Answer
The partial fraction decomposition of the rational expression is $\frac{6}{x-4}+\frac{5}{x+3}$
Work Step by Step
Factor the denominator by the method of grouping as given below:
$\begin{align}
& {{x}^{2}}-x-12={{x}^{2}}-4x+3x-12 \\
& =x\left( x-4 \right)+3\left( x-4 \right) \\
& =\left( x-4 \right)\left( x-3 \right)
\end{align}$
Set up the partial fraction decomposition by writing an unknown constant as given below:
$\frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\frac{A}{x-4}+\frac{B}{x+3}$
Multiply both sides of the equation by $\left( x-4 \right)\left( x+3 \right)$ as given below:
$\begin{align}
& \left( x-4 \right)\left( x+3 \right)\cdot \frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\left( x-4 \right)\left( x+3 \right)\left( \frac{A}{x-4}+\frac{B}{x+3} \right) \\
& \left( x-4 \right)\left( x+3 \right)\cdot \frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\left( x-4 \right)\left( x+3 \right)\left( \frac{A}{x-4} \right)+\left( x-4 \right)\left( x+3 \right)\left( \frac{B}{x+3} \right)
\end{align}$
And divide the common factors:
$\begin{align}
& \left( x+3 \right)\left( x-4 \right)\cdot \frac{11x-2}{\left( x+3 \right)\left( x-4 \right)}=\left( x-4 \right)\left( x+3 \right)\left( \frac{A}{x-4} \right)+\left( x-4 \right)\left( x+3 \right)\left( \frac{B}{x+3} \right) \\
& 11x-2=\left( x+3 \right)A+\left( x-4 \right)B \\
& 11x-2=Ax+3A+Bx-4B \\
& 11x-2=\left( A+B \right)x+\left( 3A-4B \right)
\end{align}$
Then, equate the coefficients of like terms of the equation to write a system of equations as given below:
$ A+B=11$ (I)
$3A-4B=-2$ (II)
Multiply equation (I) by 4 and add it with equation (II) to get the value of A:
$\begin{align}
& +\text{ }\underline{\begin{align}
& 4A+4B=44 \\
& 3A-4B=-2 \\
\end{align}} \\
& \text{ 7}A=42 \\
\end{align}$
Divide both sides by 7:
$\begin{align}
& \frac{7A}{7}=\frac{42}{7} \\
& A=6
\end{align}$
Hence, the value of $ A $ is $6$.
Then, substitute the value of $ A $ in equation (I) so that the value of $ B $ can be computed:
$\begin{align}
& A+B=11 \\
& 6+B=11 \\
& B=11-6 \\
& B=5
\end{align}$
Put the values of $ A $ and $ B $ in the given equation and determine the partial fraction decomposition:
$\begin{align}
& \frac{11x-2}{\left( x-4 \right)\left( x+3 \right)}=\frac{A}{x-4}+\frac{B}{x+3} \\
& =\frac{6}{x-4}+\frac{5}{x+3}
\end{align}$
Hence, the partial fraction decomposition of the rational expression is $\frac{6}{x-4}+\frac{5}{x+3}$.
