Show that $x^3+17x+36$ is irreducible over $\mathbb{Q}[x]$ Let $\sigma_m:\mathbb{Z}\to\mathbb{Z}_m$ be the natural homomorphism given by $\sigma_m(a)=($the remainder of $a$ when divided by $m)$ for $a\in\mathbb{Z}$.
(a) Show that $\overline{\sigma}_m:\mathbb{Z}[x]\to\mathbb{Z}_m[x]$ given by $$\overline{\sigma}_m(a_0+a_1x+\dots+a_nx^n)=\sigma_m(a_0)+\sigma_m(a_1)x+\dots+\sigma_m(a_n)x^n$$ is a homomorphism of $\mathbb{Z}[x]$ onto $\mathbb{Z}_m[x]$
(b) Show that if $f(x)\in\mathbb{Z}[x]$ and $\overline{\sigma_m}(f(x))$ both have degree $n$ and $\overline{\sigma_m}(f(x))$ does not factor in $\mathbb{Z}_m[x]$ into two polynomials of degree less than $n$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$.
(c) Use part (b) to show that $x^3+17x+36$ is irreducible in $\mathbb{Q}[x]$.
My problem is with part (c) (I posted parts (a) and (b) for context). Using part (b) and $\sigma_3$, I saw that, $$\overline{\sigma}_3(x^3+17x+36)=x^3+2x=x(x^2+2)$$
But would't this show that $x^3+17x+36$ IS reducible in $\mathbb{Q}[x]$? Or am I misunderstanding some fundamental concept of the problem or the concept of irreducibility?

The modular stuff is really unnecessary. A cubic is irreducible over the rationals if and only if it has no rational roots. And by the Rational Root Theorem, all we need to check is whether one of the divisors of $36$, positive or negative, is a root of our polynomial.
But we can also work modulo $5$. If the polynomial is reducible over the rationals, it is reducible over the integers, and therefore over $\mathbb{Z}_5$. We show that the polynomial $x^3+2x+1$ has no roots in $\mathbb{Z}_5$. 
Test $0,1,2,3,4$, or if you prefer $0,1,2,-2,-1$. 
