Question

Use​ Descartes' Rule of Signs to determine the possible numbers of positive and negative real zeros of f left parenthesis x right parenthesis equals x cubed plus 9 x squared plus 3 x plus 4.
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Part 1
What are the possible numbers of positive real​ zeros?
  
enter your response here  ​(Use a comma to separate answers as​ needed.)

Answer 1

To apply Descartes' Rule of Signs to the polynomial \( f(x) = x^3 + 9x^2 + 3x + 4 \), we need to analyze the signs of the coefficients.

**Step 1: Count the Sign Changes for Positive Real Zeros**

Evaluate the coefficients of \( f(x) \):
- Coefficients are \( 1, 9, 3, 4 \) (all are positive).

Since there are no sign changes in \( f(x) \):
- **Number of positive real zeros = 0.**

**Part 1 Answer:**

0

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Now, for negative real zeros, we'll evaluate \( f(-x) \).

**Step 2: Calculate \( f(-x) \)**

\[
f(-x) = (-x)^3 + 9(-x)^2 + 3(-x) + 4 = -x^3 + 9x^2 - 3x + 4
\]

**Step 3: Count the Sign Changes for Negative Real Zeros**

Now we evaluate the coefficients:
- Coefficients of \( f(-x) \) are \( -1, 9, -3, 4 \).

Identifying the signs:
- \(-1 \) (negative)
- \( 9 \) (positive)
- \(-3 \) (negative)
- \( 4 \) (positive)

Now let's count the sign changes:
1. From \(-1\) to \(9\) (1 change)
2. From \(9\) to \(-3\) (1 change)
3. From \(-3\) to \(4\) (1 change)

There are **3 sign changes**, so the possible numbers of negative real zeros are:
- 3, or 1 (subtracting by 2 for each alternate possibility).

**Part 2 Answer:**

3, 1 (Use a comma to separate answers as needed.)