Question

What are the possible number of positive, negative, and complex zeros of f(x) = -3x4 - 5x3 - x2 - 8x + 4 ?

A) Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0
B) Positive: 1; Negative: 3 or 1; Complex: 2 or 0
C) Positive: 3 or 1; Negative: 1; Complex: 2 or 0
D) Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

Answer 1

the numbers after the x's are exponents, correct?

Answer 2

yes

Answer 3

there are 2 positive and 2 complex roots

Answer 4

i've tried using descartes rule of signs but i got confused to be honest
so i solved it

Answer 5

one root lie b/w 0 and 1

Answer 6

right

Answer 7

other will be negative
as f(x) is increasing in R+

Answer 8

Am i right????

Answer 9

no 2 positive

Answer 10

yeah 1 will be positive and one will not be possitive it can also be possilbe that roots may be irrational becuase irrational roots also exist in pairs

Answer 11

there is 1 sign change so by DRS there is 1 positive root
Sub -x for x to get negative roots, 3 sign changes so 3 or 1 negative roots
Complex roots are total roots, 4, less positive and negative roots so 0 or 2 of those.

Answer is B