Answer
$\dfrac{3x^2+17x-38}{(x-3)(x-2)(x+2)}=\dfrac{8}{x-3}-\dfrac{2}{x-2}-\dfrac{3}{x+2}$
Work Step by Step
We are given the fraction:
$\dfrac{3x^2+17x-38}{(x-3)(x-2)(x+2)}$
Write the partial fraction decomposition:
$\dfrac{3x^2+17x-38}{(x-3)(x-2)(x+2)}=\dfrac{A}{x-3}+\dfrac{B}{x-2}+\dfrac{C}{x+2}$
Multiply all terms by the least common denominator $(x-3)(x-2)(x+2)$:
$(x-3)(x-2)(x+2)\cdot\dfrac{3x^2+17x-38}{(x-3)(x-2)(x+2)}=(x-3)(x-2)(x+2)\cdot\dfrac{A}{x-3}+(x-3)(x-2)(x+2)\cdot\dfrac{B}{x-2}+(x-3)(x-2)(x+2)\cdot\dfrac{C}{x+2}$
$3x^2+17x-38=A(x-2)(x+2)+B(x-3)(x+2)+C(x-3)(x-2)$
$3x^2+17x-38=A(x^2-4)+B(x^2-x-6)+C(x^2-5x+6)$
$3x^2+17x-38=Ax^2-4A+Bx^2-Bx-6B+Cx^2-5Cx+6C$
$3x^2+17x-38=(A+B+C)x^2+(-B-5C)x+(-4A-6B+6C)$
Identify the coefficients:
$\begin{cases}
A+B+C=3\\
-B-5C=17\\
-4A-6B+6C=-38
\end{cases}$
Solve the system:
$\begin{cases}
4A+4B+4C=4(3)\\
-B-5C=17\\
-4A-6B+6C=-38
\end{cases}$
$\begin{cases}
4A+4B+4C-4A-6B+6C=12-38\\
-B-5C=17
\end{cases}$
$\begin{cases}
-2B+10C=-26\\
-B-5C=17
\end{cases}$
$\begin{cases}
-B+5C=-13\\
-B-5C=17
\end{cases}$
$-B+5C-B-5C=-13+17$
$-2B=4$
$B=-2$
$-B-5C=17$
$-(-2)-5C=17$
$5C=-15$
$C=-3$
$A+B+C=3$
$A-2-3=3$
$A-5=3$
$A=8$
The partial fraction decomposition is:
$\dfrac{3x^2+17x-38}{(x-3)(x-2)(x+2)}=\dfrac{8}{x-3}-\dfrac{2}{x-2}-\dfrac{3}{x+2}$
