Questions: Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of f(x)=-9x^3+7x^2-x+5 What is the possible number of positive real zeros? (Use a comma to separate answers as needed.) What is the possible number of negative real zeros? (Use a comma to separate answers as needed.)

Solution Steps

To determine the possible number of positive and negative real zeros of the polynomial \( f(x) = -9x^3 + 7x^2 - x + 5 \) using Descartes's Rule of Signs, follow these steps:

1. Positive Real Zeros: Count the number of sign changes in \( f(x) \).
2. Negative Real Zeros: Count the number of sign changes in \( f(-x) \).

To find the possible number of positive real zeros of the polynomial \( f(x) = -9x^3 + 7x^2 - x + 5 \), we count the number of sign changes in the coefficients of \( f(x) \). The coefficients are \(-9\), \(7\), \(-1\), and \(5\). The sign changes occur as follows:

• From \(-9\) to \(7\) (change)
• From \(7\) to \(-1\) (change)
• From \(-1\) to \(5\) (change)

Thus, there are 3 sign changes, indicating that the possible number of positive real zeros is \(3\).

Next, we find the possible number of negative real zeros by evaluating \( f(-x) \). We have:

\[ f(-x) = -9(-x)^3 + 7(-x)^2 - (-x) + 5 = 9x^3 + 7x^2 + x + 5 \]

The coefficients of \( f(-x) \) are \(9\), \(7\), \(1\), and \(5\). There are no sign changes in these coefficients:

• From \(9\) to \(7\) (no change)
• From \(7\) to \(1\) (no change)
• From \(1\) to \(5\) (no change)

Thus, there are 0 sign changes, indicating that the possible number of negative real zeros is \(0\).

Final Answer