jmc

First, we can apply difference of squares, to get $$x^{12}-1=(x^6-1)(x^6+1).$$We can apply difference of squares to $x^6-1$: $$x^6-1=(x^3-1)(x^3+1).$$These factor by difference of cubes and sum of cubes: $$(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1).$$Then by sum of cubes, $$x^6+1=(x^2+1)(x^4-x^2+1).$$Thus, the full factorization over the integers is $$x^{12}-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1),$$and there are $\boxed{6}$ factors.