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:

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1. Write the function:  
$$ t(x)=\frac{1}{9} x^{6}+\frac{1}{5} x^{4}+\frac{1}{4} x^{2}+1 $$.

2. For positive real zeros, consider the original function. The coefficients are  
$$\frac{1}{9}, \frac{1}{5}, \frac{1}{4},$$ and $$1$$,  
all of which are positive. Since there is no change in sign between consecutive coefficients, by Descartes' Rule of Signs there are $$0$$ possible positive real zeros.

3. For negative real zeros, substitute $$-x$$ for $$x$$:  
$$
t(-x)=\frac{1}{9} (-x)^{6}+\frac{1}{5} (-x)^{4}+\frac{1}{4} (-x)^{2}+1.
$$  
Since all exponents are even,  
$$
(-x)^{6}=x^{6},\quad (-x)^{4}=x^{4},\quad (-x)^{2}=x^{2},
$$  
and the function becomes  
$$
t(-x)=\frac{1}{9} x^{6}+\frac{1}{5} x^{4}+\frac{1}{4} x^{2}+1.
$$  
The coefficients remain the same (all positive), so there are no sign changes. By Descartes' Rule of Signs there are $$0$$ possible negative real zeros.

Positive real zeros: 0  
Negative real zeros: 0
Positive real zeros: 0 Negative real zeros: 0

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