Answer
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
Work Step by Step
We are given the fraction:
$\dfrac{x^2-6x+3}{(x-2)^3}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}+\dfrac{C}{(x-2)^3}$
Multiply all terms by the least common denominator $(x-2)^3$:
$(x-2)^3\cdot\dfrac{x^2-6x+3}{(x-2)^3}=(x-2)^3\cdot\dfrac{A}{x-2}+(x-2)^3\cdot\dfrac{B}{(x-2)^2}+(x-2)^3\cdot\dfrac{C}{(x-2)^3}$
$x^2-6x+3=A(x-2)^2+B(x-2)+C$
$x^2-6x+3=Ax^2-4Ax+4A+Bx-2B+C$
$x^2-6x+3=Ax^2+(-4A+B)x+(4A-2B+C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A=1\\
-4A+B=-6\\
4A-2B+C=3
\end{cases}$
Solve the system:
$A=1$
$-4(1)+B=-6$
$B=-2$
$4(1)-2(-2)+C=3$
$8+C=3$
$C=-5$
The partial fraction decomposition is:
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
