Math Problem Statement

use Descartes's rule of signs to describe the number of possible zero's of f(x)=2x^4+3x^3-9x^2-17-6 and find the number of possible postive zeros, possible negative zero's

Solution

Step 1: Descartes's Rule of Signs

Descartes's rule of signs helps us determine the number of positive and negative real zeros of a polynomial based on the number of sign changes in the polynomial's terms.

Function:

f(x)=2x4+3x39x217x6f(x) = 2x^4 + 3x^3 - 9x^2 - 17x - 6

Number of Possible Positive Zeros:

To find the possible number of positive real zeros, we check the sign changes in f(x)f(x).

  1. The signs of the terms in f(x)f(x) are: 2x4(+),3x3(+),9x2(),17x(),6()2x^4 \quad (+), \quad 3x^3 \quad (+), \quad -9x^2 \quad (-), \quad -17x \quad (-), \quad -6 \quad (-)

  2. The sign changes are between:

    • From 3x33x^3 to 9x2-9x^2 (positive to negative)

Thus, there is one sign change. According to Descartes’s rule of signs, the number of possible positive zeros is 1.

Number of Possible Negative Zeros:

To find the possible number of negative real zeros, we substitute x-x into f(x)f(x) and analyze the sign changes in f(x)f(-x).

Substitute x-x into f(x)f(x): f(x)=2(x)4+3(x)39(x)217(x)6f(-x) = 2(-x)^4 + 3(-x)^3 - 9(-x)^2 - 17(-x) - 6 Simplifying: f(x)=2x43x39x2+17x6f(-x) = 2x^4 - 3x^3 - 9x^2 + 17x - 6

Now, the signs of the terms in f(x)f(-x) are: 2x4(+),3x3(),9x2(),+17x(+),6()2x^4 \quad (+), \quad -3x^3 \quad (-), \quad -9x^2 \quad (-), \quad +17x \quad (+), \quad -6 \quad (-)

The sign changes are between:

  • From 2x42x^4 to 3x3-3x^3 (positive to negative)
  • From +17x+17x to 6-6 (positive to negative)

Thus, there are two sign changes in f(x)f(-x). According to Descartes's rule of signs, the possible number of negative zeros is 2 or 0 (since the number of real zeros decreases by even numbers).

Final Conclusion:

  • Possible number of positive real zeros: 1
  • Possible number of negative real zeros: 2 or 0

Do you need further clarification on any step or details?


Here are five related questions:

  1. How does Descartes's rule of signs apply to more complex polynomials with higher degrees?
  2. What happens if there are no sign changes in f(x)f(x)?
  3. How can we verify the actual number of real zeros after finding the possible number using Descartes’s rule of signs?
  4. Does Descartes's rule give information about complex (non-real) zeros?
  5. What additional methods can be used to precisely locate the zeros of a polynomial?

Tip:

Descartes’s rule only tells you the possible number of real zeros—it doesn’t guarantee how many real zeros there are, so further analysis or graphing may be needed.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomials
Descartes's Rule of Signs

Formulas

f(-x) substitution to determine sign changes

Theorems

Descartes's Rule of Signs

Suitable Grade Level

Grades 10-12