Answer:
The number of positive real zeros is 2 or 0
Step-by-step explanation:
Given
![f(x)=x^5+3x^2-4x+2](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E5%2B3x%5E2-4x%2B2)
Required
Number of positive real zeros
Using Descartes rule of signs;
We write out the signs in front of each term;
Sign = + + - +
Count the number of times the sign alternate; i.e. from positive to negative and from negative to positive
From positive to negative, we have: 1 (i.e. + - )
From negative to positive, we have: 1 (i.e. - +)
Add up the count
![count = 1+1](https://tex.z-dn.net/?f=count%20%3D%201%2B1)
![count = 2](https://tex.z-dn.net/?f=count%20%3D%202)
<em>Hence, the number of positive real zeros is 2 or 0</em>