Question

PLEASE HELP !!!! What are the possible number of positive, negative, and complex zeros of f(x) = –2x3 – 5x2 – 6x + 4 ?

Answer 1

descartes rule of signs
f(x)=-2x^3-5x^2-6x+4
f(-x)=2x^3-5x^2+6x+4
sign changes in f(x) -> 1
sign changes in f(-x) -> 2

possible positive, negative, complex

positive 1 1
negative 2 0
complex 2

Answer 2

there will always be exactly 1 positive real zero, there can be either 2 or 0 negative real zeros