Yes, that works. Notice that you don't have to quote the fundamental theorem of algebra. If you found the zeros of the denominator you already got them. It doesn't matter if it is true or not that every polynomial has as many roots as its degree over the complex.
Even if you haven't found all of the roots you can still argue if a given (or already found) root is a simple pole. The idea is computing $B'(z_0)$ at that point. Remember that a zero $z_0$ of a polynomial $P$ is a zero of order $k$ iff $P(z_0)=...=P^{(k-1)}(z_0)=0$ and $P^{(k)}(z_0)\neq0$. So, in the process of using the formula you quoted you already are checking that the pole is simple.
We can always determine the residue of a product by multiplying the Laurent expansions of the factors. Generally, that is tedious (and doesn't lead to a closed form expression), but in the case where one of the factors has a simple pole, and the other factor is holomorphic at the point in question, the situation is quite simple: If
\begin{align}
f(z) &= \sum_{n = -1}^{+\infty} a_n \cdot (z-z_0)^n,\\
g(z) &= \sum_{k = 0}^{+\infty} b_k\cdot (z-z_0)^k,
\end{align}
then it is easy to see that the coefficient of $(z-z_0)^{-1}$ in the product is $a_{-1}\cdot b_0$.
Thus in this case, we have the simple
$$\operatorname{Res} (f\cdot g; z_0) = g(z_0) \cdot \operatorname{Res}(f; z_0).\tag{$\ast$}$$
For the product of the $\Gamma$-function and the Riemann $\zeta$-function, we are always in that situation. All poles of either factor are simple, and the two functions have no common pole. So $\operatorname{Res} (\Gamma\cdot \zeta; 1) = \Gamma(1)\cdot \operatorname{Res}(\zeta;1) = 1\cdot 1 = 1$, and for $n \in \mathbb{N}$ (where we use the convention $0\in \mathbb{N}$), $\Gamma$ has a simple pole at $-n$, and we find
$$\operatorname{Res}(\Gamma\cdot\zeta; -n) = \zeta(-n)\operatorname{Res}(\Gamma; -n).$$
Best Answer
Everything just follows from $\Gamma(z+1)=z\,\Gamma(z)$. For instance, since $\Gamma(1)=1$ and $\Gamma(z)=\frac{\Gamma(z+1)}{z}$, $z=0$ is a simple pole for the $\Gamma$ function with residue $1$. By induction, every negative integer is a simple pole for the $\Gamma$ function and
$$\text{Res}\left(\Gamma(z),z=-n\right) = \lim_{z\to -n}(z+n)\Gamma(z)=\lim_{z\to -n}\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n-1)}=\color{red}{\frac{(-1)^n}{n!}}. $$ Similarly, for any negative integer $-n$ we have $$\text{Res}\left(\Gamma(z+1),z=-n\right)=\color{red}{\frac{(-1)^{n-1}}{(n-1)!}}.$$