What are the possible number of positive, negative, and complex zeros of f(x) = -x6 + x5- x4 + 4x3 - 12x2 + 12 I am really stuck on what to do here

Question

precal · Answer

6th power tells you that you have 6 solutions

jit4won · Answer

factor theorem should be tried

precal · Answer

I would use rational thm to figure out one zero and then use synthetic division to help factor down to the second power

anonymous · Answer

Holy crap seems like a long process lol

precal · Answer

http://www.wolframalpha.com/input/?i=factor+x%5E6%2Bx%5E5-x%5E4%2B4x%5E3-12x%5E2%2B12 not a nice one to factor

precal · Answer

2 real and 4 complex solutions

precal · Answer

I think there is a sign change analysis test you can use to determine the number of solutions.  Not sure what the name is, perhaps Descartes sign change (hopefully someone who knows will come post it)