# How do you write 2/(x^3-x^2)  as a partial fraction decomposition?

Nov 8, 2016

The result is $= - \frac{2}{x} ^ 2 - \frac{2}{x} + \frac{2}{x - 1}$

#### Explanation:

Let's factorise the denominator, ${x}^{3} - {x}^{2} = {x}^{2} \left(x - 1\right)$
So, $\frac{2}{{x}^{3} - {x}^{2}} = \frac{2}{{x}^{2} \left(x - 1\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x - 1}$
$= \frac{A \left(x - 1\right) + B x \left(x - 1\right) + C {x}^{2}}{{x}^{2} \left(x - 1\right)}$

$\therefore 2 = A \left(x - 1\right) + B x \left(x - 1\right) + C {x}^{2}$
If $x = 0$$\implies$$2 = - A$$\implies$$A = - 2$
Coefficients of $x$, $0 = A - B$$\implies$$B = - 2$
Coefficients of ${x}^{2}$, 0=B+C$\implies$$C = 2$

$\frac{2}{{x}^{3} - {x}^{2}} = \frac{2}{{x}^{2} \left(x - 1\right)} = - \frac{2}{x} ^ 2 - \frac{2}{x} + \frac{2}{x - 1}$