jmc

Let $a$ and $b$ be the integer roots. Then we can write $$f(x)=k(x-a)(x-b)$$for some integer $k$. Setting $x=0$, we get $$2010=kab.$$Since $2010=2\cdot 3\cdot 5\cdot 67$, there are $3^4$ possible ways to assign the prime factors of $2010$ to $a$, $b$, and $k$; then there are four choices for the signs of $a$, $b$, and $k$ (either all positive, or two negative and one positive), giving $3^4\cdot 4=324$ triples total. Two of these triples have $a=b$ (namely, $a=b=1$ and $k=2010$, and $a=b=-1$ and $k=2010$). Of the other $324-2=322$, we must divide by $2$ because the order of $a$ and $b$ does not matter. Therefore, the final count is $$2+\frac{322}{2}=\boxed{163}.$$