A cubic or any other polynomial of odd degree has at least one real root, simply because if the leading coefficient is positive, then g(x) tends towards -∞ as x-> -∞, and g(x) tends towards +∞ as x-> + ∞. There must be at least one real root in between the two extremes.
To find if there are two more, it will be necessary calculate the critical points, namely where g'(x)=0.
If there are no solutions to g'(x)=0, it means that there are no critical points (no local maximum or minimum). Consequently there will be only one real root, and two complex roots.
g(x) = 9x³-7x²+10x-4
g'(x) = 27x²-14x+10
Since the discriminant of the quadratic is 14²-4*10*27=-884 <0
we conclude that there are no critical points where g'(x) = 0. In other words, there is one real root and two complex roots.
To find if the (real) root is positive or negative, we apply Descartes' rule of signs.
The polynomial g(x) has 3 changes of signs, so there are 3 or 1 positive roots. On the other hand, the polynomial g(-x) has no change of signs, so there are no negative roots.
Since we have determined that there are two complex roots, we conclude that there is one positive root and two complex roots.
state the number of positive real zeros, negative real zeros, and imaginary zeros for g(x) = 9x^3 - 7x^2 +10x - 4
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