If p(x) $\in F[x]$ is of degree 3, and $p(x) = a_0 + a_1*x + a_2*x^2 + a_3*x^3$, show that p(x) is irreducible oer F if there is no element $r \in F$ such that
$a_0 + a_1*r + a_2*r^2 + a_3*r^3$.
So far I've said that p(x) is not irreducible iff it is the product of two polynomials of $deg < 3$. Therefore one of these polynomials must be of degree 1, as 3= 1+ 2 = 1 + 1 + 1 and $deg(q(x)r(x)) = deg(q(x)) + deg(r(x))$.
The polynomial of degree 1 is in the form $ ax + b$, where $ a,b \in F$ and $ a \ne 0 $. Therefore iff $p(x) $ is not irreducible, $p(x) = (ax + b)(cx^2+dx+e)$, and there always exists an $ r = -b/a$ s.t. $ r\in F$ and $p(r) = 0$.
Therefore if there exists an element $r \in F$ s.t. $p(r) = 0$, p(x) is not irreducible. Thus for p(x) to be irreducible, there cannot exist an element $ r \in F$ s.t. p(r) = 0.
Hint:
(for direct proof:) Should start with $p(r)=0$ for some $r \in \mathbb{F}$, then show that $p(x)$ has a linear factor $x-r$.
(for contrapositive:) assume $p(x)$ is reducible, then should have a linear factor and then go about finding $r$.