Using the function and Descartes' Rule of Signs, what are possible combinations of positive, negative, and imaginary zeros? - 3 positive, 1 negative, 0 imaginary
- 4 positive, 0 negative, 0 imaginary
- 0 positive, 0 negative, 4 imaginary
- 2 positive, 1 negative, 1 imaginary
- 1 positive, 1 negative, 2 imaginary

3 Positive, 1 Negative, 0 Imaginary

1 Positive, 1 Negative, 2 Imaginary

General Formulas and Concepts:

Alg II

Step-by-step explanation:

Step 1: Define function

f(x) = 3x⁴ - 7x³ + 5x - 4

Step 2: Descartes Pt. 1

Find the sign changes in the function.

3x⁴ - 7x³

7x³ - 3x²

-3x₂ + 5x

5x - 4

In the positive function, we have 3 sign changes.

Step 3: Descartes' Pt. 2

Find the sign changes in the function with x as negative.

f(-x) = 3x⁴ + 7x³ - 3x² - 5x - 4

3x⁴ + 7x³

7x³ - 3x²

-3x² - 5x

-5x - 4

In the -x function, we have only 1 sign change.

Step 4: Analyze Descartes

Positive x = 3 sign changes

Negative x = 1 sign change

Possible Combinations:

1 positive, 1 negative, 2 imaginary or 3 positive 1 negative, 0 imaginary

the correct option is b

step-by-step explanation: