Answer
$$\dfrac{2x^2+1}{(x-3)^3}=\dfrac{A}{x-3}+\dfrac{B}{(x-3)^2}+\dfrac{C}{(x-3)^3}$$
Work Step by Step
The denominator of the rational expression is already factor. There is only one factor, and it is linear, so put a constant in the numerator of the partial fraction. Since this factor has a multiplicity of 3, we must have 3 terms in the partial fraction, one for each multiplicity.
$$\dfrac{2x^2+1}{(x-3)^3}=\dfrac{A}{x-3}+\dfrac{B}{(x-3)^2}+\dfrac{C}{(x-3)^3}$$
