Reduction modulo p

If $f$ is reducible, then in particular $f$ is reducible modulo every prime $p$ – since $$f(x)=g(x)h(x) \implies f(x) \equiv g(x)h(x) \pmod p $$for every $p$.

So if $f=gh$ where deg $g$ = $1$, then modulo $2$ we would expect $f$ to have a degree $1$ factor. Since this is not the case, we cannot have deg $g = 1.$

Similarly, if deg $g = 2$, then we would expect $f$ to have factors with degrees summing to $2$ modulo $3$. Since this is not the case, we cannot have deg $g=2$ - so $f$ is irreducible.

As a side point, if you haven't discovered it yet, this website has a large collection of notes based on the Cambridge Tripos that may be worth looking at!

If a polynomial is irreducible modulo $p$, then it is irreducible over the integers, for a factorization over the integers would induce a factorization modulo $p$.

A polynomial of degree $2$ or $3$ is irreducible over a field $F$ if and only if the polynomial has no zeros in the field.

By "trying everything" we can verify that $x^2+x+1$ has no zeros modulo $2$, and that $x^3+2x+1$ has no zeros modulo $3$.