Seriously! That short?

Since it has integral coefficients and roots, $f(x) = a(x-b)(x-c)$ for some integers $a,b,c$ . Thus $f(0) = a(-b)(-c) = abc = 2010$ . We have two cases:

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Case 1: $b = c$

This case reduces to $ab^2 = 2010$ . However, $b^2 \ge 0$ , so $a$ is non-negative, and $2010$ is square-free, so all prime divisors of $2010$ divide $a$ (and not $b$ ). This means $a = 2010$ . We have two choices for $b$ , either $1$ or $-1$ , giving $f(x) = 2010(x-1)^2$ and $f(x) = 2010(x+1)^2$ .

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Case 2: $b \neq c$

Note that $f(x) = a(x-b)(x-c)$ and $f(x) = a(x-c)(x-b)$ give the same polynomial, so we must halve the count at the end if we assume $b,c$ are distinguishable.

First, the four prime divisors of $2010$ (namely $2,3,5,67$ ) can go to any of the three numbers $a,b,c$ freely. That is, for each $p \in \{2,3,5,67\}$ , exactly one of $a,b,c$ is divisible by $p$ ; in addition, none of them is divisible by $p^2$ . However, we're free to select which of the three of $a,b,c$ is the one that's divisible by $p$ . This gives $3^4 = 81$ ways to choose the magnitudes.

Next, we can choose their signs as well. Whatever signs we select for $a,b$ , there is exactly one sign for $c$ that keeps the product positive. Thus there are $2^2 = 4$ ways to choose the signs, and thus $81 \cdot 4 = 324$ triples in total.

However, this counts the two triples with $b = c$ , so we're left with $322$ remaining. Furthermore, as described above, we need to halve this count since $b,c$ are indistinguishable, for a total of $161$ triples.

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In total, there are $2+161 = \boxed{163}$ triples.

$P(x) = ax^{2} + bx + 2010$ . It implies that $a|2010$ . Divide the problems by the possible values of $a$ we got 163 different polinomial $P(x)$ .