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Apply Descartes' Rule of Signs. Determine the number of possible positive and negative real zeros for the given function. \( t(x)=\frac{1}{5000} x^{6}+\frac{1}{500} x^{4}+\frac{1}{50} x^{2}+1 \) 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

Ask by Perkins Wyatt. in the United States
Mar 15,2025

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Part 1 of 2 Number of possible positive real zeros: 0 Part 2 of 2 Number of possible negative real zeros: 0

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To determine the number of possible positive real zeros using Descartes' Rule of Signs, we first analyze the function \( t(x) = \frac{1}{5000} x^{6} + \frac{1}{500} x^{4} + \frac{1}{50} x^{2} + 1 \). This function consists of all positive coefficients, meaning there are no sign changes. Hence, according to Descartes' Rule, there are no positive real zeros. For negative real zeros, we evaluate \( t(-x) \): \[ t(-x) = \frac{1}{5000} (-x)^{6} + \frac{1}{500} (-x)^{4} + \frac{1}{50} (-x)^{2} + 1 = \frac{1}{5000} x^{6} + \frac{1}{500} x^{4} + \frac{1}{50} x^{2} + 1 \] Again, the signs do not change as all coefficients remain positive. Thus, there are no negative real zeros. **Part 1:** 0 **Part 2:** 0

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