Reducible and Irreducible Quadratics in $\mathbb Z_5[x]$

Question

Determine how many reducible monic quadratics there are in $\mathbb Z_5[x]$.

Now determine how many reducible and irreducible quadratics there are in $\mathbb Z_5[x]$.

I know there are 100 quadratics in $\mathbb Z_5[x]$ with 25 being monic (I believe). I think there are 7 irreducible monic quadratics(degree 1 polynomials, $x^2+2,x^2+3, x^2+x+1$), thus there are 18 monic reducible quadratics. This number seems unreasonably high. Am I even close with this?

Answer 1

A reducible monic polynomial is of the form $x^2+2bx+c$, where $b^2-c$ is a square, that is $\Delta=b^2-c\in\{0,1,4\}$.

For any of these cases, we can assign $b$ any value and determine $c$. So $15$.

In another way: a reducible monic polynomial has the form $(x-h)(x-k)$, where $h$ and $k$ are the roots. Five cases for $h=k$; ten cases for $h\ne k$, because $$ \binom{5}{2}=10 $$ (the order of the roots is irrelevant).

Another way. The field $F_{25}$ with $25$ elements is generated by any root of an irreducible polynomial; there are $25-5=20$ elements that generate $F_{25}$, but the minimal polynomial of any of these $20$ elements has two distinct roots. So we have $10$ irreducible polynomials.

Answer 2

Here is a variant of egreg's answer.

The field $\mathbb F_{25}$ is generated by any root of an irreducible quadratic polynomial. Since all elements of $\mathbb F_{25}$ are roots of $x^{25}-x$, the irreducible quadratic polynomial correspond to the irreducible quadratic factors of $x^{25}-x$. Factoring out $x^5-x$, which is a product of $5$ linear factors, leaves a polynomial of degree $20$ that is the product of irreducible quadratic polynomials. Therefore, there are $10$ irreducible quadratic polynomials.