Answer
The partial fraction is, $\frac{3}{\left( x+2 \right)}+\frac{-2}{\left( {{x}^{2}}+4 \right)}$.
Work Step by Step
$\begin{align}
& \frac{3{{x}^{2}}-2x+8}{{{x}^{3}}+2{{x}^{2}}+4x+8}=\frac{3{{x}^{2}}-2x+8}{{{x}^{2}}\left( x+2 \right)+4\left( x+2 \right)} \\
& =\frac{3{{x}^{2}}-2x+8}{\left( x+2 \right)\left( {{x}^{2}}+4 \right)} \\
& =\frac{A}{\left( x+2 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+4 \right)}
\end{align}$
Now, take the L.C.M. on the right side:
$\frac{3{{x}^{2}}-2x+8}{{{x}^{3}}+2{{x}^{2}}+4x+8}=\frac{A\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)\left( x+2 \right)}{\left( x+2 \right)\left( {{x}^{2}}+4 \right)}$.
By eliminating the denominator from both sides, we get
$\begin{align}
& 3{{x}^{2}}-2x+8=A\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)\left( x+2 \right) \\
& =A{{x}^{2}}+4A+B{{x}^{2}}+2Bx+Cx+2C \\
& =\left( A+B \right){{x}^{2}}+\left( 2B+C \right)x+4A+2C
\end{align}$
Then, compare the coefficients of ${{x}^{2}},x$ and the constant term:
$A+B=3$ …… (I)
$2B+C=-2$ …… (II)
$4A+2C=8$
$2A+C=4$ …… (III)
And find the value of A from equation (I):
$B=3-A$ …… (IV)
Value of B is put in the equation (II):
$\begin{align}
& 2\left( 3-A \right)+C=-2 \\
& 6-2A+C=-2
\end{align}$
$-2A+C=-8$ …… (V)
And add equation (V) with equation (III):
$\begin{align}
& 2A+C=4 \\
& -2A+C=-8 \\
& 2C=-4 \\
& C=-2
\end{align}$
And put the value of C in equation (II):
$\begin{align}
& 2B-2=-2 \\
& B=0
\end{align}$
Put the value of B in equation (I):
$A=3$
Then,
$\frac{3{{x}^{2}}-2x+8}{{{x}^{3}}+2{{x}^{2}}+4x+8}=\frac{3}{\left( x+2 \right)}+\frac{-2}{\left( {{x}^{2}}+4 \right)}$.
Thus, the partial fraction of the given expression is $\frac{3}{\left( x+2 \right)}+\frac{-2}{\left( {{x}^{2}}+4 \right)}$.
