$ p \in Q[x] $ has as a root a fifth primitive root of unity, then every fifth primitive root of unity is a root of $p$. I'm extremely stuck. Can't figure it. 
The conjugate is easy: let $w$ be a primitive root of unity, then $w^{-1}$ will also be a root, that's easy. But I'm missing $w^2$ and $w^3$. Why would they be also roots of p? 

Since $\;p(x)\in\Bbb Q[x]\;$ and it vanishes on $\;\zeta:=e^{2\pi i/5}\;$ , the minimal polynomial of $\;\zeta\;$ over the rationals (also known as a cyclotomic polynomial) also divides $\;p(x)\;$ , and thus
$$\Phi_5(x)\,|\,p(x)\implies p(x)=\overbrace{(x^4+x^3+x^2+x+1)}^{=\Phi_5(x)}\cdot h(x)\;,\;\;h(x)\in\Bbb Q[x]$$
Yet 
$$\;x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}\implies$$
the roots of $\;\Phi_5(x)\;$ are precisely the four primitive roots of unity of order five, and thus these roots are also roots of $\;p(x)\;$
