## Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{2}{x}+\frac{3}{x+2}-\frac{1}{x-1}$
We know that the partial fraction decomposition can be computed as given below: Factor the denominator by the method of grouping and set up the partial fraction decomposition by writing the constants as given below: $\frac{4{{x}^{2}}-3x-4}{x\left( x+2 \right)\left( x-1 \right)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-1}$ Now, multiply both sides of the equation by $x\left( x+2 \right)\left( x-1 \right)$ as given below: \begin{align} & x\left( x+2 \right)\left( x-1 \right)\cdot \frac{4{{x}^{2}}-3x-4}{x\left( x+2 \right)\left( x-1 \right)}=x\left( x+2 \right)\left( x-1 \right)\left( \frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-1} \right) \\ & x\left( x+2 \right)\left( x-1 \right)\cdot \frac{4{{x}^{2}}-3x-4}{x\left( x+2 \right)\left( x-1 \right)}=\left\{ x\left( x+2 \right)\left( x-1 \right)\left( \frac{A}{x} \right)+x\left( x+2 \right)\left( x-1 \right)\left( \frac{B}{x+2} \right)+x\left( x+2 \right)\left( x-1 \right)\left( \frac{C}{x-1} \right) \right\} \end{align} Divide out the common factors: \begin{align} & x\left( x+2 \right)\left( x-1 \right)\cdot \frac{4{{x}^{2}}-3x-4}{x\left( x+2 \right)\left( x-1 \right)}=\left\{ x\left( x+2 \right)\left( x-1 \right)\left( \frac{A}{x} \right)+x\left( x+2 \right)\left( x-1 \right)\left( \frac{B}{x+2} \right)+x\left( x+2 \right)\left( x-1 \right)\left( \frac{C}{x-1} \right) \right\} \\ & 4{{x}^{2}}-3x-4=\left( x+2 \right)\left( x-1 \right)\left( A \right)+x\left( x-1 \right)\left( B \right)+x\left( x+2 \right)C \\ & 4{{x}^{2}}-3x-4=\left( {{x}^{2}}+2x-x-2 \right)A+\left( {{x}^{2}}-x \right)B+\left( {{x}^{2}}+2x \right)C \\ & 4{{x}^{2}}-3x-4=\left( A+B+C \right){{x}^{2}}+\left( A-B+2C \right)x+\left( -2A \right) \end{align} Now, equate the coefficients of like terms of the equation to write a system of equations as given below: $A+B+C=4$ (I) $A-B+2C=-3$ (II) $-2A=-4$ (III) And divide both sides of equation (III) by $-2$ to compute the value of A: \begin{align} & \frac{-2A}{-2}=\frac{-4}{-2} \\ & A=2 \end{align} Put the value of $A$ in equations (I) and (II), to compute the values of B and C: \begin{align} & A+B+C=4 \\ & 2+B+C=4 \\ & B+C=2 \end{align} (IV) And \begin{align} & A-B+2C=-3 \\ & 2-B+2C=-3 \\ & -B+2C=-5 \end{align} (V) Add equation (I) with equation (II) to get the value of C: \begin{align} & +\text{ }\underline{\begin{align} & B+C=2 \\ & -B+2C=-5 \end{align}} \\ & \text{ }3C=-3 \\ \end{align} Divide both sides by 3 as given below: \begin{align} & \frac{3C}{3}=\frac{-3}{3} \\ & C=-1 \end{align} Therefore, the value of $C$ is $-1$. Now, put the value of $C$ in equation (IV) to get the value of $B$: \begin{align} & B+C=2 \\ & B-1=2 \\ & B=2+1 \\ & B=3 \end{align} Put the values of $A,\,\,B$, and $C$ in the initial equation and write the partial fraction decomposition: \begin{align} & \frac{4{{x}^{2}}-3x-4}{x\left( x+2 \right)\left( x-1 \right)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-1} \\ & =\frac{2}{x}+\frac{3}{x+2}+\frac{-1}{x-1} \\ & =\frac{2}{x}+\frac{3}{x+2}-\frac{1}{x-1} \end{align} Thus, the partial fraction decomposition of the rational expression is $\frac{2}{x}+\frac{3}{x+2}-\frac{1}{x-1}$.