Showing existence of irreducible polynomial of degree 3 in $\mathbb{F}_p$

Let $a_1$ and $a_2$ be distinct elements in $\mathbb{F}_p$. Next let $c \in \mathbb{F}_p$ be such that $a^3_1-a^3_2 = c(a_1 - a_2)$, there exists such a $c$ as $a_1 \not = a_2$. Then the mapping $f: x \mapsto x^3- cx$; $x \in \mathbb{F}_p$ is not surjective mapping from $\mathbb{F}_p$ onto $\mathbb{F}_p$, as there exist two distinct $a_1,a_2 \in \mathbb{F}_p$ satisfying $f(a_1)=f(a_2)$. So for this particular $c$, there exists an $A \in \mathbb{F}_p$ such that there is no $x \in \mathbb{F}_p$ that satisfies $x^3-cx = A$. Thus the polynomial $x^3-cx-A$ is irreducible in $\mathbb{F}_p$.

For something in the same spirit as the example for $2$: $x^3-x=(x+1)x(x-1)$ has three zeros (two for $p=2$), so it can't be surjective. Thus some $x^3-x-a$ has no zeros, and a cubic with no linear factors is irreducible.

This approach won't generalize any further; we need to account for more than just linear factors to verify that a polynomial of degree $\ge 4$ is irreducible.

Oh, and your example for quadratic polynomials isn't quite universal; both $x^2=x\cdot x$ and $x^2+1=(x+1)(x+1)$ factor for $p=2$. We need a special case $(x^2+x+1)$ for that.