Answer
The simplified partial fraction expansion is $\frac{3x}{{{x}^{2}}-2x+2}+\frac{x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$.
Work Step by Step
The provided rational fraction is as follows:
$\frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$
Now, we demonstrate the steps as follows:
Step 1:
Set up the partial fraction expansion with unknown constant coefficients and then write a constant coefficient over each of the two distinct algebraic linear factors in the denominator of the expression.
$\frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}-2x+2}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$
Step 2:
Multiply both sides of the equation of the expression by ${{\left( {{x}^{2}}-2x+2 \right)}^{2}}$, considering the least common denominator.
${{\left( {{x}^{2}}-2x+2 \right)}^{2}}\times \left( \frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \right)={{\left( {{x}^{2}}-2x+2 \right)}^{2}}\times \left( \frac{Ax+B}{{{x}^{2}}-2x+2}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \right)$
Then, simplify both sides of the equation as given below:
$3{{x}^{3}}-6{{x}^{2}}+7x-2=\left( {{x}^{2}}-2x+2 \right)\left( Ax+B \right)+\left( Cx+D \right)$
$\begin{align}
& 3{{x}^{3}}-6{{x}^{2}}+7x-2=\left( {{x}^{2}}-2x+2 \right)\left( Ax+B \right)+\left( Cx+D \right) \\
& =A\left( {{x}^{3}}-2{{x}^{2}}+2x \right)+B\left( {{x}^{2}}-2x+2 \right)+Cx+D \\
& =A{{x}^{3}}+{{x}^{2}}\left( -2A+B \right)+x\left( 2A-2B+C \right)+2B+D
\end{align}$
Step 3:
And equate the coefficients of like powers of $ x $ and of the constant terms. Then we get the system of linear equations with the unknown values of $ A $, $ B $, $ C $ and $ D $.
$ A=3$ (I)
$-2A+B=-6$ (II)
$2A-2B+C=7$ (III)
$2B+D=-2$ (IV)
Step 4:
Then, solve the system for $ A $, $ B $, $ C $, and $ D $.
Put the value of $ A $ in equation (II) and simplify as follows:
$\begin{align}
& -2A+B=-6 \\
& -2\times 3+B=-6 \\
& B=0
\end{align}$
Similarly, putting the value of $ B $ in equation (IV) and simplifying as follows, we get:
$\begin{align}
& 2B+D=-2 \\
& 2\times 0+D=-2 \\
& D=-2
\end{align}$
Similarly, putting the value of $ A $,$ B $, and $ D $ in equation (III) and simplifying as follows, we get:
$\begin{align}
& 2A-2B+C=7 \\
& 2\times 3-2\times 0+C=7 \\
& C=7-6 \\
& C=1
\end{align}$
Step 5:
By putting the values of $ A $,$ B $, $ C $ and $ D $ and writing the partial function expression, we get:
$\begin{align}
& \frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}-2x+2}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \\
& =\frac{3\times x+0}{{{x}^{2}}-2x+2}+\frac{1\times x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}} \\
& =\frac{3x}{{{x}^{2}}-2x+2}+\frac{x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}
\end{align}$
Thus, $\frac{3x}{{{x}^{2}}-2x+2}+\frac{x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$ is the required partial fraction expansion of the rational expression $\frac{3{{x}^{3}}-6{{x}^{2}}+7x-2}{{{\left( {{x}^{2}}-2x+2 \right)}^{2}}}$.
