Answer
The partial fraction decomposition of the rational expression is
$-\frac{4}{\left( x-1 \right)}+\frac{4}{x-2}-\frac{2}{{{\left( x-2 \right)}^{2}}}$
Work Step by Step
We know that the linear factor $ x-2$ occurs multiple times in the denominator of the rational expression; therefore, for each power of $ x-2$, assign an undefined constant factor in the denominator by the method of grouping as shown below:
$\frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\frac{A}{\left( x-1 \right)}+\frac{B}{x-2}+\frac{C}{{{\left( x-2 \right)}^{2}}}$
Multiply both sides of the equation by $\left( x-1 \right){{\left( x-2 \right)}^{2}}$:
$\begin{align}
& \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\left( x-1 \right){{\left( x-2 \right)}^{2}}\left( \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{{{\left( x-2 \right)}^{2}}} \right) \\
& \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\left\{ \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{A}{x-1}+\left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{B}{x-2}+\left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{C}{{{\left( x-2 \right)}^{2}}} \right\}
\end{align}$
And divide the common factors:
$\begin{align}
& \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\left\{ \left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{A}{x-1}+\left( x-1 \right)\left( x-2 \right)\left( x-2 \right)\cdot \frac{B}{x-2}+\left( x-1 \right){{\left( x-2 \right)}^{2}}\cdot \frac{C}{{{\left( x-2 \right)}^{2}}} \right\} \\
& 2x-6={{\left( x-2 \right)}^{2}}\cdot A+\left( x-1 \right)\left( x-2 \right)\cdot B+\left( x-1 \right)C \\
& 2x-6=\left( {{x}^{2}}+4-4x \right)A+\left( {{x}^{2}}-2x-x+2 \right)B+\left( x-1 \right)C \\
& 2x-6=\left( A+B \right){{x}^{2}}+\left( -4A-3B+C \right)x+\left( 4A+2B-C \right)
\end{align}$
Then, equate the coefficients of like terms of the equation to write a system of equations as given below:
$ A+B=0$ (I)
$-4A-3B+C=2$ (II)
$4A+2B-C=-6$ (III)
From equation (I), it is inferred that $ A=-B $.
Put $ A=-B $ in equation (II) to obtain the equation:
$\begin{align}
& -4A-3B+C=2 \\
& -4\left( -B \right)-3B+C=2 \\
& 4B-3B+C=2 \\
& B+C=2
\end{align}$ (IV)
Put $ A=-B $ in equation (III) to obtain the equation:
$\begin{align}
& 4A+2B-C=-6 \\
& 4\left( -B \right)+2B-C=-6 \\
& -4B+2B-C=-6 \\
& -2B-C=-6
\end{align}$ (V)
Add equation (IV) with equation (V) to get the value of B:
$\begin{align}
& +\text{ }\underline{\begin{align}
& B+C=2 \\
& -2B-C=-6
\end{align}} \\
& \text{ }-B=-4 \\
\end{align}$
Multiply both sides by $-1$:
$\begin{align}
& -1\left( -B \right)=\left( -4 \right)\left( -1 \right) \\
& B=4
\end{align}$
Therefore, the value of B is $4$.
Now, substitute the value of B in equation (IV) to get the value of B:
$\begin{align}
& B+C=2 \\
& 4+C=2 \\
& C=2-4 \\
& C=-2
\end{align}$
As $ A=-B $, therefore, $ A=-4$
Put the values of $ A,\,\,B,\,\,\text{and }C $ in the given equation, and determine the partial fraction decomposition:
$\begin{align}
& \frac{2x-6}{\left( x-1 \right){{\left( x-2 \right)}^{2}}}=\frac{A}{\left( x-1 \right)}+\frac{B}{x-2}+\frac{C}{{{\left( x-2 \right)}^{2}}} \\
& =\frac{-4}{\left( x-1 \right)}+\frac{4}{x-2}+\frac{-2}{{{\left( x-2 \right)}^{2}}} \\
& =-\frac{4}{\left( x-1 \right)}+\frac{4}{x-2}-\frac{2}{{{\left( x-2 \right)}^{2}}}
\end{align}$
Thus, the partial fraction decomposition of the rational expression is $-\frac{4}{\left( x-1 \right)}+\frac{4}{x-2}-\frac{2}{{{\left( x-2 \right)}^{2}}}$.
