To determine the number of vertical asymptotes for the function
[tex]\[ F(x) = \frac{2}{(x-1)(x+3)(x+8)} \][/tex]
we need to identify where the function's denominator equals zero, because vertical asymptotes occur at these points, provided the numerator is non-zero at those points.
1. We begin by setting the denominator equal to zero:
[tex]\[ (x-1)(x+3)(x+8) = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex] by setting each factor equal to zero:
[tex]\[
\begin{align*}
x - 1 &= 0, & &\Rightarrow x = 1 \\
x + 3 &= 0, & &\Rightarrow x = -3 \\
x + 8 &= 0, & &\Rightarrow x = -8
\end{align*}
\][/tex]
Now, we have three distinct x-values where the denominator is zero: [tex]\( x = 1 \)[/tex], [tex]\( x = -3 \)[/tex], and [tex]\( x = -8 \)[/tex].
3. Check if the numerator is non-zero at these points, which it is (since it is just 2, a constant non-zero value).
Therefore, the function [tex]\( F(x) \)[/tex] has vertical asymptotes at the points [tex]\( x = 1 \)[/tex], [tex]\( x = -3 \)[/tex], and [tex]\( x = -8 \)[/tex].
Thus, the number of vertical asymptotes for the function [tex]\( F(x) \)[/tex] is:
[tex]\[ \boxed{3} \][/tex]
