how i make an demonstration that a polynomial is irreducible?

Question

$2+4X+2X^2+X^3$ is irreducible $\Bbb{F}_5$. I know that the divisors of 2: +-1, +-2 not are root of the polynomial and now?

Answer 1

Since our polynomial is third degree, it's enough to show that $$x^3+2x^2+4x+2\equiv 0\pmod 5$$ has no solutions.

By considering five cases : $x\in\{0,\pm1,\pm2\}$ we can get this.

Now, let our polynomial is reducible.

Thus, let it be a product of two polynomials with degree $m$ and $n$, where $m$ and $n$ are naturals.

Thus, $m+n=3$, which gives that $\{m,n\}=\{1,2\}$ and we got that our polynomial has root from $\mathbb F_5,$ which is a contradiction.

Answer 2

A polynomial $p(x)$ is reducible if there are polynomials $q(x)$ and $r(x)$, both of degree at least $1$, such that $$p(x) = q(x)r(x).$$ The degrees of $q(x)$ and $r(x)$ must sum to the degree of $p(x)$, hence one must be degree $1$, and the other must be degree $2$.

A degree $1$ polynomial takes the form $x - \alpha$, for some $\alpha$ in the field (in this case $\Bbb{F}_5$). So, if the given polynomial is reducible, we should be able to find $q(x)$ such that $$x^3 + 2x^2 + 4x + 2 = (x - \alpha)q(x).$$ Note that, if we substitute $x = \alpha$ into both sides of the equation, $$\alpha^3 + 2\alpha^2 + 4\alpha + 2 = (\alpha - \alpha) q(\alpha) = 0.$$ In particular, $\alpha$ must be a root of the given polynomial.

However, as you've verified, $\alpha = \pm 1$ and $\alpha = \pm 2$ don't work. There's only one other element of $\Bbb{F}_5$, which is $0$. Substituting that into the polynomial yields $$0^3 + 2 \cdot 0^2 + 4 \cdot 0 + 2 = 2 \neq 0,$$ so $0$ is not a root either. But, that means that $\alpha$ cannot exist; we've literally tried every possible value. Hence, we have a contradiction, and the polynomial is indeed irreducible.

Note that this method works with quadratics too. In that case, we'd have to have two degree $1$ polynomials $q(x)$ and $r(x)$, each with a root. If there is no root, then there can be no such factorisation.

However, it does fall apart with polynomials of degree $4$ or more. If the degree of $p(x)$ were $4$, then it would be possible for $p(x)$ and $q(x)$ to be degree $2$ and irreducible. Then, neither $q(x)$ nor $r(x)$ would have roots, and hence, neither would $p(x)$, despite $p(x)$ being reducible.

It's lucky you were only given a cubic!