Answer
Positive real zeros = 1 or 3
Negative real zeros = 0 or 2
Imaginary zeros = 4, 2 or 0
Work Step by Step
In the given polynomial of degree $5$, there are $5$ zeros.
Here, we have $h(x)=x^5-3x^3+8x-10$
By Descartes' Rule, we have $3$ sign changes, thus we have either $3$ or $1$ positive real roots.
Now, $h(-x)=-x^5+3x^3-8x+10$
The above polynomial shows that there are $2$ sign changes.
By Descartes' Rule, we have $2$ or $0$ negative real roots.
Thus, the other two roots must be imaginary.
Positive real zeros = 1 or 3
Negative real zeros = 0 or 2
Imaginary zeros = 4, 2 or 0
