Obviously, we can factor out an $x$ first to get $x(x^8-1)$.
Next, we repeatedly factor differences of squares:
$$x(x^4+1)(x^4-1)$$
$$x(x^4+1)(x^2+1)(x^2-1)$$
$$x(x^4+1)(x^2+1)(x+1)(x-1)$$
None of these 5 factors can be factored further, so the answer is
$\boxed{\textbf{(B) } 5}$.