Answer
Positive real zeros = 1 or 3
Negative real zeros= 1 or 3
Imaginary zeros = 4, 2 or 0
Work Step by Step
In the given polynomial of degree $6$ , there are $6$ zeros.
Here, we have $g(x)=x^6+x^5-3x^4+x^3+5x^2+9x-18$
By Descartes' Rule, we have $3$ sign changes, thus we have either $3$ or $1$ positive real roots.
Now, $g(-x)=x^6-x^5-3x^4-x^3+5x^2-9x-18$
The above polynomial shows that there are $3$ sign changes.
By Descartes' Rule, we have $3$ or $1$ negative real roots.
Thus, the other two roots must be imaginary.
Hence, Positive real zeros = 1 or 3
Negative real zeros= 1 or 3
Imaginary zeros = 4, 2 or 0
