What are the possible numbers of positive real, negative real, and complex zeros of f(x) = 8x^4 + 13x^3 − 11x^2 + x + 9?

By Descartes' Rules of Signs, you need to find the sign changes for f(x) and f(-x) [Positive and Negative Case - respectively]

f(x) = 8x^4 + 13x³ - 11x² + x + 9

Since there are two sign changes between terms, we can show that there are possibly 2 positive roots.

f(-x) = 8x^4 - 13x³ - 11x² - x + 9

For the negative case, you let x = -x. Then, we have 8x^4 - 13x³ - 11x² - x + 9. There are 2 possible negative roots since there are 2 sign changes. Hence, the possible choice is:

 

Positive Real: 2 or 0
Negative Real: 2 or 0
Complex: 4, 2 or 0


i know i am late and i hope you still need the help ! :D

Ithink you’re answering is b