To apply Descartes' Rule of Signs for determining the possible number of positive and negative real zeros of a polynomial, we examine the signs of the coefficients in two different forms of the polynomial:

1. For positive real zeros: Look at $f(x)$, the original polynomial.
2. For negative real zeros: Look at $f(-x)$, by substituting $-x$ for $x$ and examining the sign changes.

Step 1: Analyze $f(x)$ for positive real zeros

The given polynomial is:

$f(x) = 3x^5 - 13x^4 + 19x^3 - 5x^2 - 9x + 11$

Now, examine the signs of each term:

• $3x^5$ (positive)
• $-13x^4$ (negative)
• $19x^3$ (positive)
• $-5x^2$ (negative)
• $-9x$ (negative)
• $11$ (positive)

The sign changes between consecutive terms are:

1. From $3x^5$ to $-13x^4$ (positive to negative): 1 sign change.
2. From $-13x^4$ to $19x^3$ (negative to positive): 1 sign change.
3. From $19x^3$ to $-5x^2$ (positive to negative): 1 sign change.
4. From $-5x^2$ to $-9x$ (negative to negative): no sign change.
5. From $-9x$ to $11$ (negative to positive): 1 sign change.

So, there are 4 sign changes in $f(x)$.

According to Descartes' Rule of Signs, the number of possible positive real zeros is either the number of sign changes or that number minus an even integer. Therefore, the possible number of positive real zeros is:

• 4,
• 2, or

Step 2: Analyze $f(-x)$ for negative real zeros

Now, substitute $-x$ for $x$ in $f(x)$:

$f(-x) = 3(-x)^5 - 13(-x)^4 + 19(-x)^3 - 5(-x)^2 - 9(-x) + 11$ Simplifying: $f(-x) = -3x^5 - 13x^4 - 19x^3 - 5x^2 + 9x + 11$

Now, examine the signs of each term:

• $-3x^5$ (negative)
• $-13x^4$ (negative)
• $-19x^3$ (negative)
• $-5x^2$ (negative)
• $+9x$ (positive)
• $11$ (positive)

The sign changes between consecutive terms are:

1. From $-3x^5$ to $-13x^4$ (negative to negative): no sign change.
2. From $-13x^4$ to $-19x^3$ (negative to negative): no sign change.
3. From $-19x^3$ to $-5x^2$ (negative to negative): no sign change.
4. From $-5x^2$ to $9x$ (negative to positive): 1 sign change.
5. From $9x$ to $11$ (positive to positive): no sign change.

So, there is 1 sign change in $f(-x)$.

According to Descartes' Rule of Signs, the number of possible negative real zeros is either the number of sign changes or that number minus an even integer. Therefore, the possible number of negative real zeros is:

• 1.

Final Answer:

• The possible number of positive real zeros is 4, 2, or 0.
• The possible number of negative real zeros is 1.

Would you like further details or explanations? Here are some related questions to explore:

1. How does Descartes' Rule of Signs apply to polynomials with repeated factors?
2. How can we use synthetic division to confirm the number of real zeros?
3. What other methods can be used to approximate the roots of a polynomial?
4. How can we determine the number of complex roots of a polynomial?
5. How do changes in the degree of a polynomial affect the number of possible real zeros?

Tip: Remember that Descartes' Rule of Signs doesn't give the exact number of real zeros but only the possible numbers, so it should be used alongside other methods like synthetic division or graphing.