Answer
The simplified partial fraction expansion is:
$\frac{x-2}{{{x}^{2}}-2x+3}+\frac{2x+1}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$.
Work Step by Step
The provided rational fraction expression is as follows:
$\frac{{{x}^{3}}-4{{x}^{2}}+9x-5}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$
Then, we demonstrate the steps as follows:
Step 1:
Set up up the partial fraction expansion with unknown constant coefficients and then write a constant coefficients over each of the two distinct algebraic linear factors in the denominator of the expression.
$\frac{{{x}^{3}}-4{{x}^{2}}+9x-5}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}-2x+3}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$
Step 2:
Multiply both sides of the equation by the expression ${{\left( {{x}^{2}}-2x+3 \right)}^{2}}$, considering the least common denominator.
${{\left( {{x}^{2}}-2x+3 \right)}^{2}}\times \left( \frac{{{x}^{3}}-4{{x}^{2}}+9x-5}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}} \right)={{\left( {{x}^{2}}-2x+3 \right)}^{2}}\times \left( \frac{Ax+B}{{{x}^{2}}-2x+3}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}} \right)$
Now, simplify both sides of the equation as follows:
${{x}^{3}}-4{{x}^{2}}+9x-5=\left( {{x}^{2}}-2x+3 \right)\left( Ax+B \right)+\left( Cx+D \right)$
$\begin{align}
& {{x}^{3}}-4{{x}^{2}}+9x-5=\left( {{x}^{2}}-2x+3 \right)\left( Ax+B \right)+\left( Cx+D \right) \\
& =A\left( {{x}^{3}}-2{{x}^{2}}+3x \right)+B\left( {{x}^{2}}-2x+3 \right)+Cx+D \\
& =A{{x}^{3}}+{{x}^{2}}\left( -2A+B \right)+x\left( 3A-2B+C \right)+3B+D
\end{align}$
Step 3:
Equate the coefficients of like-powers of $ x $ and of the constant terms.
Get the system of linear equations with the unknown values of $ A $, $ B $, $ C $ and $ D $.
$ A=1$ (I)
$-2A+B=-4$ (II)
$3A-2B+C=9$ (III)
$3B+D=-5$ (IV)
Step 4:
Solve the system for $ A $,$ B $, $ C $ and $ D $,
Substitute the value of $ A $ in the above equation (II) and simplify as given below:
$\begin{align}
& -2A+B=-4 \\
& -2\times 1+B=-4 \\
& B=-2
\end{align}$
Similarly, substitute the value of $ B $ in the above equation (IV) and simplify the equations as given below:
$\begin{align}
& 3B+D=-5 \\
& 3\times \left( -2 \right)+D=-5 \\
& D=1
\end{align}$
Similarly, substitute the values of $ A $,$ B $ and $ D $ in equation (III) and simplify the equations as given below:
$\begin{align}
& 3A-2B+C=9 \\
& 3\times 1-2\times \left( -2 \right)+C=9 \\
& C=9-7 \\
& C=2
\end{align}$
Step 5:
Replace the values of $ A $,$ B $, $ C $ and $ D $, writing the partial function expression as given below:
$\begin{align}
& \frac{{{x}^{3}}-4{{x}^{2}}+9x-5}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}=\frac{Ax+B}{{{x}^{2}}-2x+3}+\frac{Cx+D}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}} \\
& =\frac{1\times x+\left( -2 \right)}{{{x}^{2}}-2x+3}+\frac{2x+1}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}} \\
& =\frac{x-2}{{{x}^{2}}-2x+3}+\frac{2x+1}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}
\end{align}$
Thus, $\frac{x-2}{{{x}^{2}}-2x+3}+\frac{2x+1}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$ is the required partial fraction expansion of the rational expression $\frac{{{x}^{3}}-4{{x}^{2}}+9x-5}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$.
