Answer
$\dfrac{2x-6}{(x-1)(x-2)^2}=-\dfrac{4}{x-1}+\dfrac{4}{(x-2)}-\dfrac{2}{(x-2)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{2x-6}{(x-1)(x-2)^2}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{2x-6}{(x-1)(x-2)^2}=\dfrac{A}{x-1}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-2)^2}$
Multiply all terms by the least common denominator $(x-1)(x-2)^2$:
$(x-1)(x-2)^2\cdot\dfrac{2x-6}{(x-1)(x-2)^2}=(x-1)(x-2)^2\cdot\dfrac{A}{x-1}+(x-1)(x-2)^2\cdot\dfrac{B}{(x-2)}+(x-1)(x-2)^2\cdot\dfrac{C}{(x-2)^2}$
$2x-6=A(x-2)^2+B(x-1)(x-2)+C(x-1)$
$2x-6=A(x^2-4x+4)+B(x^2-3x+2)+C(x-1)$
$2x-6=Ax^2-4Ax+4A+Bx^2-3Bx+2B+Cx-C$
$2x-6=(A+B)x^2+(-4A-3B+C)x+(4A+2B-C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=0\\
-4A-3B+C=2\\
4A+2B-C=-6
\end{cases}$
Solve the system:
$\begin{cases}
A+B=0\\
-4A-3B+C+4A+2B-C=2-6
\end{cases}$
$\begin{cases}
A+B=0\\
-B=-4
\end{cases}$
$B=4$
$A+B=0$
$A+4=0$
$A=-4$
$4A+2B-C=-6$
$4(-4)+2(4)-C=-6$
$-8-C=-6$
$C=-2$
The partial fraction decomposition is:
$\dfrac{2x-6}{(x-1)(x-2)^2}=-\dfrac{4}{x-1}+\dfrac{4}{(x-2)}-\dfrac{2}{(x-2)^2}$
