# How do you write the partial fraction decomposition of the rational expression  x/(x^3-x^2-6x)?

Jul 29, 2016

$\frac{x}{{x}^{3} - {x}^{2} - 6 x} = \frac{1}{5 \left(x - 3\right)} - \frac{1}{5 \left(x + 2\right)}$

with exclusion $x \ne 0$

#### Explanation:

Both the numerator and the denominator are divisible by $x$, so we can simplify (noting $x = 0$ as an excluded value):

$\frac{x}{{x}^{3} - {x}^{2} - 6 x}$

$= \frac{1}{{x}^{2} - x - 6}$

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

$= \frac{A}{x - 3} + \frac{B}{x + 2}$

with exclusion $x \ne 0$

Use Heaviside's cover-up method to find:

$A = \frac{1}{\left(3\right) + 2} = \frac{1}{5}$

$B = \frac{1}{\left(- 2\right) - 3} = - \frac{1}{5}$

So:

$\frac{x}{{x}^{3} - {x}^{2} - 6 x} = \frac{1}{5 \left(x - 3\right)} - \frac{1}{5 \left(x + 2\right)}$

with exclusion $x \ne 0$