An important point is that a "pole" is actually the same thing as a removable singularity, if we think of our function as a map that takes values on the Riemann sphere (which is the complex plane with a point at $\infty$ added; the complex structure near $\infty$ comes from the map $z\mapsto 1/z$).
So a function that has a removable singularity or a pole at $z_0$ doesn't have a "real" singularity there at all; rather, we can extend the function to an analytic or meromorphic function in $z_0$. If we cannot extend the function in this way, the singularity is indeed "essential"; i.e., we cannot get rid of it. Thus the terminology is not one that is merely used for convenience or pedagogical purposes; rather, it is extremely natural.
As has already been mentioned, the magic of complex numbers results in many beautiful facts about essential singularities: functions with these singularities are very far from extending continuously.
The simplest of these facts is the Casorati-Weierstraß Theorem: The image of a neighborhood of an essential singularity is dense in the complex plane.
This is just a consequence of the removable singularities theorem. (If $f$ omitted a neighborhood of $a$, we could postcompose $f$ with a Möbius transformation that takes $a$ to infinity and see that the resulting function has a removable singularity.)
The most well-known result of this type is Picard's theorem which was already mentioned.
There are various beautiful strengthenings of Picard's theorem that arise from Nevanlinna theory, and Ahlfors's theory of covering surfaces.
So all essential singularities have some things in common, but on the other hand this should not lead us to believe that they are all the same. What they have in common is complicated behaviour, but they can be complicated in very different ways! Indeed, different transcendental entire functions (those that have an essential singularity at infinity; i.e. are not polynomials) can vary very much with respect to their behavior near infinity. Just for example, for some such functions, such as $z\mapsto e^z$, there exist curves tending to infinity on which the function is bounded, while for others this is not the case.
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.
Best Answer
The point $z_{0}$ is an isolated singularity of $f(z)$ if $f(z)$ is analytic in $0 \lt |z-z_{0}| \lt r$ (a circle of radius r centered at $z_{0}$ with the point $z_{0}$ punched out). If one expands a function $f(z)$ in a Laurent series about the point $z_{0}$, $$f(z) = \sum\limits_{k=-\infty}^{\infty} a^{k} (z-z_{0})^{k}$$ we can classify isolated singularties into 3 cases:
If there are no negative powers of $z-z_{0}$, then $z_{0}$ is a removable singularity and the Laurent series is a power series.
$f(z)$ has a pole of order m at $z_{0}$ if m is the largest positive integer such that $a_{-m} \ne 0$. A pole of order one is a simple pole. A pole of order two is a double pole, etc.
If there are an infinite number of negative powers of $z-z_{0}$, then $z_{0}$ is an essential singularity.