# How do you write the partial fraction decomposition of the rational expression (-3x^2 +3x - 5) / (3x^3 +x^2 + 6x +2)?

##### 1 Answer
Aug 23, 2017

$\frac{- 3 {x}^{2} + 3 x - 5}{3 {x}^{3} + {x}^{2} + 6 x + 2} = - \frac{3}{3 x + 1} + \frac{1}{{x}^{2} + 2}$

#### Explanation:

Given:

$\frac{- 3 {x}^{2} + 3 x - 5}{3 {x}^{3} + {x}^{2} + 6 x + 2}$

Note that the denominator factors as:

$3 {x}^{3} + {x}^{2} + 6 x + 2 = \left(3 x + 1\right) \left({x}^{2} + 2\right)$

The quadratic factor $\left({x}^{2} + 2\right)$ is irreducible over the reals, so we are (probably) looking for a partial fraction decomposition of the form:

$\frac{- 3 {x}^{2} + 3 x - 5}{3 {x}^{3} + {x}^{2} + 6 x + 2} = \frac{A}{3 x + 1} + \frac{B x + C}{{x}^{2} + 2}$

$\textcolor{w h i t e}{\frac{- 3 {x}^{2} + 3 x - 5}{3 {x}^{3} + {x}^{2} + 6 x + 2}} = \frac{A \left({x}^{2} + 2\right) + \left(B x + C\right) \left(3 x + 1\right)}{3 {x}^{3} + {x}^{2} + 6 x + 2}$

$\textcolor{w h i t e}{\frac{- 3 {x}^{2} + 3 x - 5}{3 {x}^{3} + {x}^{2} + 6 x + 2}} = \frac{\left(A + 3 B\right) {x}^{2} + \left(B + 3 C\right) x + \left(2 A + C\right)}{3 {x}^{3} + {x}^{2} + 6 x + 2}$

Equating coefficients, this gives us a system of equations:

$\left\{\begin{matrix}A + 3 B = - 3 \\ B + 3 C = 3 \\ 2 A + C = - 5\end{matrix}\right.$

Subtracting $3$ times the third equation from the second, we get:

$- 6 A + B = 18$

Subtracting $3$ times this equation from the first equation, we get:

$19 A = - 57$

Dividing both sides by $19$, we find:

$A = - 3$

Putting this value of $A$ into the first equation, we get:

$- 3 + 3 B = - 3$

Hence:

$B = 0$

Putting this value of $B$ into the second equation, we get:

$0 + 3 C = 3$

Hance:

$C = 1$

So:

$\frac{- 3 {x}^{2} + 3 x - 5}{3 {x}^{3} + {x}^{2} + 6 x + 2} = - \frac{3}{3 x + 1} + \frac{1}{{x}^{2} + 2}$