Why don't quaternions contradict the Fundamental Theorem of Algebra? I don't pretend to know anything much about the Fundamental Theorem of Algebra (FTA), but I do know what it states: for any polynomial with degree $n$, there are exactly $n$ solutions (roots). 
Well, when it comes to quaternions, apparently $i^2=j^2=k^2=-1$, but $i\ne j\ne k\ne i$. So now, we have apparently found three solutions to the second-degree polynomial $x^2=-1$.
I'm not aware of the justification of the FTA, nor I am I aware of Hamilton's justification for quaternions. However, I know a contradiction when I see one. What am I missing here?

dxiv answered the question perfectly in the comment; but let me elaborate.
The fundamental theorem of algebra states that every polynomial of the form
$$p(z) = a_{n}z^{n} + a_{n-1}z^{n-1} + \dots + a_{1}z + a_{0}$$ where each $a_{n}$ is a complex number.
can be factored as
$$p(z) = a_{n}(z - z_{1})\cdot(z - z_{2})\cdots(z - z_{n-1})\cdot(z - z_{n})$$
for some choice of complex numbers $z_{1},\cdots,z_{n}$.
Now, this is equivalent to saying there precisely $n$ solutions which lie in the complex numbers to the equation $p(z) = 0$, so long as we agree to count solutions in terms of their multiplicity.
Your example shows there are other solutions which lie within the Quaternion field.  Since not all of your solutions are complex numbers, there is no contradiction.
Hope that helps.
