How can I prove the polynomial f is irreducible We have $f\in \mathbb{Z}_{3}\left[X\right],\:\:f=x^3+2x^2+a,\:\:a\in \mathbb{Z}_{3}$ and we need to find $a$ for which polynomial $f$ is irreducible. 

I looked on google but I don't understand very well when a polynomial is reducible or irreducible, so can give me an example when is reducible and when is irreducible? What mean each other?
 A: A polynomial is reducible if it can be factored into nonconstant, lower degree polynomials with coefficients from the same field as the original polynomial. So in this case $a \neq 0$, otherwise $f(x) = x^2(x+2)$ and is thus reducible over $\mathbb{Z}_3$. So what are the other options for $a$?

Hint: Consider $a=1$, then $f(x) = x^3 + 2x^2 + 1$. Now we wish to find out if $f(x)$ can be factored into lower degree polynomials over $\mathbb{Z}_3$. Note that $\deg(f(x)) = 3$, so if $f(x)$ can be factored, one of the polynomial factors will be degree 1. So a degree 1 polynomial factor means that $f(x)$ has a root in $\mathbb{Z}_3$. So in this case you can just check for roots of $f(x)$. Note that $f(0) = 1 \neq 0$, $f(1) = 1 + 2 + 1 = 1 \neq 0$, 
and $f(2) = 2 + 2 + 1 = 2 \neq 0$. Thus $f(x)$ has no roots when $a=1$ and is therefore irreducible over $\mathbb{Z}_3$. Now you only have one case left to consider.
A: Since this polynomial is of degree three it's either irreducible or it has a linear factor, so all you need to do is try $a = 0, 1, 2$, and test which of these choices of $a$ give you a polynomial that has none of $\{0, 1, 2\}$ as roots. There are few enough cases here that you can just do it by hand.
