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

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**Step 1: Examine the polynomial in standard form**

The function is
$$
t(x)=\frac{1}{5000}x^{6}+\frac{1}{500}x^{4}+\frac{1}{50}x^{2}+1.
$$
The coefficients are, in order of descending powers:
$$
\frac{1}{5000},\ 0,\ \frac{1}{500},\ 0,\ \frac{1}{50},\ 0,\ 1.
$$
Ignoring the zero coefficients, we have:
$$
\frac{1}{5000},\ \frac{1}{500},\ \frac{1}{50},\ 1.
$$

**Step 2: Apply Descartes' Rule of Signs for positive zeros**

Examine the signs of the coefficients in the polynomial:
$$
\frac{1}{5000} > 0,\quad \frac{1}{500} > 0,\quad \frac{1}{50} > 0,\quad 1 > 0.
$$
There are no sign changes. Therefore, by Descartes' Rule of Signs, the number of possible positive real zeros is 
$$
0.
$$

**Step 3: Apply Descartes' Rule of Signs for negative zeros**

To find the possible negative real zeros, substitute $$-x$$ for $$x$$ in the polynomial:
$$
t(-x)=\frac{1}{5000}(-x)^{6}+\frac{1}{500}(-x)^{4}+\frac{1}{50}(-x)^{2}+1.
$$
Since the exponents are even, we have:
$$
(-x)^6=x^6,\quad (-x)^4=x^4,\quad (-x)^2=x^2.
$$
Thus,
$$
t(-x)=\frac{1}{5000}x^{6}+\frac{1}{500}x^{4}+\frac{1}{50}x^{2}+1.
$$
The coefficients are the same as in $$t(x)$$:
$$
\frac{1}{5000} > 0,\quad \frac{1}{500} > 0,\quad \frac{1}{50} > 0,\quad 1 > 0.
$$
There are no sign changes. Therefore, by Descartes' Rule of Signs, the number of possible negative real zeros is 
$$
0.
$$

**Answer:**

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

Part 2 of 2  
Number of possible negative real zeros: $$0$$
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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