## Precalculus (6th Edition) Blitzer

The partial fraction decomposition contains a term of the form: $\frac{Ax+B}{a{{x}^{2}}+bx+c}$, where $a{{x}^{2}}+bx+c\ne 0$ when the denominator of a rational function is with a prime quadratic factor of the form of $\left( a{{x}^{2}}+bx+c \right)$.
We know the denominator is in the form of a prime quadratic factor. Consider an example, $\frac{3x}{x\left( {{x}^{2}}+5 \right)}$ And the partial fraction decomposition of the rational function is: $\frac{3x}{x\left( {{x}^{2}}+5 \right)}=\frac{A}{x}+\frac{B}{\left( {{x}^{2}}+5 \right)}$ And by multiplying both sides by $x\left( {{x}^{2}}+5 \right)$: \begin{align} & 3x=A\left( {{x}^{2}}+5 \right)+Bx \\ & 3x=A{{x}^{2}}+5A+Bx \end{align} Equating the like terms of the equation: \begin{align} & A=0 \\ & B=3 \end{align} Put these values in $\frac{3x}{x\left( {{x}^{2}}+5 \right)}=\frac{A}{x}+\frac{B}{\left( {{x}^{2}}+5 \right)}$. Thus, $\frac{3x}{x\left( {{x}^{2}}+5 \right)}=\frac{3}{\left( {{x}^{2}}+5 \right)}$.