Precalculus (6th Edition) Blitzer

Let us consider, $\frac{7{{x}^{2}}+9x+3}{\left( x+5 \right)\left( {{x}^{2}}-3x+2 \right)}=\frac{A}{\left( x+5 \right)}+\frac{Bx+C}{\left( {{x}^{2}}-3x+2 \right)}$ Where, $P\left( x \right)=7{{x}^{2}}+9x+3$ and $Q\left( x \right)=\left( x+5 \right)\left( {{x}^{2}}-3x+2 \right)$ \begin{align} & \frac{7{{x}^{2}}+9x+3}{\left( x+5 \right)\left( {{x}^{2}}-3x+2 \right)}=\frac{A}{\left( x+5 \right)}+\frac{Bx+C}{\left( {{x}^{2}}-3x+2 \right)} \\ & 7{{x}^{2}}+9x+3=A\left( {{x}^{2}}-3x+2 \right)+Bx\left( x+5 \right)+C\left( x+5 \right) \end{align} Therefore, the partial fraction does not make sense because the degree of $P\left( x \right)$ is less than the degree of $Q\left( x \right)$. Thus, this partial fraction does make sense.