Apply Descartes' Rule of Signs. Determine the number of possible positive and negative real zeros for the given function. \( t(x)=\frac{1}{9} x^{6}+\frac{1}{5} x^{4}+\frac{1}{4} x^{2}+1 \) 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 Number of possible negative real zeros:

To determine the number of possible positive real zeros using Descartes’ Rule of Signs, we look at the function \( t(x) = \frac{1}{9} x^{6}+\frac{1}{5} x^{4}+\frac{1}{4} x^{2}+1 \). The coefficients are all positive, so there are no sign changes. This means there are 0 possible positive real zeros. For negative real zeros, we can analyze \( t(-x) = \frac{1}{9} (-x)^{6}+\frac{1}{5} (-x)^{4}+\frac{1}{4} (-x)^{2}+1 = \frac{1}{9} x^{6}+\frac{1}{5} x^{4}+\frac{1}{4} x^{2}+1 \). Again, all coefficients are positive with no sign changes. Therefore, there are also 0 possible negative real zeros. So the final answers are: Part 1: 0 Part 2: 0