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.
 A: 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.
