Question

what are the possible number of positive negative and complex zeros of f(x)= -x6+x5-x4+4x3-12x2+12

Answer 1

Counting sign changes, since the exponents are in the very helpful descending order and all the exponents are integers.

... 5! Okay, there are 5, 3, or 1 positive Real zeros. There MUST be a positive Real zero.

Counting sign changes in f(-x)...1 ! There MUST be a negative Real zero.

Where does that leave us?

Answer 2

4 and 2 complex

Answer 3

Maybe. We know for sure there are 2 Real. There may be 4 Real. There may be 6 Real. We have to prove it.

The next step, in my view, wuld be to find any Real Rational zeros. These can be ONLY +/- 1, 2, 3, 4, 6, 12. Do you know why?

Answer 4

Sure about this statement? "We know for sure there are 2 Real. There may be 4 Real. There may be 6 Real."

Answer 5

(Also, good luck finding rational roots - this problem doesn't ask for that, which is a good thing).

Answer 6

It's a matter of looking.

f(0) > 0
f(1) > 0
f(2) < 0 so there is one on (1,2) and finding this result also suggests there cannot be one greater than x = 2 so we are done on the positive side.
f(-1) < 0 so there is one on (-1,0)

In any case, we've long since answered the qeustion.

Answer 7

Indeed.

Answer 8

^ doesn't answer the question.

Answer 9

"what are the possible number of...?"