Question

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

Answer 1

$f(x) = -x^6 -x^5 - x^4 - 4x^3 - 12x^2 + 12$

There is only one change of sign, so there is only one possible positive root.

$f(-x) = -(-x)^6 -(-x)^5 - (-x)^4 - 4(-x)^3 - 12(-x)^2 + 12\\ f(-x) = -x^6 +x^5 - x^4 + 4x^3 - 12x^2 + 12$

There are five changes of signs, so there are 5,3 or 1 possible negative roots.

The number of complex roots can be equal to 4,2 or 0 (degree of a polynomial - possible positive roots - possible negative roots)

Answer 2

Answer:

Positive: 2 or 0; negative: 4, 2, or 0; complex: 6, 4, 2, or 0

Step-by-step explanation:

I am pretty sure this is right and I know that there are 2 sigh changes so there are for sure 2 or 0 positive