What are the possible numbers of positive, negative, and complex zeros of f (x) = x6 - x5 - x4 + 4x3 - 12x2 + 12?

Using Descartes' rule of signs.

There are 4 changes of signs of coefficients:

+ to - (+x^6-x^5)

- to + (-x^4+4x^3)

+ to - (+4x^3-12x^2)

- to + (-12x^2+12)

Therefore, there are 4,2 or 0 positive roots at most. So it must be the third answer.