How to factorize the polynomial in the ring $\mathbb{Z}_5[x]$ I need to factor the polynomial $x^5+3x^4+x^3+x^2+3$ into the product of the irreducible ones in the ring $\mathbb{Z}_5[x]$.
The problem is I don't see any whole roots (I tried every possible divisor of 3) in the given polynomial. Does it mean that it is already irreducible?

You need to make it more aggressively.
Check that $2$, $3$ and $4$ are roots, which gives the factorization: 
$$x^5+3x^4+x^3+x^2+3\equiv x^5-2x^4+x^3+x^2-12=$$
$$=x^5-2x^4+x^3-2x^2+3x^2-6x+6x-12=(x-2)(x^4+x^2+3x+6)\equiv$$
$$\equiv(x-2)(x^4-9x^2-2x+6)=(x-2)(x^2(x-3)(x+3)-2(x-3))=$$
$$=(x-2)(x-3)(x^3+3x^2-2)\equiv(x-2)(x-3)(x^3-2x^2-32)=$$
$$=(x-2)(x-3)(x^3-4x^2+2x^2-32)=(x-2)(x-3)(x-4)(x^2+2x+8)=$$
$$=(x-2)(x-3)(x-4)((x+1)^2-3).$$
Easy to see that the quadratic factor is irreducible. 
