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 $f(x)=x^5+7x^4-4x^3-3x^2+9x-15$
By Descartes' Rule, we have $3$ sign changes, thus we have either $3$ or $1$ positive real root.
Now, $f(-x)=-x^5+7x^4+4x^3-3x^2-9x-15$
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.
Hence, positive real zeros = 1 or 3
Negative real zeros= 0 or 2
Imaginary zeros = 4, 2 or 0
