Answer
$\dfrac{11x-2}{(x-4)(x+3)}=\dfrac{6}{x-4}+\dfrac{5}{x+3}$
Work Step by Step
We are given the fraction:
$\dfrac{11x-2}{x^2-x-12}$
First we factor the denominator:
$\dfrac{11x-2}{x^2-x-12}=\dfrac{11x-2}{x^2-4x+3x-12}=\dfrac{11x-2}{x(x-4)+3(x-4)}=\dfrac{11x-2}{(x-4)(x+3)}$
We can write the partial fraction decomposition:
$\dfrac{11x-2}{(x-4)(x+3)}=\dfrac{A}{x-4}+\dfrac{B}{x+3}$
Multiply each term by the least common denominator $(x-4)(x+3)$:
$(x-4)(x+3)\cdot\dfrac{11x-2}{(x-4)(x+3)}=(x-4)(x+3)\cdot\dfrac{A}{x-4}+(x-4)(x+3)\cdot\dfrac{B}{x+3}$
$11x-2=A(x+3)+B(x-4)$
$11x-2=Ax+3A+Bx-4B$
$11x-2=(A+B)x+(3A-4B)$
Identify the coefficients:
$\begin{cases}
A+B=11\\
3A-4B=-2
\end{cases}$
Solve the system:
$\begin{cases}
4A+4B=44\\
3A-4B=-2
\end{cases}$
$4A+4B+3A-4B=44-2$
$7A=42$
$A=6$
$A+B=11$
$6+B=11$
$B=5$
The partial fraction decomposition is:
$\dfrac{11x-2}{(x-4)(x+3)}=\dfrac{6}{x-4}+\dfrac{5}{x+3}$
