## Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{4{{x}^{3}}+5{{x}^{2}}+7x-1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{4x+1}{{{x}^{2}}+x+1}+\frac{2x-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$.
Factor the denominator by the method of grouping as the expression ${{x}^{2}}+x+1$ occurs multiple times in the denominator of the rational expression, so, for each power of ${{x}^{2}}+x+1$ assign an undefined constant given below: $\frac{4{{x}^{3}}+5{{x}^{2}}+7x-1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}+x+1}+\frac{Cx+D}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$ Multiply both sides by ${{\left( {{x}^{2}}+x+1 \right)}^{2}}$ as given below: \begin{align} & 4{{x}^{3}}+5{{x}^{2}}+7x-1=\left( Ax+B \right)\left( {{x}^{2}}+x+1 \right)+Cx+D \\ & 4{{x}^{3}}+5{{x}^{2}}+7x-1=A{{x}^{3}}+A{{x}^{2}}+Ax+B{{x}^{2}}+Bx+B+Cx+D \\ & 4{{x}^{3}}+5{{x}^{2}}+7x-1=A{{x}^{3}}+\left( A+B \right){{x}^{2}}+\left( A+B+C \right)x+\left( B+D \right) \\ \end{align} Then, equate the coefficients of like terms of the equation to write a system of equations as given below: $A=4$ (I) $A+B=5$ (II) $A+B+C=7$ (III) $B+D=-1$ (IV) And from equation (II), obtain the value of B: \begin{align} & B=5-4 \\ & =1 \end{align} From the equation (III), obtain the value of C: \begin{align} & C=7-5 \\ & =2 \end{align} From the equation (IV), obtain the value of D: \begin{align} & D=-1-1 \\ & =-2 \end{align} So, $A=4,B=1,C=2,D=-2$. Put the values of $A,\,\,B,\,\,C,\,\,\text{and }D$ in the given equation, and define the partial fraction decomposition: $\frac{Ax+B}{{{x}^{2}}+x+1}+\frac{Cx+D}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{4x+1}{{{x}^{2}}+x+1}+\frac{2x-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$ Thus, the partial fraction decomposition of the rational expression is $\frac{4{{x}^{3}}+5{{x}^{2}}+7x-1}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}=\frac{4x+1}{{{x}^{2}}+x+1}+\frac{2x-2}{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}$.