# How do you write the partial fraction decomposition of the rational expression  1/(x^3-6x^2+9x) ?

Dec 14, 2015

1/(9x)-1/(9(x-3))+1/(3(x-3)^2

#### Explanation:

Factor the denominator.

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

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

$1 = A {x}^{2} - 6 A x + 9 A + B {x}^{2} - 3 B x + C x$

$1 = {x}^{2} \left(A + B\right) + x \left(- 6 A - 3 B + C\right) + 1 \left(9 A\right)$

Thus, $\left\{\begin{matrix}A + B = 0 \\ - 6 A - 3 B + C = 0 \\ 9 A = 1\end{matrix}\right.$

Solve to see that $\left\{\begin{matrix}A = \frac{1}{9} \\ B = - \frac{1}{9} \\ C = \frac{1}{3}\end{matrix}\right.$

Therefore,

1/(x^3-6x^2+9x)=1/(9x)-1/(9(x-3))+1/(3(x-3)^2