Assistance with polynomial factorization Is it possible to factorize  $x^6 + 4x + 8$?
If so what is the fully simplified form, and what would be the steps to get there?
 A: No, it,s not possible. Suppose $x^6 + 4x + 8=(x^n+f(x))(x^{6-n}+g(x))$, where $n\in\{1,2,3\}$ and $\deg f<n,\ \deg g<6-n$. Then since $x^6 + 4x + 8\equiv x^6(\mod2)$, we have $f\equiv 0(\mod4)$ and $g\equiv 0(\mod2)$. Write $f=2f^\prime$, $g=2g^\prime$. Since $x^6 + 4x + 8\equiv x^6(\mod4)$, we have $2x^ng^\prime+2x^{6-n}f^\prime\equiv 0(\mod4)$, or equivalently, $x^ng^\prime+x^{6-n}f^\prime\equiv 0(\mod2)$, which implies $g^\prime\equiv -x^{6-2n}f^\prime(\mod2)$. If $n=1,2$, the coefficients of $x^0$ and $x$ in $g$ is divisible by $4$, which implies the coefficient of $x$ in $(x^n+f(x))(x^{6-n}+g(x))$ is divisible by $8$, a contradiction; If $n=3$, then $g^\prime\equiv f^\prime(\mod2)$, thus $g\equiv f(\mod4)$, which also implies that the coefficient of $x$ in $(x^n+f(x))(x^{6-n}+g(x))$ is divisible by $8$.
A: Here’s a quicker, but much more advanced argument for irreducibility: The Newton Polygon of your $x^6 + 4x + 8$ as a $2$-adic polynomial has its vertices at $(0,3)$ (for the constant term), $(1,2)$ (for the linear) and $(6,0)$. The segments have slope $-1$ and $-2/5$. So there is a $2$-adic factorization into linear times quintic, and the linear is of the form $x-2u$, where $u$ is a unit $2$-adic integer. The other segment passes through no integer points, so its quintic polynomial is irreducible. Thus the Newton factorization is the factorization over $\Bbb Q_p$, and would be the factorization over $\Bbb Q$ if there were one. But neither of $\pm2$ is a root, so the original is irreducible.
