Write $f(x) = a(x - r)(x - s),$ so $a \cdot rs = 2010 = 2 \cdot 3 \cdot 5 \cdot 67;$ count choices of $a$ and the unordered root pair

When $|r| \ne |s|,$ split the four primes among $|a|,$ $|r|,$ $|s|$ and check that all four sign patterns of $(r, s)$ work. Treat $|r| = |s| = 1$ separately.

Write $f(x) = a(x - r)(x - s)$ with integer roots $r$ and $s;$ such a polynomial is determined by $a$ and the unordered pair $\{r, s\}.$ The condition $f(0) = 2010$ says $a \cdot rs = 2010 = 2 \cdot 3 \cdot 5 \cdot 67.$ Since $2010$ is squarefree, each of the four primes goes entirely to one of $|a|,$ $|r|,$ $|s|.$

First suppose $|r| \ne |s|.$ Choosing which $k \ge 1$ of the four primes divide the roots ($\binom{4}{k}$ ways) and splitting those primes between the two roots ($2^{k-1}$ unordered ways) gives $\sum_{k=1}^{4} \binom{4}{k} 2^{k-1} = 4 + 12 + 16 + 8$ $= 40$ choices of magnitudes. For each, the four sign patterns $(+,+),$ $(+,-),$ $(-,+),$ $(-,-)$ of $(r, s)$ are distinct and each forces the sign of $a,$ giving $4 \cdot 40 = 160$ polynomials.

If $|r| = |s|,$ squarefreeness forces $|r| = |s| = 1,$ so $|a| = 2010:$ the options are roots $1, 1$ or $-1, -1$ with $a = 2010,$ or roots $1, -1$ with $a = -2010,$ adding $3$ more. In total $160 + 3 = 163.$