An irreducible polynomial in $\mathbb Q[x]$ that has all zeros with multiplicity $2$ I'm reading an Abstract Algebra textbook written by Jb-Fraleigh. When learning the Splitting field. I should first give the definition of a perfect field: 
Definition: Perfect field A field is perfect if every finite extension is a separable extension.
Now I encounter this problem: 
Problem: Give an example of an $f(x)\in \mathbb Q[x]$ that has no zeros in $\mathbb Q$ but whose zeros in $\mathbb C$ are all of multiplicity $2$. Explain how this is consistent with Theorem 51.13, which shows that $\mathbb Q$ is perfect.
This is Theorem 51.13: 
Theorem 51.13 Every field of characteristic zero is perfect. 
I notice that in the proof of this Theorem, we first assume $E$ is a finite extension of the field of characteristic zero $F$, then let $\alpha\in E$ and $f(x)=\operatorname{irr}(\alpha,F)$ be an irreducible. Then we write $$f(x)=\prod_{i}(x-\alpha_i)^v$$
where $\alpha_i$ are distinct zeros and $v\in \mathbb Z^+$. In the end we can show that $v=1$. 
Because of notice this, I wonder in the question is $f(x)$ an irreducible or a general (i.e., reducible) polynomial? If it can be reducible then I could give an example $$f(x)=(x^2-2)^2$$
but if what it means is finding an irreducible has all zeros multiplicity $2$ when factors in $\mathbb C[x]$, I really have no idea. Notice that I knew $\mathbb C$ is an infinite extension of $\mathbb Q$, so the proof of Theorem 51.13 may not be applied in this case, but I couldn't link everything together in order to give a reasonable proof for this problem. 
 A: If $F$ is a field of characteristic zero and $f\in F[x]$ is irreducible then $f$ is separable. The easiest way to see this is define the formal derivative $f'$. (which looks exactly like the derivative from calculus but it is formal, there is no convergence here). It is easy to see that $\alpha$ is a root of $f$ with multiplicity higher than $1$ if and only if $f'(\alpha)=0$. It follows that $f$ is separable if and only if $\gcd(f,f')=1$. 
Now, if $f$ is irreducible then $\gcd(f,f')\in\{1,f\}$. Since we are working in a field of characteristic zero we have $deg(f')=def(f)-1$ and hence the $\gcd$ can't be $f$. So it must be $1$ and hence $f$ is separable. This is actually a way to prove that any algebraic extension of a field of characteristic zero is separable. 
So the point is you can't find an irreducible polynomial in $\mathbb{Q}[x]$ which has roots with multiplicity $2$. 
A: There is none;  only reducible ones.  
If there were an irreducible one, $f$, we could adjoin a root, and we'd have a finite extension of $\Bbb Q $ (of $\operatorname {deg}f$), namely $\Bbb Q[x]/(f)$, which wasn't separable. 
This violates theorem $51.13$, since $\operatorname {char}\Bbb Q=0$.
