Showing $x^4+x^3+2x+15$ is irreducible in $\mathbb{Q}[x]$

Question

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?

Answer 1

If I recall my algebra correctly, there's a theorem that says that if $p(x) \in \mathbb{Z}[x]$ is irreducible over $\mathbb{Z}[x],$ then it's irreducible over $\mathbb{Q}[x].$ Therefore, if $p(x)$ has no rational roots but is reducible over $\mathbb{Z}[x],$ then it'll be the product of two quadratics with integer coefficients, but then it should still be the product of two quadratics when going mod $3.$

Answer 2

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.