Questions: a (Fall 2024) CRN 18067 Question 11 of 30 This test: 30 point(s) possible This question: 1 point(s) possible Test Use Descartes' rule of signs to determine the possible number of positive and negative real zeros for the following function. Then, use a graph to determine the actual numbers of positive and negative real zeros. P(x)=7x^4+2x^3+3x-1 What is the possible number of positive real zeros? A. 1 B. 0 C. 3 or 1 D. 2 or 0 What is the possible number of negative real zeros? A. 3 or 1 B. 1 C. 0 D. 2 or 0 What is the actual number of positive real zeros? A. 1 B. 0 C. 3 or 1 D. 2 or 0 What is the actual number of negative real zeros? A. 3 or 1

Solution Steps

To solve this problem, we will use Descartes' rule of signs to determine the possible number of positive and negative real zeros of the polynomial \( P(x) = 7x^4 + 2x^3 + 3x - 1 \). Descartes' rule of signs states that the number of positive real zeros of a polynomial is equal to the number of sign changes between consecutive non-zero coefficients, or less than that by an even number. For negative real zeros, we apply the rule to \( P(-x) \).

1. Determine the possible number of positive real zeros:

   • Count the sign changes in \( P(x) \).

2. Determine the possible number of negative real zeros:

   • Substitute \( x \) with \( -x \) in \( P(x) \) to get \( P(-x) \).
   • Count the sign changes in \( P(-x) \).

3. Determine the actual number of positive and negative real zeros:

   • Use a graphing tool to plot the polynomial and count the actual number of positive and negative real zeros.

To determine the possible number of positive real zeros, we use Descartes' rule of signs on \( P(x) \).

\[ P(x) = 7x^4 + 2x^3 + 3x - 1 \]

We count the number of sign changes in the polynomial \( P(x) \):

• From \( 7x^4 \) to \( 2x^3 \): no sign change (both positive)
• From \( 2x^3 \) to \( 3x \): no sign change (both positive)
• From \( 3x \) to \( -1 \): one sign change (positive to negative)

There is 1 sign change, so the possible number of positive real zeros is 1.

To determine the possible number of negative real zeros, we use Descartes' rule of signs on \( P(-x) \).

\[ P(-x) = 7(-x)^4 + 2(-x)^3 + 3(-x) - 1 = 7x^4 - 2x^3 - 3x - 1 \]

We count the number of sign changes in the polynomial \( P(-x) \):

• From \( 7x^4 \) to \( -2x^3 \): one sign change (positive to negative)
• From \( -2x^3 \) to \( -3x \): no sign change (both negative)
• From \( -3x \) to \( -1 \): no sign change (both negative)

There is 1 sign change, so the possible number of negative real zeros is 1.

To determine the actual number of positive real zeros, we would graph the polynomial \( P(x) \) and observe the x-intercepts in the positive x-axis region. However, since we are not graphing here, we will use the information provided in the question.

The actual number of positive real zeros is given as one of the multiple-choice options. Based on the possible number of positive real zeros (1), the actual number of positive real zeros is:

\[ \boxed{\text{A. 1}} \]

Final Answer