## Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{3}{5\left( x-3 \right)}+\frac{2}{5\left( x+2 \right)}$
Factor the denominator by the method of grouping as given below: $\frac{x}{\left( x-3 \right)\left( x+2 \right)}=\frac{A}{x-3}+\frac{B}{x+2}$ Multiply both sides of the equation by $\left( x-3 \right)\left( x+2 \right)$ as given below: \begin{align} & \left( x-3 \right)\left( x+2 \right)\cdot \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\left( x-3 \right)\left( x+2 \right)\left( \frac{A}{x-3}+\frac{B}{x+2} \right) \\ & \left( x-3 \right)\left( x+2 \right)\cdot \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\left( x-3 \right)\left( x+2 \right)\left( \frac{A}{x-3} \right)+\left( x-3 \right)\left( x+2 \right)\left( \frac{B}{x+2} \right) \end{align} Divide the common factors: \begin{align} & \left( x-3 \right)\left( x+2 \right)\cdot \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\left( x-3 \right)\left( x+2 \right)\left( \frac{A}{x-3} \right)+\left( x-3 \right)\left( x+2 \right)\left( \frac{B}{x+2} \right) \\ & x=\left( x+2 \right)A+\left( x-3 \right)B \\ & x=Ax+2A+Bx-3B \\ & x=\left( A+B \right)x+\left( 2A-3B \right) \end{align} Equate the coefficients of like terms of the equation to write a system of equations as given below: $A+B=1$ (I) $2A-3B=0$ (II) Multiply equation (I) by 3 and add with equation (II) to get the value of A: \begin{align} & +\text{ }\underline{\begin{align} & 3A+3B=3 \\ & 2A-3B=0 \\ \end{align}} \\ & \text{ }5A=3 \\ \end{align} Hence, the value of A is $\frac{3}{5}$. Now, put the value of A in equation (I) so that the value of B can be computed as given below: \begin{align} & A+B=1 \\ & \frac{3}{5}+B=1 \\ & B=1-\frac{3}{5} \\ & B=\frac{2}{5} \end{align} Now, put the value of $A$ and $B$ in the initial equation and write the partial fraction decomposition as given below: \begin{align} & \frac{x}{\left( x-3 \right)\left( x+2 \right)}=\frac{A}{x-3}+\frac{B}{x+2} \\ & =\frac{\frac{3}{5}}{x-3}+\frac{\frac{2}{5}}{x+2} \\ & =\frac{3}{5\left( x-3 \right)}+\frac{2}{5\left( x+2 \right)} \end{align} Thus, the partial fraction decomposition of the rational expression is $\frac{3}{5\left( x-3 \right)}+\frac{2}{5\left( x+2 \right)}$.