## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{3}{\left( x+2 \right)}+\frac{-2}{\left( {{x}^{2}}+4 \right)}$.
\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 $\frac{3}{\left( x+2 \right)}+\frac{-2}{\left( {{x}^{2}}+4 \right)}$.