Use Decartes Rule of signs to determine the possible number of postive and negative real zeros for the given function.
1. f(x)=-7x^9+x^5-x^2+6
This is what I did:
There are 2 sign changes
f(-x)=-7(-x)^9+(-x)^5-(-x)^2+6
There are 2 sign changes
So, are there 2 or 0 positive zeros, 2 or 0 negative zeros?
2. f(x)10x^3-8x^2+x+5
There are 2 sign changes
f(-x)=10(-x)^3-8(-x)^2+(-x)+5
There are 2 sign changes
So, are there 2 or 0 positive zeros, no tive zeros?
To use the Descartes' Rule of Signs, follow these steps:
1. Count the number of sign changes in the coefficients of the original polynomial function f(x).
- In the case of f(x) = -7x^9 + x^5 - x^2 + 6, there are 2 sign changes.
2. Count the number of sign changes in the coefficients of the polynomial obtained by replacing x with -x in f(x), i.e., f(-x).
- For f(-x) = -7(-x)^9 + (-x)^5 - (-x)^2 + 6, there are also 2 sign changes.
3. Determine the possible number of positive and negative real zeros of the function based on the number of sign changes:
- For f(x), there can be 2 or 0 positive zeros.
- For f(x), there can also be 2 or 0 negative zeros.
Applying the same steps to the second function, f(x) = 10x^3 - 8x^2 + x + 5, we get:
1. There are 2 sign changes in the coefficients of f(x).
2. For f(-x) = 10(-x)^3 - 8(-x)^2 + (-x) + 5, there are 2 sign changes as well.
3. Thus, for f(x), there can be 2 or 0 positive zeros, and no negative zeros.
Remember that Descartes' Rule of Signs provides an estimate of the number of positive and negative real zeros, but it does not give the exact values or locations of the zeros.