smc

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 $\boxed{5}$.