Using Reduction Modulo $p$ to Prove Irreducibility of Polynomial

Question

This is a problem from Ronald Irving's Book, "Integers, Polynomial, and Rings" page 190.

I tried to prove

Polynomial $f(x)=x^5+x^2+1$ is irreducible in $\mathbb{Z}[x]$ or does not factor in $\mathbb{Z}[x]$ as a product of lower-degree polynomials.

So, hints in the book are:

1. Explain why it suffices to show simply that there exist a prime $p$ such that the reduction of $x^5+x^2+1$ modulo that prime $p$ is irreducible in $\mathbb{F}_p[x]$.
2. Try $p=2$, and write the reduced polynomial also as $x^5+x^2+1$. Show that $x^5+x^2+1$ has no root in $\mathbb{F}_2$ and conclude that $x^5+x^2+1$ has no degree-one factor in $\mathbb{F}_2[x]$.
3. Show that $x^2+x+1$ does not divide $x^5+x^2+1$ and conclude that $x^5+x^2+1$ has no degree-two factor in $\mathbb{F}_2[x]$

My answer:

1. ??
2. Suppose $a(x)$ is reduced modulo $2$ of $f(x)$, that is also $a(x)=x^5+x^2+1$. Since $\mathbb{F}_2=\{[0],[1]\}$ then $a(0)=1$ and $a(1)=1$ mod 2. This means $a(x)$ has no root in $\mathbb{F}_2$. So that $(x)$ and $(x-1)$ are not factors of a(x)? (I'm not sure factor of $a(x)$ or $f(x)$?
3. ?? (I wonder why we have to do this? What relationship between $x^2+x+1$ so this polynomial divide $x^5+x+1$?)

I appreciate any help. Thank you.

Answer 1

For point 1, it is because if $f(x)$ factored in $\mathbf Z$, its reduction modulo $p$ would factor in $\mathbf F_p$.

As to point 3, as $f(x)$ cannot have a linear factor by point 2, if it factored, it would be as the product of a quadratic and a cubic factors, both irreducible in $\mathbf F_2$. It happens that the only irreducible quadratic polynomial in $\mathbf F_2$ is $\:x^2+x+1$.