Why $x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$? 
Why $f(x) = x^2 + 7$ is the minimal polynomial for $1 + 2(\zeta + \zeta^2 + \zeta^4)$ (where $\zeta = \zeta_7$ is a primitive root of the unit) over $\mathbb{Q}$?

Of course it's irreducible by the Eisenstein criterion, however it apparently does not satisfies $1 + 2(\zeta + \zeta^2 + \zeta^4)$ as a root, I tried to calculate several times however I couldn't get $f(1 + 2(\zeta + \zeta^2 + \zeta^4))$ = 0$.
Thanks in advance.

Compute:
$$\begin{align*}
(1+2(\zeta+\zeta^2+\zeta^4))^2+7&=\bigg[1^2+4(\zeta+\zeta^2+\zeta^4)+4(\zeta+\zeta^2+\zeta^4)^2\bigg]+7\\[0.1in]
&=\bigg[1+4(\zeta+\zeta^2+\zeta^4)+4(\zeta^2+\zeta^4+\zeta^8+2\zeta^3+2\zeta^5+2\zeta^6)\bigg]+7\\[0.1in]
(\mathsf{\text{because }}\zeta^8=\zeta)\quad&=\bigg[1+4(\zeta+\zeta^2+\zeta^4)+4(\zeta^2+\zeta^4+\zeta+2\zeta^3+2\zeta^5+2\zeta^6)\bigg]+7\\[0.1in]
&=\bigg[1+8\zeta+8\zeta^2+8\zeta^3+8\zeta^4+8\zeta^5+8\zeta^6\bigg]+7\\[0.1in]
&=(-7)+8+8\zeta+8\zeta^2+8\zeta^3+8\zeta^4+8\zeta^5+8\zeta^6+7\\[0.2in]
&=(-7)+8\Phi_7(\zeta)+7\\[0.1in]
\left(\begin{array}{c}\mathsf{\text{because }}\Phi_7\mathsf{\text{ is the}}\\ \mathsf{\text{minimal poly of }}\zeta\end{array}\right)\quad&=(-7)+7\\[0.1in]
&=0
\end{align*}$$
