I have a proof quickly sketched for this problem that I believe makes sense, the only trouble is that I feel as though I'm not addressing the specific fact that $f$ is a $3^{rd}$ degree polynomial and am hoping someone can point out where I'm missing that or how the degree, in this case, is relevant to the rest of the statement.
The only facts I refer to in the proof are the following Proposition and Corollary:
Proposition: For any field $\mathbb{F}$ and $f,g \in \mathbb{F}[x]$, we have $deg(fg) = deg(f) + deg(g)$
Corollary: For every $f \in \mathbb{F}[x]$ and $\alpha \in \mathbb{F}$, there exists a polynomial $q \in \mathbb{F}[x]$ so that $f = (x -\alpha)q + f(\alpha)$. In particular, $\alpha$ is a root of $f$ iff $(x-\alpha) \mid f$
My proof
We proceed with proof by contrapositive. Assume $f$ has a root $\alpha \in \mathbb{F}$. Then the Corollary states $(x-\alpha) \mid f$, hence $f = (x-\alpha)q$ where $q \in \mathbb{F}[x]$. We see that $f$ is a product of $2$ factors: a linear polynomial ($x-\alpha)$ and another polynomial $q$ whose order must be $2$ by the Proposition: $3 = deg(f) = deg((x-\alpha)q) = deg(x-\alpha) + deg(q) = 1 + 2$. This shows that neither factor is a constant and hence $f$ is not irreducible. $\square$
It feels like my inclusion of the degree of $f$ is merely incidental and not a core aspect of the proof. I welcome anyone to please enlighten me. Thanks for your time.