# How do you express 1/((x+7)(x^2+9)) in partial fractions?

Apr 15, 2016

$\frac{1}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{1}{58 \left(x + 7\right)} + \frac{7 - x}{58 \left({x}^{2} + 9\right)}$

#### Explanation:

Working with Real coefficients, the factor ${x}^{2} + 9$ is irreducible over $\mathbb{R}$ since ${x}^{2} + 9 \ge 9$ for all $x \in \mathbb{R}$.

So we want to find a decomposition of the form:

$\frac{1}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{A}{x + 7} + \frac{B x + C}{{x}^{2} + 9}$

$= \frac{A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x + 7\right)}{\left(x + 7\right) \left({x}^{2} + 9\right)}$

$= \frac{\left(A + B\right) {x}^{2} + \left(7 B + C\right) x + \left(9 A + 7 C\right)}{\left(x + 7\right) \left({x}^{2} + 9\right)}$

Equating coefficients, we get the following system of simultaneous linear equations:

$\left\{\begin{matrix}A + B = 0 \\ 7 B + C = 0 \\ 9 A + 7 C = 1\end{matrix}\right.$

Hence:

$\left\{\begin{matrix}A = \frac{1}{58} \\ B = - \frac{1}{58} \\ C = \frac{7}{58}\end{matrix}\right.$

So:

$\frac{1}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{1}{58 \left(x + 7\right)} + \frac{7 - x}{58 \left({x}^{2} + 9\right)}$