When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is: 
$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$
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 
$\boxed{\textbf{(B) } 5}$.