### Instruction: Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010.$.### Response: To find a second-degree polynomial with integer coefficients and integer zeros, I can start by writing it in factored form: $f(x) = (x-a)(x-b)$, where $a$ and $b$ are the zeros.
If I expand this product, I get $f(x) = x^2 - (a+b)x + ab$.
Now, I know that $f(0) = 2010$, which means that $ab = 2010$.
So I need to find all the possible pairs of integers $(a,b)$ that multiply to 2010.
I can use prime factorization to help me with this.
I know that $2010 = 2 \times 3 \times 5 \times 67$.
So any pair of factors $(a,b)$ must be of the form $a = 2^p 3^q 5^r 67^s$ and $b = 2^{1-p} 3^{1-q} 5^{1-r} 67^{1-s}$, where $p,q,r,s$ are either 0 or 1.
This gives me 16 possible pairs of factors, since I have two choices for each exponent.
However, some of these pairs are repeated, since swapping $a$ and $b$ does not change the polynomial.
For example, $(a,b) = (2, 1005)$ and $(a,b) = (1005, 2)$ are the same pair.
So I only need to count the distinct pairs of factors.
One way to do this is to assume that $a \leq b$, so that I avoid swapping.
Then I can list the pairs of factors in increasing order of $a$: $(1, 2010), (2, 1005), (3, 670), (5, 402), (6, 335), (10, 201), (15, 134), (30, 67)$.
I see that there are 8 distinct pairs of factors, so there are 8 second-degree polynomials with the given properties.