What are the possible numbers of positive, negative, and complex zeros of f(x)= -x^6 + x^5 - x^4 + 4x^3 - 12x^2 + 12?

Using Descartes' Rule of Signs:
The signs are: - + - + - +
There are 5 signs changes in this sequence, so there could be either 5, 3, or 1 positive roots.
If we negate the terms with odd numbers (x^5, x^3), we end up with the signs: - - - - - +
Since there is 1 sign change, there can be only 1 negative root.
This means the positive and negative roots can either be 6, 4, or 2.
Since the total number of roots cannot exceed 6, there are either 0, 2, or 4 complex roots.

Ibelieve its 0 but i am not sure