# How do you write the partial fraction decomposition of the rational expression  3/(x ^(2) + 16x +5)?

Dec 14, 2015

$\frac{3}{{x}^{2} + 16 x + 5}$

#### Explanation:

Notice how the denominator cannot be factored (within the real numbers).

This fraction is already as simple as it can get.

Dec 14, 2015

$\frac{3}{{x}^{2} + 16 x + 5} = \frac{3}{2 \sqrt{59} \left(x + 8 - \sqrt{59}\right)} - \frac{3}{2 \sqrt{59} \left(x + 8 + \sqrt{59}\right)}$

#### Explanation:

${x}^{2} + 16 x + 5 = {x}^{2} + 16 x + 64 - 59$

$= {\left(x + 8\right)}^{2} - {\left(\sqrt{59}\right)}^{2}$

$= \left(x + 8 - \sqrt{59}\right) \left(x + 8 + \sqrt{59}\right)$

So to find the partial fraction decomposition, solve:

$\frac{3}{\left(x + 8 - \sqrt{59}\right) \left(x + 8 + \sqrt{59}\right)}$

$= \frac{A}{x + 8 - \sqrt{59}} + \frac{B}{x + 8 + \sqrt{59}}$

$= \frac{A \left(x + 8 + \sqrt{59}\right) + B \left(x + 8 - \sqrt{59}\right)}{\left(x + 8 - \sqrt{59}\right) \left(x + 8 + \sqrt{59}\right)}$

$= \frac{\left(A + B\right) x + 8 \left(A + B\right) + \left(A - B\right) \sqrt{59}}{\left(x + 8 - \sqrt{59}\right) \left(x + 8 + \sqrt{59}\right)}$

Equating the coefficient of $x$, we find $A + B = 0$. So $B = - A$.

Equating the remaining constant term we find $2 A \sqrt{59} = 3$

Hence $A = \frac{3}{2 \sqrt{59}}$ and:

$\frac{3}{{x}^{2} + 16 x + 5} = \frac{3}{2 \sqrt{59} \left(x + 8 - \sqrt{59}\right)} - \frac{3}{2 \sqrt{59} \left(x + 8 + \sqrt{59}\right)}$