Finding roots of the polynomial $x^4+x^3+x^2+x+1$ In general, how could one find the roots of a polynomial like $x^4+x^3+x^2+x^1+1$? I need to find the complex roots of this polynomial and show that $\mathbb{Q (\omega)}$ is its splitting field, but I have no idea of how to proceed in this question. Thanks in advance.

Since $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$, the roots are the fifth roots of $1$, excluding $1$. Note that the set of fifth roots of $1$ is a group of prime order, so it is cyclic and any element is a generator. Thus, if $\omega$ is any of the roots of the polynomials, the full set of roots is given by $\omega,\omega^2,\omega^3,\omega^4$. In particular all roots belong to $\mathbb{Q}[\omega]$.
The roots are
$$
\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\\
\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}\\
\cos\frac{6\pi}{5}+i\sin\frac{6\pi}{5}\\
\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}
$$
However, you can also determine them “more explicitly”. Rewrite the equation as
$$
x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0
$$
and recall that
$$
x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2
$$
so, setting $t=x+\dfrac{1}{x}$, the equation is
$$
t^2+t-1=0
$$
Find the two roots $t_1$ and $t_2$; next solve the equations
$$
x+\dfrac{1}{x}=t_1,\qquad x+\dfrac{1}{x}=t_2
$$
