Can an irreducible polynomial of $\mathbb{Q}[x]$ have more than two real roots? The title is pretty much self explanatory; Can an irreducible polynomial of $\mathbb{Q}[x]$ have more than two real roots? And if so, what is an example of a polynomial with more than two real root? All the polynomials I've seen, had maximum two real roots, but from what I've learned of Galois Theory it seems to indicate that there could be more than two.
 A: A cubic polynomial has three real roots if and only if its discriminant is positive. If it is a depressed cubic,
$$f(x) = x^3+px+q$$
then the discriminants is
$$\Delta(f) = -4p^3 - 27q^2.$$
Now, if there is a prime $\ell$ such that $\ell$  divides both $p$ and $q$, and $\ell^2$ does not divide $q$, then by Eisenstein's Criterion we would have that $f(x)$ is irreducible over $\mathbb{Q}$. So all we need is to pick appropriate $p$ and $q$ to get this done. For example,
$$ f(x) = x^3 - 9x + 3$$
will do, since it is Eisenstein at $3$, hence irreducible, and
$$\Delta(f) = 4(9)^3 - 27(9) = 2673\gt 0.$$
A: Have you heard about totally real number fields? If $K$ is a totally real number field of degree $n$, then there is an irreducible polynomial $f$ over $\mathbb Q$ of degree $n$ such that all roots are real and generate $K$. This post https://math.stackexchange.com/a/2645829 implies that for every $n \in \mathbb N$ you find such a totally real number field of degree $n$. So for each $n$ you find an irreducible polynomial $f$ of $\mathbb Q$ with $n$ real roots.
