## Precalculus (6th Edition) Blitzer

Equate the expression with $\frac{Ax+B}{a{{x}^{2}}+bx+c}+\frac{Cx+D}{{{\left( a{{x}^{2}}+bx+c \right)}^{2}}}$ and find the values of A and B by comparing the terms
In order to find partial fraction decomposition of a rational function with a repeated, prime quadratic factor in the denominator, equate the expression with $\frac{Ax+B}{a{{x}^{2}}+bx+c}+\frac{Cx+D}{{{\left( a{{x}^{2}}+bx+c \right)}^{2}}}$ and find the values of A and B by comparing the terms. Let us take an example Consider $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$ And the partial fraction decomposition of the rational function is: $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}+5}+\frac{Cx+D}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$ And multiply both sides by ${{\left( {{x}^{2}}+5 \right)}^{2}}$: \begin{align} & 4x=\left( Ax+B \right)\left( {{x}^{2}}+5 \right)+\left( Cx+D \right) \\ & 4x=A{{x}^{3}}+5Ax+B{{x}^{2}}+5B+Cx+D \\ & 4x=A{{x}^{3}}+B{{x}^{2}}+\left( 5A+C \right)x+D \end{align} And equating the equation: \begin{align} & A=0 \\ & B=0 \\ & 5A+C=4 \\ & D=0 \end{align} After that solving the above equation, $A=0,B=0,C=4,D=0$ Put these values in $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}+5}+\frac{Cx+D}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$. Thus, $\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}=\frac{4x}{{{\left( {{x}^{2}}+5 \right)}^{2}}}$.