How many such polynomial exist? 
Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$.

I got:
$$P(x) = ax^2 + bx + c \implies P(0) = c = 2010$$
Let $P(r_1, r_2) = 0$ thus, $r_1 + r_2 = -\frac{b}{a}$ and $r_1 \cdot r_2 = \frac{c}{a} = \frac{2010}{a}$. 
Since: $2010 = 1 \cdot 2 \cdot 3 \cdot 5 \cdot 67$ there are $10$ values for $a$ so $r_1r_2$ is an integer. That is:
$a \in (-67, -5, -3, -2, -1, 1, 2, 3, 5, 67)$. 
But then since $r_1 + r_2 = -\frac{b}{a}$, $b$ has infinite values. so I cant solve the problem.
HINTS ONLY PLEASE!
 A: Hint: You've done most of the (non-combinatorial) thinking necessary here.  For any given $a$, it suffices to find all pairs of integers $r_1$ and $r_2$ such that
$$
r_1 \cdot r_2 = \frac{2010}{a}
$$
Or, in other words, find all triples of integers $r_1,r_2,a$ satisfying
$$
r_1 \cdot r_2 \cdot a = 2010
$$
From there, set
$$
b = -a(r_1 + r_2)
$$
Also, you're missing quite a few valid values for $a$; you should find that there are $2^5 = 32$ possibilities.

Further Hint
For any pair $r_1,r_2$ such that $r_1r_2$ divides $2010 = 2 \cdot 3 \cdot 5 \cdot 67$, there is a unique suitable polynomial with these roots.
The number of ways to divide the (multi)set 
$$
\{-1,-1,2,3,5,67\}
$$
into $3$ subsets is given by $\frac{1}{2}(3^6 + 3^4)$.
We can have a repeated root if $r_1 = r_2 = 1$ or $r_1 = r_2 = -1$, and exactly two associated polynomials. In all other cases, the two roots are distinct.  The corresponding polynomials were counted once in the above consideration, but all others were counted twice. 
A: Hint: $b$ cannot take infintely many values, since $r_1r_2=\frac{2010}{a}\leq2010$. 
If $b$ gets too large, then $r_1+r_2$ gets to large. 
If you want another hint, feel free to comment. 
