2. Let $a$, $b$, $c$ be integers, and for all real numbers $x$, we have
$$
(x-a)(x-2009)-2=(x-b)(x-c)
$$

holds. Then the number of all such ordered triples $(a, b, c)$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6
2. B.

From the problem, we have
$$
\begin{array}{l}
x^{2}-(a+2009) x+2009 a-2 \\
=(x-b)(x-c)=0 . \\
\text { Hence } \Delta=(a+2009)^{2}-4(2009 a-2) \\
=(a-2009)^{2}+8
\end{array}
$$

is a perfect square, let it be $n^{2}\left(n \in \mathbf{N}_{+}\right)$. Then
$$
(n-a+2009)(n+a-2009)=8 \text {. }
$$

Since $n-a+2009$ and $n+a-2009$ have the same parity and are both greater than 0, we have
$$
\left\{\begin{array} { l } 
{ n - a + 2 0 0 9 = 2 , } \\
{ n + a - 2 0 0 9 = 4 }
\end{array} \text { or } \left\{\begin{array}{l}
n-a+2009=4 , \\
n+a-2009=2 .
\end{array}\right.\right.
$$

Solving these, we get $(a, n)=(2010,3),(2008,3)$.
When $a=2008$, $\frac{(a+2009) \pm 3}{2}$ are the roots of the equation, i.e.,
$$
(b, c)=(2010,2007),(2007,2010) ;
$$

When $a=2010$, $\frac{(a+2009) \pm 3}{2}$ are the roots of the equation, i.e.,
$$
(b, c)=(2011,2008),(2008,2011) .
$$

Therefore, the ordered triples $(a, b, c)$ that satisfy the conditions are as follows:
$$
\begin{array}{l}
(2008,2010,2007),(2008,2007,2010), \\
(2010,2011,2008),(2010,2008,2011) .
\end{array}
$$