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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.

user167000
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    in spirit you have the right idea but your proof is not written correctly. You should be clear which way you want to proceed: direct proof or proof by contradiction or proof by contrapositive. Once you decide that then you will be able to make your proof more coherent. – Anurag A Aug 16 '14 at 19:08

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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 $r$.

Anurag A
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    Proof by contrapositive:let p(x) be reducible. Then there exists an ax+ b s.t. $a,b \in F$ and $a \ne 0$ and $p(x) = (ax + b)(cx^2+dx+e)$. Then there exists an r s.t. p(r) = 0. – user167000 Aug 16 '14 at 19:20