$\omega^2+\omega+1$divides a polynomial The question is 
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?

See that $\omega$ is a root of both $f(x)=x^5+x^4+1$ and $g(x)=x^2+x+1$. Now $g(x)$ has all coefficients real, and it is a polynomial of degree 2. Hence it has two roots and the second root must be the complex conjugate of $\omega$, that is $\omega^2$. Hence $g(x)=(x-\omega)(x-\omega^2)$. Also note that $f(\omega^2)=0$. Hence $f(x)=(x-\omega)(x-\omega^2)h(x)=g(x)h(x)$ for some real polynomial $h(x)$. Now you can easily show by equating coefficients that $h(x)$ has integer coefficients. Now put $x=n$ on both sides. Then since $h(n)$ turns out to be an integer $n^2+n+1|n^5+n^4+1$.
