Fundamental theorem of algebra in an algebraically closed field Let $K$ be a field, and let $f(x)\in K[x]$. 
Is it true that $f(x) = 0$ has $\deg f$ solutions in $\overline K$? 
In particular, I am interested in solving $x^6 - 1=0$ in a field $K$ with $\operatorname{char}(K)\neq 2,3$. I'm not sure why I've been given the characteristic here, or how it can help me.
I believe that the same argument of FTA applies, i.e., we may write $f$ as $a_{\deg f}\prod_{k=1}^{\deg f}(x-\alpha_k)$ with $\alpha_k\in \overline K$, and that since we are in an integral domain, the result follows.
 A: First note that
$$\begin{align*}
x^6-1 &= (x^3-1)(x^3+1)\\
&=(x-1)(x^2+x+1)(x+1)(x^2-x+1).
\end{align*}$$
Note that we cannot assume that $\overline{K}=\Bbb{C}$. However, we can use what we know about the roots of $x^6-1$ in $\Bbb{C}$ to make a guess about what will happen in $\overline{K}$, and then we just have to prove that our guess is correct.
Let $\theta$ be a root of $x^2-x+1$. Then we will show that
$$x^6-1=\prod_{i=1}^6(x-\theta^i).$$
Since $\theta$ is a root of $x^2-x+1$, we have that $\theta^2=\theta-1$.
Also, since $\theta$ is a root of $x^2-x+1$, and $(x+1)(x^2-x+1)=x^3+1$, we have that $\theta$ is also a root of $x^3+1$. Hence $\theta^3=-1$.
From the fact that $\theta^2=\theta-1$ and $\theta^3=-1$, we get the following (as you can check):
$$\theta^4=-\theta,$$
$$\theta^5=1-\theta,$$
$$\theta^6=1.$$
With these relations, you can check that each of the following hold:
$$(x-\theta)(x-\theta^5)=x^2-x+1,$$
$$(x-\theta^3)=x+1,$$
$$(x-\theta^2)(x-\theta^4)=x^2+x+1,$$
$$(x-\theta^6)=x-1,$$
and having verified the above equations, we have verified that
$$x^6-1=\prod_{i=1}^6(x-\theta^i).$$
The restrictions on $\text{char}(K)$ are used to show that the roots $\theta,\theta^2,\ldots,\theta^6$ are distinct.
For example, $\theta=\theta^4$ iff $\text{char}(K)=2$. Note that if $\text{char}(K)=2$, then $x^6-1=(x^3-1)^2$.
Also, note that the following are equivalent:
$$\theta=\theta^3.$$
$$\theta=-1.$$
$$-1\text{ is a root of }x^2-x+1.$$
$$\text{char}(K)=3.$$
If $\text{char}(K)=3$, then $x^6-1=(x^2-1)^3=(x-1)^3(x+1)^3$.
If $\text{char}(K)\ne2,3$, then we can show that the roots of $x^6-1$ are distinct by using the following:
Fact: Let $F$ be a field, let $p(x)\in F[x]$ and let $\lambda$ be a root $p(x)$. Then $\lambda$ is a multiple root of $p(x)$ iff $\lambda$ is a root of $p'(x)$, where $p'(x)$ is the derivative of $p(x)$ (defined formally).
Let $p(x)=x^6-1$, then $p'(x)=6x^5$. So if $\text{char}(K)\ne2,3$, then $GCD\left(p(x),p'(x)\right)=1$, so the roots of $p(x)$ are distinct.
