## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{2}{\left( x-3 \right)}+\frac{2x+5}{{{x}^{2}}+3x+3}$
Let us consider the equation: $\frac{4{{x}^{2}}+5x-9}{{{x}^{3}}-6x-9x}$ $\frac{4{{x}^{2}}+5x-9}{\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)}$ Firstly, set up the partial fraction decomposition with the unknown constant: $\frac{4{{x}^{2}}+5x-9}{\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)}=\frac{A}{\left( x-3 \right)}+\frac{Bx+C}{{{x}^{2}}+3x+3}$ And multiply both sides by: $\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)$: $4{{x}^{2}}+5x+9=A\left( {{x}^{2}}+3x+3 \right)+Bx\left( x-3 \right)+C\left( x-3 \right)$ …… (I) Then for the value of A, put $x=3$ in equation (I): \begin{align} & 4\left( {{3}^{2}} \right)+5\left( 3 \right)-9=A\left( 9+9+3 \right) \\ & 36+15-9=21A \\ & 42=21A \\ & A=2s \end{align} Put $x=0$ in equation (I): \begin{align} & -9=3A-3C \\ & -9=3\left( 2 \right)-3C \\ & -9=6-3C \\ & 3C=15 \end{align} $C=5$ Then, put $x=1$ in equation (I): \begin{align} & 4+5-9=7A-B-2C \\ & 0=14-2B-10 \\ & 2B=4 \\ & B=2 \end{align} Now, put the value of $A=2$ $B=2\text{ and }C=5$ in equation (I). $\frac{4{{x}^{2}}+5x-9}{\left( x-3 \right)\left( {{x}^{2}}+3x+3 \right)}=\frac{2}{\left( x-3 \right)}+\frac{2x+5}{{{x}^{2}}+3x+3}$ Thus, the partial fraction is $\frac{2}{\left( x-3 \right)}+\frac{2x+5}{{{x}^{2}}+3x+3}$