f(x)=x^6+4x^5+4x^4+3x^3-x^2-5x+4
Find the maximum number of real zeros, and Use Descartes's rule of signs to determine how many positive and how many negative zeros the function has. You do not need to find the zeros.
find only rational zeros f(x)=x^5-9x64+21x^3+11x^2-100x=100
To find the maximum number of real zeros of a polynomial, we need to count the number of sign changes in the coefficients of the polynomial and apply Descartes's rule of signs.
The given polynomial is f(x) = x^6 + 4x^5 + 4x^4 + 3x^3 - x^2 - 5x + 4.
Step 1: Count the number of sign changes in the coefficients of the polynomial.
We arrange the coefficients in decreasing order of their exponents:
4, 4, 3, -1, -5, 4.
We observe the following sign changes: + to + (no change), + to + (no change), + to + (no change), + to - (1 change), - to + (1 change), and + to + (no change).
So, there are 2 sign changes.
Step 2: Determine the number of positive and negative zeros using Descartes's rule of signs.
The number of positive zeros is equal to the number of sign changes in the polynomial f(x) or less than that by an even number (i.e., 0 or 2 positive zeros).
The number of negative zeros is equal to the number of sign changes in the polynomial f(-x) or less than that by an even number (i.e., 0 or 2 negative zeros).
Since we have 2 sign changes, the number of positive zeros can be 0 or 2.
To find the number of negative zeros, we evaluate f(-x) replacing 'x' with '-x' in the given polynomial:
f(-x) = (-x)^6 + 4(-x)^5 + 4(-x)^4 + 3(-x)^3 - (-x)^2 - 5(-x) + 4
= x^6 - 4x^5 + 4x^4 - 3x^3 - x^2 + 5x + 4.
We repeat the sign change procedure for f(-x):
The coefficients of f(-x) are: 4, -4, 4, -3, -1, 5, 4.
We observe the following sign changes: + to - (1 change), - to + (1 change), + to - (1 change), - to + (1 change), + to - (1 change), and - to + (1 change).
So, there are 6 sign changes.
Since there are 6 sign changes in f(-x), the number of negative zeros can be 0, 2, 4, or 6.
Therefore, using Descartes's rule of signs, we can conclude that the given polynomial f(x) = x^6 + 4x^5 + 4x^4 + 3x^3 - x^2 - 5x + 4 has at most 2 positive real zeros and at most 6 negative real zeros.