Answer
The partial fraction decomposition of the rational expression is $\frac{2}{x}+\frac{3}{x+2}-\frac{1}{x-1}$
Work Step by Step
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 $\frac{2}{x}+\frac{3}{x+2}-\frac{1}{x-1}$.
