Question 30357
f(x) = x^6 - 2x^5 + 4x^4 + 4x^3 + 5x^2 - 5x + 3

Number of variations in sign is four
=> Number of positive real zeros is 4, 2, or none

f(-x) =  x^6 + 2x^5 + 4x^4 - 4x^3 + 5x^2 + 5x - 3
Number of variations in sign is three 
=> Number of negative real zeros is 3 or 1 


The correct answer would be d (4, 2, or no positive real zeros; 3 or 1 negative real zero)