# How do you write the partial fraction decomposition of the rational expression (9x^2+9x+40)/(x(x^2+5))?

Sep 10, 2016

$\frac{9 {x}^{2} + 9 x + 40}{x \left({x}^{2} + 5\right)} = \frac{8}{x} + \frac{9 x + 1}{{x}^{2} + 5}$

#### Explanation:

$\frac{9 {x}^{2} + 9 x + 40}{x \left({x}^{2} + 5\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 5}$

$\textcolor{w h i t e}{\frac{9 {x}^{2} + 9 x + 40}{x \left({x}^{2} + 5\right)}} = \frac{A \left(x + 5\right) + \left(B x + C\right) x}{x \left({x}^{2} + 5\right)}$

$\textcolor{w h i t e}{\frac{9 {x}^{2} + 9 x + 40}{x \left({x}^{2} + 5\right)}} = \frac{B {x}^{2} + \left(A + C\right) x + 5 A}{x \left({x}^{2} + 5\right)}$

Equating coefficients, we find:

$\left\{\begin{matrix}B = 9 \\ A + C = 9 \\ 5 A = 40\end{matrix}\right.$

Hence:

$\left\{\begin{matrix}A = 8 \\ B = 9 \\ C = 1\end{matrix}\right.$

So:

$\frac{9 {x}^{2} + 9 x + 40}{x \left({x}^{2} + 5\right)} = \frac{8}{x} + \frac{9 x + 1}{{x}^{2} + 5}$