Question: 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?

Answer: 
It is easy to check that $-1$ is a root, and by *Horner's algorithm*, you obtain 
$$x^5+3x^4+x^3+x^2+3=(x+1)(x^4+2x^3-x^2+2x-2).$$
Similarly, the quotient has $2$ as a root and
$$ x^4+2x^3-x^2+2x-2=(x-2)(x^3-x^2+2x+1).$$
The latter quotient has $-2$ as a root, and
$$x^3-x^2+2x+=(x+2)(x^2+2x+3),$$
so that $$x^5+3x^4+x^3+x^2+3=(x+1)(x-2)(x+2)(x^2+2x+3). $$
Now the quadratic factor is irreducible because it has no root. Indeed
$$ x^2+2x+3=(x+1)^2+2,$$
and the non-zero squares in $\mathbf Z/5\mathbf Z$ are $\pm 1$.
Actually, $2$, $3$, and $4$ are roots of that polynomial in $\mathbb Z\_5[x]$. So, your polynomial can be written as$$(x^2+ax+b)(x-2)(x-3)(x-4).$$Since $b\times(-4)=3$, $b=3$. It is not hard now to see that $a=2$.