# How do you write the partial fraction decomposition of the rational expression [((3x^2)+3x+12) / ((x-5)(x^2+9))]?

Jan 27, 2017

The answer is $= \frac{3}{x - 5} + \frac{3}{{x}^{2} + 9}$

#### Explanation:

Let's work out the decomposition into partial fractions

$\frac{3 {x}^{2} + 3 x + 12}{\left(x - 5\right) \left({x}^{2} + 9\right)} = \frac{A}{x - 5} + \frac{B x + C}{{x}^{2} + 9}$

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

Since the denominators are the same, we can compare the numerators

$3 {x}^{2} + 3 x + 12 = A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 5\right)$

Let $x = 5$, $\implies$, $75 + 15 + 12 = \left(25 + 9\right) A$

$\implies$, $34 A = 102$, $\implies$, $A = 3$

Coefficients of ${x}^{2}$, $3 = A + B$

$\implies$, $B = 3 - A = 0$

Coefficients of $x$, $\implies$, $3 = C$

So,

$\frac{3 {x}^{2} + 3 x + 12}{\left(x - 5\right) \left({x}^{2} + 9\right)} = \frac{3}{x - 5} + \frac{3}{{x}^{2} + 9}$