## Precalculus (6th Edition) Blitzer

number of positive real zeros: $3,1$ number of negative real zeros: $2, 0$
Step 1. Given the function $f(x)=2x^5-3x^3-5x^2+3x-1$ we can identify that there are 3 sign changes and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be $3,1$ Step 2. Examine: $f(-x)=2(-x)^5-3(-x)^3-5(-x)^2+3(-x)-1=-2x^5+3x^3-5x^2-3x-1$ We can identify that there are 2 sign changes and based on Descartes’s Rule of Signs, we can determine the possible number of negative real zeros to be $2, 0$