To apply Descartes's Rule of Signs and determine the number of positive and negative real zeros for the polynomial $P(x) = x^4 + x^3 + x^2 + x + 13$, follow these steps:

1. Positive Real Zeros:

Descartes's Rule of Signs states that the number of positive real zeros is determined by the number of sign changes in $P(x)$.

The polynomial is: $P(x) = x^4 + x^3 + x^2 + x + 13$

• All the terms are positive, and there are no sign changes.

Thus, there are 0 positive real zeros.

2. Negative Real Zeros:

To determine the possible number of negative real zeros, we evaluate $P(-x)$. This is done by substituting $-x$ for $x$ in the polynomial.

$P(-x) = (-x)^4 + (-x)^3 + (-x)^2 + (-x) + 13$

This simplifies to: $P(-x) = x^4 - x^3 + x^2 - x + 13$

Now, let's count the sign changes in $P(-x)$: $P(-x) = x^4 \ (-) \ x^3 \ (+) \ x^2 \ (-) \ x \ (+) \ 13$

There are 4 sign changes in $P(-x)$. According to Descartes's Rule of Signs, the number of negative real zeros could be 4, 2, or 0, since the actual number of real zeros is reduced by an even number from the number of sign changes.

3. Total Number of Real Zeros:

The degree of the polynomial $P(x) = x^4 + x^3 + x^2 + x + 13$ is 4, which means it has a total of 4 real or complex zeros.

Summary:

• Number of positive real zeros: 0
• Number of negative real zeros: 4, 2, or 0
• Total number of real zeros (including both positive, negative, and complex zeros): 4

Let me know if you'd like any more details!

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Follow-up questions:

1. What does Descartes's Rule of Signs tell us about complex zeros?
2. How does the degree of a polynomial determine the total number of zeros?
3. Can a polynomial with no sign changes still have real zeros?
4. Why does Descartes's Rule of Signs reduce the number of zeros by an even number?
5. How do we confirm the actual number of real and complex zeros?

Tip:

The total number of real and complex zeros of a polynomial always equals its degree (counting multiplicity).