8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five distinct integers satisfying the condition
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=9
$$

If $b$ is an integer root of the equation
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)=2009
$$

then the value of $b$ is $\qquad$.
8. 10 .

Notice
$$
\left(b-a_{1}\right)\left(b-a_{2}\right)\left(b-a_{3}\right)\left(b-a_{4}\right)\left(b-a_{5}\right)=2009,
$$

and $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five different integers, so all $b-a_{1}, b-a_{2}, b-a_{3}, b-a_{4}, b-a_{5}$ are also five different integers.
$$
\begin{array}{l}
\text { Also, } 2009=1 \times(-1) \times 7 \times(-7) \times 41, \text { then } \\
b-a_{1}+b-a_{2}+b-a_{3}+b-a_{4}+b-a_{5}=41 .
\end{array}
$$

From $a_{4}+a_{2}+a_{3}+a_{4}+a_{5}=9$, we get $b=10$.