Prove that $f(x)=x^4+8x^3+x^2+2x+5$ is irreducible in $\mathbb Q[x]$.
I've tried many methods:
Eiseinstein's criterion doesn't apply here. I've tried to project the polynomial over $\mathbb Z_2=\{\overline{0}, \overline{1}\}$, getting $f(x)=x^4+x^2+1$, but it is reducible: $f(x)=(x^2+x+1)^2$. Over $\mathbb Z_3$ the polynomial is also reducible: $f(x)=x^4+2x^3+x^2+2x+2$ has $2$ as a root. Over $\mathbb Z_5$ it is evidently reducible. Please, give me an idea!
Put the pieces that you have together!
Modulo $2$ you have, indeed, $f(x)=(x^2+x+1)^2$. This doesn't tell us that $f$ is irreducible, but it does tell us that the only possible factorization (over $\Bbb{Z}$) is into a product of two quadratic factors - both congruent to $x^2+x+1$ modulo two.
Modulo $3$ we can use the root you found and factor $$ f(x)=(x+1)(x^3+x^2+2). $$ Here the latter factor $p(x)$ is seen to be irreducible. It is cubic, so it suffices to check the absence of roots in $\Bbb{Z}_3$, which is easy. We also observe that its reciprocal is $$ \tilde{p}(x):=x^3p(\frac1x)=-(x^3-x-1). $$ This is of the "standard" irreducible type $x^p-x-a,a\neq0,$ over $\Bbb{Z}_p$ (search the site if you haven't seen that result yet).
Whichever way you conclude irreducibility of $p(x)$ doesn't matter. Modulo three information tells us that the only way $f$ can factor (again over $\Bbb{Z}$) is into a product of a linear and a cubic factor.
But, of course, the modulo two and modulo three bits of information are incompatible in any way other than $f$ being irreducible in $\Bbb{Z}[x]$ and hence also in $\Bbb{Q}[x]$.