what does Descartes rule of signs tell you about the possible number of positive real number zeros and the possible number of negative real zeros of the function?
1. f(x)=2x^5-5x^2+x-1
2. f(x)=5x^6+7x^4-x-6
please show work
What percent of the total population is found between the mean and a z-score of -0.50?
f(x) = 2x3 - 7x2 + 8x + 2
To determine the possible number of positive real zeros and negative real zeros of a polynomial function using Descartes' rule of signs, follow these steps:
1. For f(x) = 2x^5 - 5x^2 + x - 1:
Count the number of sign changes in the coefficients of the terms when writing the terms in descending order of powers. In this case, there are two changes (from positive to negative):
- Positive to negative: 2x^5 --> - 5x^2
- Negative to positive: - 5x^2 --> + x
Therefore, there are two possible positive real zeros.
2. For f(x) = 5x^6 + 7x^4 - x - 6:
Count the number of sign changes in the coefficients of the terms when writing them in descending order of powers. In this case, there are three changes (from positive to negative):
- Positive to negative: 5x^6 --> + 7x^4
- Negative to positive: + 7x^4 --> - x
- Positive to negative: - x --> - 6
Therefore, there are three possible positive real zeros.
To find the number of negative real zeros:
3. Replace x with -x in the given polynomial, f(x).
For f(-x) = 2(-x)^5 - 5(-x)^2 + (-x) - 1, simplify it to:
f(-x) = -2x^5 - 5x^2 - x - 1
4. Repeat steps 1 and 2 for f(-x) to find the number of possible negative real zeros.
- Number of sign changes for f(-x) = 3.
Therefore, there are three possible negative real zeros.