How many irreducible monic quadratic polynomials are there in $\mathbb{F}_p[X]$? Can some of you help me with my homework?

I had to count the irreducible, quadratic, monic polynomials in $\mathbb{F}_p[X]$ for arbitrary $p$. 

I will show you what I tried myself.
Research effort
First of all the set $\{X^ 2 + bX +c : b, c \in \mathbb{F}_p \}$ contains $p^2$ elemenst. We need to substract the amount of elements that is reducible. These elements are exactly the set $A = \{(X+c)(X+d) : c, d \in \mathbb{F}_p \}$. There are $p(p-1)$ pairs of $(a,b) \in \mathbb{F}_p \times \mathbb{F}_p$ such that $a \neq b$, and there are $p$ pairs such that $a=b$.   
When $a$ and $b$ are unequal, whe know that $a$ and $b$ can be swapped without changing anything to the polynomial $(X+a)(X+b)$, so I obtained that $|A| \leq p +(p-1)- \frac{1}{2}(p-1)=p-\frac{1}{2}(p-1)$.
In other to find out these are the elements, or that there are still less, I thought I had to assume that $(X+c)(X+d) = (X+\gamma)(X+\delta)$, to deduce some information about the realation between $(c,d)$ and $(\gamma, \delta)$, but this led to taking square roots in $\mathbb{F}_p$, and I am not familiar with that.
Could you please provide me some more information?
 A: Monic reducible degree two polynomials have a root, so they are of the form $(X-a)(X-b)$.
There are $p$ polynomials of the form $(X+a)^2$ and $\binom{p}{2}$ of the form $(X+a)(X+b)$ with $a\ne b$ (as you computed). So
$$
p^2 - p - \binom{p}{2}=\frac{p(p-1)}{2}
$$
is the correct number.
Examples. For $p=2$ the reducible polynomials are $X^2$, $X(X+1)$ and $(X+1)^2$; so you get $4-3=1$, indeed the only one is $X^2+X+1$.
For $p=3$ we have $X^2$, $(X+1)^2$, $(X+2)^2$, $X(X+1)$, $X(X+2)$, $(X+1)(X+2)$. The three irreducible polynomials are $X^2+1$, $X^2+X+2$, $X^2+2X+2$.
A: Hint: the product of all monic irreducible polynomials of degree dividing $n$ with coefficients in $\mathbf F_p$ is $x^{p^n}-x$.
So, with $n=2$, you just have to exclude the linear monic polynomials...
A: I am sorry that I "answer" my question now, but it's quite unpractical to react on hints by comments only.

Oke, this is what I've got. The polynomial $x^{p^2}-x$ contains monic irreduible polynomials of degree $1$ as you said. I eliminated them by dividing: 
$$ \frac{x^{p^2}-x}{x^{p^1}-x} \quad = \quad \frac{x^ {p^2-1}-1}{x^{p-1}-1} \quad = \quad \frac{[x^{p-1}]^{p+1}-1}{x^{p-1}-1} \quad = \quad \sum_{k=0}^{p+1}x^k.$$
This polynomial can only have $\frac{1}{2}(p+1)$ quadratic factors in the case that $p\neq 2$. If $p=2$ we only have $X^2+X+1$. Is there something I forgot to include?
