Note: I'm not asking how many monic irreducible quadratics there are.
How many reducible quadratics are there in Zmod5[X]? How many irreducible? Generalize: How many irreducible quadratics are there over a finite field of n elements?
I found that there are 60 reducible quadratics in Z mod 5 [X]. Since f(x) has the form of a(x-b)(x-c), a not equal to 0, then we can use the unique factorization theorem. a has 4 possible choices, b has 5, and when b = 0, c has 5 choices. b = 1, c has 4 choices and so on. Then I found the sum for the choices of c and multiplied this by 4 (choices for a). I know there are 15 monic reducible quadratics in Zmod5[Z]. I understand that I need to find how many total possible quadratics there are in Zmod5[X], but I'm not sure how to go about doing this. We haven't specifically learned this in class, and I'm wondering if there is intuitive in some way? Once I find how many total quadratics there are, I know I subtract 60 from this value to find the number of irreducible quadratics.
To generalize, we take the total number of quadratics and subtract the number of reducible quadratics there are. Equation thus far: (n-1)n^2 - (n-1)[1+2+3+...+(n-1)+(n)]
I'm not sure if this generalized formula is correct. I created it from when I was trying to find the "choices" for variables a, b and c. a has (n-1) choices and b has n choices. When b=0, c has n choices, b=1, c has (n-1) choices, b=2, c has (n-2) choices, and this continues until you reach b=(n-1).
Thanks in advance!
You've actually already solved the hard part of this. Finding the total number of quadratics can be done similarly. A quadratic is $$ax^2+bx+c$$ where $a\neq 0$, hence there are $4\cdot 5\cdot 5=100$.