# How do you express (6x^2+8x+30)/(x^3-27) in partial fractions?

Oct 2, 2016

$\frac{6 {x}^{2} + 8 x + 30}{{x}^{3} - 27} = \frac{6}{x - 3} - \frac{10}{{x}^{2} + 3 x + 9}$

#### Explanation:

Note that:

${x}^{3} - 27 = \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$

Sticking with Real coefficients, we are looking for a partial fraction decomposition of the form:

$\frac{6 {x}^{2} + 8 x + 30}{{x}^{3} - 27} = \frac{A}{x - 3} + \frac{B x + C}{{x}^{2} + 3 x + 9}$

$\textcolor{w h i t e}{\frac{6 {x}^{2} + 8 x + 30}{{x}^{3} - 27}} = \frac{A \left({x}^{2} + 3 x + 9\right) + \left(B x + C\right) \left(x - 3\right)}{{x}^{3} - 27}$

$\textcolor{w h i t e}{\frac{6 {x}^{2} + 8 x + 30}{{x}^{3} - 27}} = \frac{\left(A + B\right) {x}^{2} + \left(3 A - 3 B + C\right) x - 3 C}{{x}^{3} - 27}$

So equating coefficients, we get the following system of linear equations:

$\left\{\begin{matrix}A + B = 6 \\ 3 A - 3 B + C = 8 \\ - 3 C = 30\end{matrix}\right.$

From the third equation we find:

$C = - 10$

Substituting this value of $C$ into the second equation, we get:

$3 A - 3 B - 10 = 8$

Hence:

$3 A - 3 B = 18$

So:

$A - B = 6$

Combining this with the first equation, we find:

$A = 6$

$B = 0$

So:

$\frac{6 {x}^{2} + 8 x + 30}{{x}^{3} - 27} = \frac{6}{x - 3} - \frac{10}{{x}^{2} + 3 x + 9}$