## Precalculus (6th Edition) Blitzer

The simplified partial fraction expansion is: $\frac{x-2}{{{x}^{2}}-2x+3}+\frac{2x+1}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$.
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 $\frac{{{x}^{3}}-4{{x}^{2}}+9x-5}{{{\left( {{x}^{2}}-2x+3 \right)}^{2}}}$.