It's not the roots, it's the "$2$"!
A polynomial is irreducible over a ring if it cannot be written as a product of two non-invertible polynomials. In $\mathbb{Z}$, "$2$" is noninvertible, so $(x^2+2)2$ is an appropriately "nontrivial" factorization.
Meanwhile, over in $\mathbb{Q}$, the polynomial "$2$" is invertible, since ${1\over 2}$ is rational (proof: exercise :P). So the factoriztion $(x^2+2)2$ is "trivial" in the context of $\mathbb{Q}$, since we can always extract a factor of $2$ from any polynomial.
EDIT: Think of it this way: saying that a polynomial is irreducible over a ring means it has no "nontrivial" factorizations. Now, when we make the ring bigger (e.g. pass from $\mathbb{Z}$ to $\mathbb{Q}$) two things happen:
So even though your first instinct might be "polynomials will only go from "irreducible" to "reducible" as the ring gets bigger," actually the opposite can happen!
In fact, here's a good exercise:
Can you find a polynomial $p\in\mathbb{Z}[x]$ which is irreducible over $\mathbb{Z}$ but reducible over $\mathbb{Q}$?
Note that the definition of reducibility over a field may sound different:
For $F$ a field, a polynomial $p\in F[x]$ is irreducible if $p$ cannot be written as the product of two nonconstant polynomials.
But this is actually equivalent to the definition I gave above, in case we're over a field: the noninvertible elements of $F[x]$ are precisely the nonconstant polynomials!
Let $\beta$ be a root of $g$. It is not necessarily the case that $F(\beta)$ is contained in $F(\alpha)$, so we should instead look at the extension $F(\alpha, \beta)$ of $F$, which contains both $F(\alpha)$ and $F(\beta)$. We can write the degree of $F(\alpha,\beta)$ over $F$ in two ways:
$$ [F(\alpha,\beta) : F(\alpha)][F(\alpha) : F] = [F(\alpha, \beta) : F(\beta)][F(\beta) : F]$$
or in other words,
$$ m[F(\alpha,\beta) : F(\alpha)] = n[F(\alpha,\beta) : F(\beta)]$$
Clearly, $n$ divides the product of $m$ (to which it is relatively prime) and $[F(\alpha,\beta) : F(\alpha)]$. If $g$, whose degree is $n$, is not irreducible over $F(\alpha)$, then the extension $[F(\alpha,\beta) : F(\alpha)]$ must be an integer which is strictly less than $n$. See if you can get a contradiction from here.
Best Answer
The function $f(x) = x^3-2$ has only one real-valued zero, namely $\sqrt[3]{2}$. (It has only one, since $f'(x) = 3x^2 > 0$). As a polynomial of degree $3$ it has two more zeros, which have to be in $\mathbb{C} \setminus \mathbb{R}$.
We can write \begin{align} f(x) = (x- \sqrt[3]{2})\cdot h(x) \end{align} where $h$ is a polynomial of degree $2$. Due to that, the function $h(x)$ is irreducible (over $\mathbb{Q}(\sqrt[3]{2})$) if it has no zeros in $\mathbb{Q}(\sqrt[3]{2})$. As mentioned, the two other zeros are in $\mathbb{C} \setminus \mathbb{R}$. But $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$. Hence, the two other zeros are not contained in the field extension $\mathbb{Q}(\sqrt[3]{2})$. As a result $h(x)$ has to be irreducible.