Apply Descartes' Rule of Signs. Determine the number of possible positive and negative real zeros for the given function. \[ p(x)=5 x^{6}-7 x^{5}+7 x^{4}+4 x^{3}+6 x^{2}+8 x+2 \] Each part may have more than one possibility. Use commas to separate multiple responses. Part 1 of 2 Number of possible positive real zeros: Part 2 of 2

To determine the number of possible positive real zeros using Descartes' Rule of Signs, we look at the sign changes of \( p(x) \). The function is \( p(x) = 5x^6 - 7x^5 + 7x^4 + 4x^3 + 6x^2 + 8x + 2 \). Analyzing the signs, we have: - \( 5 \) (positive) - \( -7 \) (negative) - \( +7 \) (positive) - \( +4 \) (positive) - \( +6 \) (positive) - \( +8 \) (positive) - \( +2 \) (positive) There is 1 sign change (from \( 5 \) to \( -7 \)). Therefore, according to Descartes' Rule, there is exactly **1 possible positive real zero**. Now for the negative real zeros, we analyze \( p(-x) \): \[ p(-x) = 5(-x)^6 - 7(-x)^5 + 7(-x)^4 + 4(-x)^3 + 6(-x)^2 + 8(-x) + 2 \] This simplifies to: \[ p(-x) = 5x^6 + 7x^5 + 7x^4 - 4x^3 + 6x^2 - 8x + 2 \] The signs are 5 (positive), 7 (positive), 7 (positive), -4 (negative), 6 (positive), -8 (negative), and 2 (positive). We observe: - From \( +5 \) to \( +7 \): no change - From \( +7 \) to \( -4 \): 1 change - From \( -4 \) to \( +6 \): 1 change - From \( +6 \) to \( -8 \): 1 change - From \( -8 \) to \( +2 \): 1 change Totaling 4 sign changes, we conclude there are **4 or 2 or 0 negative real zeros**. So, summarizing: Part 1: **1** Part 2: **4, 2, 0**