Polynomial degree 3 Equation Question:

If $p(x) \in F[x]$ is of degree $3$, and $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, show that $p(x)$ is irreducible over $F$ if there is no element $r\in F$ such that $a_0+a_1r+a_2r^2+a_3r^3 =0$.

My understanding:
If $p(x)$ is reducible, then there exists $ax + b$ such that $a, b \in F$ and $a\ne 0$. And $p(x) = (ax + b)(cx^2 + dx + e)$. Then an $r$ exists such that $p(r) = 0$.
Is that how you're supposed to tackle it?
I understand that I am supposed to first explain the argument for a polynomial of degree 1, but I do not understand how to go about that, and why that helps.
 A: Well, your theorem says that, if no $r \in F$ such that $a_0 + a_1r + a_2r^2 + a_3r^3=0$ exists (the polynomial has no roots in $F$), then $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ is irreducible over $F$.
And after all, if that polynomial were reducible, you'd be able to write it as a product of two polynomials, none of which is a constant. So, according to your example, you could maybe write $p(x)$ as $(n_o + n_1x)(m_0 + m_1x + m_2x^2)$. But then, $r = -n_0/n_1 \in F$ would be a root of such polynomial, since it yields $0 \cdot (m_0 + m_1x + m_2x^2) = 0$, which goes against the fact that no $r$ with that property existed. This works also if you wrote the polynomial as a product of three other polynomials of degree $1$, that is the only other possibility left.
A: You have what looks like a reasonable outline but I'm guessing you don't understand parts of it since you're asking this question.

If $p(x)$ is reducible, then there exists $ax + b$ such that $a, b \in F$ and $a\ne 0$. And $p(x) = (ax + b)(cx^2 + dx + e)$.

Why?  (Note that this is only true for a cubic, not for anything of higher degree.  What's the difference?)

Then an $r$ exists such that $p(r) = 0$.

Yes, but what, specifically, is $r$?
