Confusion about a way to compute a residue at a pole Suppose I have a function of the form $$f(x)=\frac{1}{(x-a)(x-b)^2(x-c)^3}$$
Clearly, I have a simple pole at $a$, and poles of order 2,3 at $b,c$, respectively. 
By definition, the residue at $x=\alpha$ is the coeffecient of the term $(x-\alpha)^{-1}$ in the Laurent expansion of $f(x)$ centered at $x=\alpha$. From my experience, this residue at $x=\alpha$ is equivalent to the coefficient of the term $(x-\alpha)^{-1}$ in the partial fraction decomposition of $f(x)$. However, the partial fraction of $f(x)$ will not be a Laurent series since it has no center (I.e., there are denominators of the form $x-\beta$ with $\beta \not = \alpha$). Is there a rule that says that the coefficient of $(x-\alpha)^{-1}$ in the partial fraction decompositon is the residue at $x=\alpha$, even if the partial fraction decompoition is not a Laurent expansion?

Theorem: If $f(x) = g(x)h(x)$ and $g$ is holomorphic near $a$ and $h$ has a simple pole at $a$, then the residue of $f$ at $a$ is $g(a)$ times the residue of $h$ at $a$.
