## Precalculus (6th Edition) Blitzer

The partial fraction is $\frac{1}{\left( x-2 \right)}-\frac{2}{{{\left( x-2 \right)}^{2}}}-\frac{5}{{{\left( x-2 \right)}^{3}}}$
$\frac{{{x}^{2}}-6x+3}{{{\left( x-2 \right)}^{3}}}=\frac{A}{\left( x-2 \right)}+\frac{B}{{{\left( x-2 \right)}^{2}}}+\frac{C}{{{\left( x-2 \right)}^{3}}}$ Now, multiply both sides by ${{\left( x-2 \right)}^{3}}$; then we get, \begin{align} & {{x}^{2}}-6x+3=A{{\left( x-2 \right)}^{2}}+B\left( x-2 \right)+C \\ & =A\left( {{x}^{2}}+4-4x \right)+Bx-2B+C \\ & =A{{x}^{2}}+4A-4Ax+Bx-2B+C \\ & =A{{x}^{2}}+\left( -4A+B \right)x+4A-2B+C \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A=1$ …… (1) $-4A+B=-6$ …… (2) $4A-2B+C=3$ …… (3) And put the value of A in equation (2): \begin{align} & -4\left( 1 \right)+B=-6 \\ & B=-6+4 \\ & =-2 \end{align} …… (4) Also, put the value of A and B in equation (3): \begin{align} & 4\left( 1 \right)-2\left( -2 \right)+C=3 \\ & C=3-8 \\ & =-5 \end{align} Therefore, $\frac{{{x}^{2}}-6x+3}{{{\left( x-2 \right)}^{3}}}=\frac{1}{\left( x-2 \right)}-\frac{2}{{{\left( x-2 \right)}^{2}}}-\frac{5}{{{\left( x-2 \right)}^{3}}}$ Thus, the partial fraction of the provided expression is $\frac{1}{\left( x-2 \right)}-\frac{2}{{{\left( x-2 \right)}^{2}}}-\frac{5}{{{\left( x-2 \right)}^{3}}}$.