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

I know that if the coefficients and roots are integers then for every $r \in \Bbb Z$ with $f(r) = 0$ we have that $r \mid 2010$ since for $f(x)=ax^2+bx+c$ we have that $f(0)=c = 2010$.
So every root $r$ of $f$ divides $2010$. Now $2010=2\cdot3\cdot5\cdot67$ so it has $16$ divisors.
What kind of consdierations I should do know? Since $\deg f=2$ from the $16$ possible choices for the roots only $2$ are applicable.
I don't quite know how I can turn this data to figure out the possible coefficients for $f.$

Hint
Let $x_1$ and $x_2$ the interger roots. You know that
$$x_1x_2=\frac{2010}{a}$$
$$x_1+x_2=-\frac ba$$
It means that $a$ divide $b$ and $2010$.
Can you finish?
