Use Descartes Rule of Signs to determine the possible number

Use Descartes\' Rule of Signs to determine the possible numbers of positive and negative real zeros of f(x) = x^3 + 8x^2 + 4x + 1. What are the possible numbers of positive real zeros? Square box (Use a comma to separate answers as needed.) What are the possible numbers of negative real zeros? square box (Use a comma to separate answers as needed.)

Solution

f(x)=x³+8x²+4x+1

there is no change of sign between the terms , so no positive real zeroes

possible numbers of positive real zeroes =0

f(-x)=-x³+8x²-4x+1

there is 3 changes of sign between the terms ,

possible numbers of negative real zeroes =3,1