## Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{7{{x}^{2}}-7x+23}{\left( x-3 \right)\left( {{x}^{2}}+4 \right)}=\frac{5}{x-3}+\frac{2x-1}{{{x}^{2}}+4}$.
By putting the constant $A$ over the linear factor $x-3$ and a linear expression $Bx+C$ over the quadratic factor: $\frac{7{{x}^{2}}-7x+23}{\left( x-3 \right)\left( {{x}^{2}}+4 \right)}=\frac{A}{x-3}+\frac{Bx+C}{{{x}^{2}}+4}$ And multiply both sides by $\left( x-3 \right)\left( {{x}^{2}}+4 \right)$ as shown below: \begin{align} & 7{{x}^{2}}-7x+23=A\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)\left( x-3 \right) \\ & 7{{x}^{2}}-7x+23=A{{x}^{2}}+4A+B{{x}^{2}}-3Bx+Cx-3C \\ & 7{{x}^{2}}-7x+23=\left( A+B \right){{x}^{2}}+\left( C-3B \right)x+\left( 4A-3C \right) \end{align} By equating the terms of equal powers, we get: $A+B=7$ (I) $C-3B=-7$ (II) $4A-3C=23$ (III) And from equation (II), $B=\frac{1}{3}\left( C+7 \right)$ (IV) Put the value of B from equation (IV) in equation (I): \begin{align} & A+\frac{1}{3}\left( C+7 \right)=7 \\ & 3A+C+7=21 \\ & 3A+C=14 \end{align} Multiplying both sides by 3 as given below: $9A+3C=42$ (V) Add equations (III) and (V): \begin{align} & 13A=65 \\ & A=5 \end{align} From equation (I), compute the value of B: \begin{align} & B=7-5 \\ & =2 \end{align} From equation (II), compute the value of C: \begin{align} & C=3B-7 \\ & =6-7 \\ & =-1 \end{align} So, the values are $A=5,B=2,C=-1$. Put the values of $A,\,\,B,\text{ and }C$ in the given equation, and define the partial fraction decomposition: $\frac{A}{x-3}+\frac{Bx+C}{{{x}^{2}}+4}=\frac{5}{x-3}+\frac{2x-1}{{{x}^{2}}+4}$ Thus, the partial fraction decomposition of the rational expression is $\frac{7{{x}^{2}}-7x+23}{\left( x-3 \right)\left( {{x}^{2}}+4 \right)}=\frac{5}{x-3}+\frac{2x-1}{{{x}^{2}}+4}$.