With a little calculus is not hard to show this polynomial has three different real roots.
If $\,\alpha\,$ is one of the roots, then
The quadratic's discriminant is $\,\Delta:=12-3\alpha^2=3(2-\alpha)(2+\alpha)\,$
Now, again using a little a calculus, we can see the roots are in the following intervals:
and since all the roots are in $\,(-2,2)\,$ we get that $\,\Delta>0\,$ no matter what root $\,\alpha\,$ we chose from the
three existing ones.
Thus, since the roots of the quadratic are
We can see that $\,\Bbb Q(\alpha)\,$ contains them all and, thus, this is also true for any field containing $\,\alpha\,$.
Of course, if the polynomial has no roots in $\,\Bbb K\supset\Bbb Q\,$ then it must remain irreducible there (why?)