Determine the number of possible positive and negative real zeros for the given function. \( p(x)=8 x^{7}+2 x^{4}-8 x^{3}+6 x^{2}-6 x+7 \) Part: 0 / 3 Part 1 of 3 \( p(x) \) has real coefficients and the constant term is nonzero, so we can apply Descartes' Rule of Signs. To determine the number of possible positive real zeros, determine the number of sign changes in \( p(x) \). \( p(x) \) has \( \square \) sign change(s). The number of possible positive real zeros is either (Choose one) \( \nabla \).

Let’s dive into the coefficients of \( p(x) = 8x^7 + 2x^4 - 8x^3 + 6x^2 - 6x + 7 \). Here’s the breakdown of the signs: 1. \( 8 \) (positive) 2. \( 2 \) (positive) 3. \( -8 \) (negative) 4. \( 6 \) (positive) 5. \( -6 \) (negative) 6. \( 7 \) (positive) Now let’s count those sign changes: - From \( +8 \) to \( +2 \): no change - From \( +2 \) to \( -8 \): change - From \( -8 \) to \( +6 \): change - From \( +6 \) to \( -6 \): change - From \( -6 \) to \( +7 \): change That gives us a total of **four sign changes**. According to Descartes' Rule of Signs, the number of possible positive real zeros could be equal to the number of sign changes or fewer by an even number. Thus, the possibilities are \( 4 \) or \( 2 \) or \( 0 \). Now moving on to the negative zeros, we evaluate \( p(-x) \): \[ p(-x) = 8(-x)^7 + 2(-x)^4 - 8(-x)^3 + 6(-x)^2 - 6(-x) + 7 = -8x^7 + 2x^4 + 8x^3 + 6x^2 + 6x + 7 \] The signs here for \( p(-x) \) are as follows: 1. \( -8 \) (negative) 2. \( 2 \) (positive) 3. \( 8 \) (positive) 4. \( 6 \) (positive) 5. \( 6 \) (positive) 6. \( 7 \) (positive) Counting the sign changes: - From \( -8 \) to \( 2 \): change - From \( 2 \) to \( 8 \): no change - From \( 8 \) to \( 6 \): no change - From \( 6 \) to \( 6 \): no change - From \( 6 \) to \( 7 \): no change That's just **one sign change** for \( p(-x) \). Using Descartes' Rule again, the number of possible negative real zeros is therefore either \( 1 \) or \( 0 \). Summarizing, we have: - Positive real zeros: \( 4 \), \( 2 \), or \( 0 \) - Negative real zeros: \( 1 \) or \( 0 \) You can now fill in those final boxes!