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.

\[ t(x) = 2x^4 - 5x^3 + 5x^2 - 5x + 3 \]

Answer

How to enter your answer (opens in new window)

Separate multiple answers with commas.

Number of Positive Real Zeros: _____

Number of Negative Real Zeros: _____

Solution

To determine the possible number of positive and negative real zeros for the function \( f(x) = 2x^4 - 5x^3 + 5x^2 - 5x + 3 \) using Descartes’ Rule of Signs, follow these steps: Step 1: Count the number of sign changes in \( f(x) \) to determine the possible number of positive real zeros. \[ f(x) = 2x^4 - 5x^3 + 5x^2 - 5x + 3 \] There are 4 sign changes in \( f(x) \): - \( 2x^4 \) to \( -5x^3 \) (positive to negative) - \( -5x^3 \) to \( 5x^2 \) (negative to positive) - \( 5x^2 \) to \( -5x \) (positive to negative) - \( -5x \) to \( 3 \) (negative to positive) According to Descartes’ Rule of Signs, the number of positive real zeros is either equal to the number of sign changes or less than it by an even number. Therefore, the possible number of positive real zeros is 4, 2, or 0. Step 2: Count the number of sign changes in \( f(-x) \) to determine the possible number of negative real zeros. \[ f(-x) = 2(-x)^4 - 5(-x)^3 + 5(-x)^2 - 5(-x) + 3 = 2x^4 + 5x^3 + 5x^2 + 5x + 3 \] There are 0 sign changes in \( f(-x) \). Therefore, there are 0 negative real zeros. Number of Positive Real Zeros: 4, 2, 0 Number of Negative Real Zeros: 0