we are given
![3x^5-x^4-5x^2+1=0](https://tex.z-dn.net/?f=3x%5E5-x%5E4-5x%5E2%2B1%3D0)
We can use Descartes rule of sign
This is rule is used to find number positive real roots and negative real roots
Positive real roots:
![3x^5-x^4-5x^2+1=0](https://tex.z-dn.net/?f=3x%5E5-x%5E4-5x%5E2%2B1%3D0)
We will find number of sign changes in this equation
We can it changes from +3 to -1 and -5 to +1
so, number of sign changes =2
so, total number of positive real roots =2
Negative real roots:
We can replace x as -x
![3(-x)^5-(-x)^4-5(-x)^2+1=0](https://tex.z-dn.net/?f=3%28-x%29%5E5-%28-x%29%5E4-5%28-x%29%5E2%2B1%3D0)
now, we can simplify it
![-3x^5-x^4-5x^2+1=0](https://tex.z-dn.net/?f=-3x%5E5-x%5E4-5x%5E2%2B1%3D0)
now, we will find number of sign changes
we can see that it changes from -5 to +1 only
so, number of sign changes =1
so, number of negative real roots =1
Complex roots:
We are given
degree of polynomial =5
so, total number of roots=5
we know that total root is sum of all roots
Total number of roots = number of positive real roots + number of negative real roots + number of complex roots
now, we can plug values
5=2+1+number of complex roots
so, we get
Number of complex roots =2.........Answer