# How do you express  x/(x^3-x^2-6x) in partial fractions?

Feb 5, 2016

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

#### Explanation:

begin by factorising the denominator

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

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

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

multiply through the equation by (x-3)(x+2)

hence 1 = A(x+2) + B(x-3) ............(*)

[note that x=-2 and x=3 will make the terms with A and B equal to zero.]

let x = -2 in (*): 1 = -5B $\Rightarrow B = - \frac{1}{5}$

let x = 3 in(*) : 1 = 5A $\Rightarrow A = \frac{1}{5}$

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