Question

My answer is C


What are the possible number of positive real, negative real, and complex zeros of f(x) = 8x4 + 13x3 - 11x2 + x + 9? (1 points)
Answer

Positive Real: 1
Negative Real: 3 or 1
Complex: 2 or 0

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

Positive Real: 3 or 1
Negative Real: 1
Complex: 2 or 0

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

Answer 1

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:

[D]

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

Answer 2

Thank you so much for the explanation. Very thorough! Thanks!!!!!

Answer 3

Np ^