Factoring in terms of Irreducibles Factor the polynomial $x^5 + 2x^3 + 3x^2 + 1$ as a product of irreducible polynomials in $\mathbb{Z}_5[x]$.
My thoughts: I know what the definition of an irreducible function is but as far as methods to find the correct answer as easy as possible is difficult to understand.

You found the linear factors already, leaving the quartic $x^4 + 3x^3 + x^2 + x + 3$. The only possibility is that this factors into two quadratics, thus:
$$
(x^2+ax+b)(x^2+cx+d)\equiv x^4 + 3x^3 + x^2 + x + 3\pmod5
$$
which you can expand as
$$
x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd\equiv x^4 + 3x^3 + x^2 + x + 3\pmod5
$$
which you can solve as 5 simultaneous equations.

The equations are:
$$1\equiv1,\ a+c\equiv3,\ ac+b+d\equiv1,\ ad+bc\equiv1,\ bd\equiv3$$
You can drop the first one, which is already solved:
$$a+c\equiv3,\ ac+b+d\equiv1,\ ad+bc\equiv1,\ bd\equiv3$$
Now there are only two ways to get 3 mod 5 as a product of two numbers: 1 times 3 or 2 times 4 (in either order, of course). Since the problem is symmetrical with respect to switching (a,b) and (c,d), you can assume without loss of generality that $b\le d$ if you like, so there are two possibilities: $b=1,d=3$ and $b=2,d=4$. The first:
$$a+c\equiv3,\ ac+4\equiv1,\ 3a+c\equiv1,\ 3\equiv3$$
and the second:
$$a+c\equiv3,\ ac+1\equiv1,\ 4a+2c\equiv1,\ 3\equiv3$$
Simplifying both you get
$$a+c\equiv3,\ ac\equiv2,\ 3a+c\equiv1$$
and
$$a+c\equiv3,\ ac\equiv0,\ 4a+2c\equiv1$$
Solve both, using $c\equiv3-a$, and see if you can find any solutions.
