There are a number of ways to do this with calculus, but I imagine that it may be a bit "high brow" of a way of doing things. You can compute where the local maxima/minima are, then use that information to pinpoint where roots could be. At that point you can guess and check.
For your example, though, there is a very easy factorization (which takes far less effort than other approaches). terms 1+2, 3+4, and 5+6 together have a factor of x-1, so you get (x\^4 - 2x\^2 + 1)(x-1). The first term in this factorization is just (x\^2 - 1)\^2, which further factors into (x-1)\^2 \* (x+1)\^2. Thus you get a complete factorization of (x-1)\^3 \* (x+1)\^2, so this polynomial has a root of multiplicity 3 at 1, and a root of multiplicity 2 at -1.