state the number of positive real zeros, negative real zeros, and imaginary zeros for g(x) = 9x^3 - 7x^2 +10x - 4

To determine the number of positive real zeros, negative real zeros, and imaginary zeros for the function g(x) = 9x^3 - 7x^2 + 10x - 4, we can use Descartes' Rule of Signs and the complex conjugate theorem.

1. Descartes' Rule of Signs:
- Count the number of sign changes in the coefficients of g(x).
- The number of sign changes represents the maximum number of positive real zeros.

Coefficient Sign Pattern: + - + +
Number of Sign Changes: 2
Maximum Number of Positive Real Zeros: 2

2. Apply Descartes' Rule of Signs to g(-x):
- Substitute -x for x in g(x) and count the number of sign changes.
- The number of sign changes represents the maximum number of negative real zeros.

Coefficient Sign Pattern: + + + +
Number of Sign Changes: 0
Maximum Number of Negative Real Zeros: 0

3. Complex Conjugate Theorem:
- If a polynomial has complex coefficients and a + bi is a zero, then a - bi is also a zero.
- Since g(x) has real coefficients, complex zeros occur in conjugate pairs.

Since all the coefficients in g(x) are real, the number of imaginary zeros will be equal to the difference between the total number of zeros and the sum of positive and negative real zeros.

In this case, since we haven't found any real zeros, the number of imaginary zeros will be equal to the number of zeros of g(x).

Therefore, the number of positive real zeros is 2, the number of negative real zeros is 0, and the number of imaginary zeros is 3.

In summary:
- Number of positive real zeros: 2
- Number of negative real zeros: 0
- Number of imaginary zeros: 3

To determine the number of positive real zeros, negative real zeros, and imaginary zeros for the function g(x) = 9x^3 - 7x^2 + 10x - 4, we can use Descartes' Rule of Signs and the Fundamental Theorem of Algebra.

1. Begin by arranging the terms of the function in descending order:
g(x) = 9x^3 - 7x^2 + 10x - 4

2. Next, count the number of sign changes in the coefficients of the polynomial when you write it in its standard form:
The sign changes in the coefficients of g(x) are: 0 times from positive to negative.

3. According to Descartes' Rule of Signs, this means that the number of positive real zeros of g(x) is either 0 or an even number.

4. Calculate the number of sign changes in the coefficients when you substitute -x for x in the polynomial:
g(-x) = 9(-x)^3 - 7(-x)^2 + 10(-x) - 4

Simplify:
g(-x) = -9x^3 - 7x^2 - 10x - 4

The sign changes in the coefficients of g(-x) are: 2 times from negative to positive.

5. According to Descartes' Rule of Signs, this means that the number of negative real zeros of g(x) is either 2 or an even number.

6. Finally, using the Fundamental Theorem of Algebra, which states that a polynomial of degree n has exactly n complex zeros (counting multiplicity), we can conclude that the number of imaginary zeros of g(x) is equal to the total number of zeros (positive real, negative real, and complex) minus the number of real zeros.

To find the total number of zeros, we need to use alternative methods like factoring the polynomial or using numerical techniques such as graphing or using a calculator.

Therefore, based on Descartes' Rule of Signs, we know that g(x) has either 0 or 2 positive real zeros and either 2 or 0 negative real zeros. The number of imaginary zeros is equal to the total number of zeros minus the number of real zeros, but without further information, we cannot determine the exact number of imaginary zeros.

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.