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 factor out the lowest power of \( x \) from the polynomial \( x^8 - 6x^6 - 9x^4 + x^3 \). Factoring out \( x^3 \), we rewrite the polynomial as \( x^3(x^5 - 6x^3 - 9x + 1) \). Next, we consider the remaining polynomial \( x^5 - 6x^3 - 9x + 1 \) for positive real zeros. To do this, we look for sign changes in the coefficients: 1. Positive: \( 1 \) (no sign change) 2. Negative: \( 6 \) (sign change) 3. Negative: \( -9 \) (sign change) 4. Positive: \( 1 \) (sign change) Counting the sign changes, we find three sign changes in total, which indicates there could be 3, 1, or no positive real zeros. Now, let's sum it up: There are **3 possible positive real zeros** from our analysis using Descartes' rule of signs!