Question: Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros…

Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros. Be sure to include all possibilities.

\[ D(x) = 6x^4 + 13x^3 + 36x^2 + 65x + 30 \]

Answer: How to enter your answer (opens in new window) 4 Points

Separate multiple answers with commas.

Number of Positive Real Zeros:

Number of Negative Real Zeros:

Solution

Let’s solve the problem step by step using Descartes’ Rule of Signs. First, consider the polynomial function: \[ \text{D}(x) = 6x^4 + 13x^3 + 36x^2 + 65x + 30 \] Number of Positive Real Zeros: Count the number of sign changes in \( \text{D}(x) \). All coefficients are positive, so there are no sign changes. \[ \text{Number of sign changes in } \text{D}(x) = 0 & \] 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: \[ \text{Number of Positive Real Zeros} = 0 & \] Number of Negative Real Zeros: Replace \( x \) with \( -x \) in \( \text{D}(x) \): \[ \text{D}(-x) = 6x^4 - 13x^3 + 36x^2 - 65x + 30 \] Count the number of sign changes in \( \text{D}(-x) \). \[ \text{Number of sign changes in } \text{D}(-x) = 4 & \] According to Descartes’ Rule of Signs, the number of negative real zeros is equal to the number of sign changes or less than that by an even number. Therefore: \[ \text{Number of Negative Real Zeros} = 4, 2, \text{ or } 0 & \] Final Answer: 1. Number of Positive Real Zeros: 0 2. Number of Negative Real Zeros: 4, 2, or 0