Constructibility of roots of a polynomial I`m trying to decide if the roots of the polynomial $f(x) = x^4+x^3-2x^2 +x +1$ is constructible.
My first thought was to show that the polynomial f is irreducible in $\mathbb{Q}$ then for any root $\alpha$ of $f$ we will have $[\mathbb{Q}(\alpha): \mathbb{Q}] = 4$  and then it will be constructible.
The problem is that I was not able to show that $f$ is irreducible, because eisenstein is not aplicable, and neither any other technique tha I know (note that is reductible on $\mathbb{Z}_2$ for instance).
Thank you for any advance.
 A: It is fairly easy to check that $f(\pm 1) \neq 0$. This implies that $f$ doesn't have any factors of degree $3$, so it is either irreducible or the product of two (irreducible) quadratic polynomials.
In both cases the degree of the roots of $f$ is a power of $2$, hence they are constructible.
A: Indeed $f$ is irreducible: other wise, it would be reducible moodulo $3$? But modulo $3$, $f(x)=x^4+x^3+x^2+x+1$, which cannot factor as $(x^2+ax+b)(x^2+a'x+b')$. Identifying coefficients, one obtains first $bb'=1$, which implies $b=b'=\pm1$,then $a+a'=1$. 
The remaining equations are $aa'+2b=1$ and $b(a+a')=1$. So $b+b'=-1$ is incompatible with the second equation, and finally we have to solve
$$s=a+a'=1,\enspace p= aa'=-1.$$
This is  a classical problem about quadratic equations. We know if there aree solutions, they are the roots of the quadratic equation $x^2-sx+p=x^2-x-1=0$, which has no root over $\mathbf F_3$.
Thus $f(x)$ is irreducible modulo $3$, and * fortiori*, it is irreducible in $\mathbf Z$.
