Degenerate zeroes, fundamental theorem of algebra. 
The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n roots. The equivalence of the two statements can be proven through the use of successive polynomial division.

How do I intuitively understand multiplicity in the above statement? For example the equation $y=x^2$ is said to have a zero at $0$ and this has multiplicity $2$. In general when can I say that a certain root has a multiplicity $2$ while a different root has multiplicity $4$ vs a root with multiplicity $1$ without factoring? 
Are there any prominent generalizations that are used?
 A: Without factoring you can tell something about the multiplicity of the roots by graphing the function.  The graph will bounce off the $x$-axis at roots with even multiplicity will go through the $x$-axis at roots of odd multiplicity.  If the root is a simple root (multiplicity $1$) it will not be tangent to the $x$-axis.  This may be enough to determine the multiplicities of all the roots through a counting argument.
Complex roots come in conjugate pairs for polynomials with real coefficients.  If you find a complex root with nonzero imaginary parts, there will be another one with the same real part and opposite sign on the imaginary part.
Additionally, you can take derivatives of the function if you find a root (you needn't find all of them) and determine the multiplicity by the number of derivatives that equal zero.  For example, let's say that you have
$$f(x) = x^4 + x^3 - 3x^2 -5x -2$$
This factors to $f(x) = (x+1)^3(x-2)$ but let's say you didn't know that.  You find that $x=-1$ works. Then you can take
$$f'(x) = 4x^3 + 3x^2 - 6x - 5.$$
Hey, $f'(-1)=0$, so it's at least multiplicity $2$.  Let's do it again:
$$f''(x) = 12x^2 + 6x - 6.$$
Wow, $f''(-1)=0$ too!  It has multiplicity $3$.  This is kinda-sorta factoring, but we didn't know the multiplicity until we took the derivatives.
