Use Descartes' rule of signs to determine the total number of real zeros and the number of positive and negative real zeros. (Hint: First factor out \( x \) to its lowest power.) Part 1 of 3 Number of possible positive real zeros: N \( x^{8}-6 x^{6}-9 x^{4}+x^{3} \)

To apply Descartes' rule of signs, we first need to factor out \( x \) from the polynomial \( x^{8} - 6x^{6} - 9x^{4} + x^{3} \). 1. **Factor out \( x \)**: \[ x^{8} - 6x^{6} - 9x^{4} + x^{3} = x^{3}(x^{5} - 6x^{3} - 9x + 1) \] Now we will analyze the polynomial \( x^{5} - 6x^{3} - 9x + 1 \) to determine the number of positive real zeros. 2. **Count the sign changes for positive real zeros**: - The coefficients of \( x^{5} - 6x^{3} - 9x + 1 \) are: \( 1, 0, -6, -9, 1 \). - The signs of the coefficients are: \( +, 0, -, -, + \). Now, we will list the signs: - \( +1 \) (for \( x^{5} \)) - \( 0 \) (for \( x^{4} \)) - \( -6 \) (for \( x^{3} \)) - \( -9 \) (for \( x^{2} \)) - \( +1 \) (for \( x^{0} \)) The sign changes occur as follows: - From \( +1 \) to \( -6 \): 1 change - From \( -6 \) to \( -9 \): 0 changes - From \( -9 \) to \( +1 \): 1 change Thus, there are a total of 2 sign changes. According to Descartes' rule of signs, the number of positive real zeros is equal to the number of sign changes or less than that by an even number. Therefore, the possible number of positive real zeros is: - 2 or 0. Thus, the number of possible positive real zeros is \( N = 2 \) or \( N = 0 \).