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I'm sure this is a terrible question but I'd like to make sure. Given a polynomial, will its roots always be complex numbers? What else could they even be?

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    If the polynomial has complex coefficients, yes. This is a corollary of the fundamental theorem of algebra. – 2017-01-27
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    What other kinds of numbers could the roots be? – 2017-01-27
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    Definitions are your friends. What is a "polynomial"? The complex numbers are very nice in giving us roots of polynomials when the coefficients are themselves complex numbers. But there are other number systems (finite fields, quaternions, etc.) for which you might construct polynomials. – 2017-01-27
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    @SimplyBeautifulArt: [matrices.](https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem). – 2017-01-27
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    @MartinArgerami Given the wording of the question, I almost doubt the OP would be thinking about such things, not that they are wrong. – 2017-01-27
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    I agree, but I think it is worth clarifying :) – 2017-01-27
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    What if it's a polynomial with integer coefficients? Will the roots always be rationals? Reals? Complex numbers? – 2017-01-27

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Usually one considers polynomials within a (number) system. Then you ask yourself whether there are roots within that system.

The prototypical example is with the usual number systems, where one enlarges the system (integers, rationals, reals, complex) until within the complex numbers we get the fundamental theorem of algebra: the every polynomial with complex coefficients has a complex root.

But, in more generality, one can consider polynomials with whatever objects can be added and multiplied: this applies to the coefficients and also to the "variable": and they don't have to be the same. As an example, since square matrices can be added and multiplied, and also multiplied by scalars, one can consider polynomials with complex coefficients but where the variable is a square matrix: things like $$ p(A)=3A^3-2A^2-A. $$

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It's easy enough to "immerse" the complex numbers inside a larger number system - say, quaternions. Doing so can create new roots for existing polynomials, just by adding a new number. For example, the complex numbers were built by tacking on a root of the polynomial $x^2 + 1$. There's nothing stopping us from adding two new, unrelated, roots (say $i$ and a new number $a$). But that seems... silly, because we're just adding numbers for no good reason.

We added $i$ because "most" quadratics had two roots in the reals, so it made sense to expect that they all had two roots hidden somewhere. So maybe the more reasonable version of your question might be "might we want to do this again?" The answer is no. It turns out (by the Fundamental Theorem of Algebra) that every $n$th-degree polynomial has exactly $n$ roots in the complex numbers (counting with multiplicity). What that means is that there's no "unfairness" to level out; if we added more roots to a polynomial, it would have to be something frivolous like what I outlined earlier.

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No. Give a polynomial with coefficients in a field $K$, the roots are in the algebraic closure $\overline K$ which is unique. By definition it is the field containing all the solutions of all the polynomials with coefficients in $K$. For instance $\mathbb C$ is algebrically closed thanks to the Fundamental Theorem of Algebra, thus every polynomial with coefficients in $\mathbb C$ has roots in $\mathbb C$. On the other hand for example $F_2$ the field with two elements is not algebrically closed, in fact $X^2+ X+1$ doesn't vanish in $0$ nor $1$, so the algebraic closure of $F_2$ is bigger than $F_2$.