Showing $x^4+x^3+2x+15$ is irreducible in $\mathbb{Q}[x]$ Specifically, I'm trying to solve this problem:

Prove that $p(x)=x^4+x^3+2x+15$ is an irreducible polynomial in $\mathbb{Q}[x]$ by considering $p(x)$ mod $3$ and showing that $p(x)$ has no rational roots.

I'm able to show this is irreducible by applying the rational root theorem to eliminate the possibility of a linear factor and then brute force eliminating the possible quadratic factors, but I don't see how to do this in the way the problem states. Taking $p(x)$ mod $3$, we have
$$x^4+x^3+2x+15\equiv x(x^3+x^2+2)\bmod 3.$$
Then, this cubic term is irreducible mod $3$, but how does this help me derive the desired conclusion?

I believe it is simpler to check that $p(x)$ is irreducible over $\mathbb{F}_2$, thus it is irreducible over $\mathbb{Q}$. Indeed
$$ p(x) \equiv x^4+x^3+1 $$
clearly has no root in $\mathbb{F}_2$, but the only quadratic irreducible polynomial over $\mathbb{F}_2$ is $x^2+x+1$. Since $p(x)\neq (x^2+x+1)^2=x^4+x^2+1$, $p(x)$ has no quadratic factor and it is irreducible.
