
$\therefore \frac{x^3 - 8}{3x^2 - 8x + 4} = \frac{(x-2)(x^2+2x+4)}{(3x-2)(x-2)} = \frac{x^2+2x+4}{3x-2}$ | **(9)**\begin{enumerate}[label=(\alph*)]
\item Express
$$\frac{4x}{x^2 - 9} - \frac{2}{x + 3}$$
as a single fraction in its simplest form. [4]
\item Simplify
$$\frac{x^3 - 8}{3x^2 - 8x + 4}.$$ [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q2 [9]}}