What?!
Let's put it another way. Take any polynomial p(z), where the coefficients are complex numbers. There is some complex number r that is a root of p(z). In other words, some r satisfies p(r) = 0. Divide through by z-r and find the next root, and so on, until p is the product of monomials z-r. This can be done for every polynomial p(z). This is the fundamental theorem of algebra.
There is a beautiful proof using Galois theory, but for those familiar with analytic functions, Liouville's theorem does the trick.
Note that p(z) is dominated by its leading term. If p(z) has degree 4, then z4 dominates everything for large enough z, even if the coefficient on z4 is small. As z approaches infinity, far from the origin, p(z) approaches infinity.
Let g(z) = 1/p(z). When z lies outside some circle of radius r, g(z) is always less than 1. (We really mean |g(z)| < 1.) If p(z) has no roots, i.e. p(z) is never 0, then p(z) attains a minimum somewhere on the closed disk of radius r, and this minimum is greater than 0. Therefore g(z) has a maximum on the disc of radius r, namely the reciprocal of the minimum of p(z). So g is bounded inside and outside the circle of radius r, hence g is bounded. Now g is also analytic, hence g is constant, and p is constant. The polynomial is 3, or something like that.
Every nontrivial polynomial has a root in the complex numbers.