PLEASE HELP! For f(x) = -2x^5 + 3x^4 – 6x^2 + 1 what does the Descartes rule?
For f(x) = -2x^5 + 3x^4 – 6x^2 + 1 what does the Descartes rules of signs tell about the number of positive real zeros and the number of negative real zeros?
Descartes' rule is based on the number of sign changes in the non-zero coefficients when the powers of x are in descending order. There are 3 such changes (-2 to +3, +3 to -6, and -6 to +1).
The number of positive real zeroes is either equal to the number of sign changes, or less than that by a multiple of 2. That means there can either be 3 positive real zeroes or 1 positive real zero.
To get the number of negative real zeroes, you have to reverse the sign on the odd powers of x (or equivalently, look at P(-x) instead of P(x)). x^5 is the only odd power of x with a non-zero coefficient in our polynomial, so we're looking at the coefficients 2 3 -6 1. There are two sign changes, so there can either be 2 negative real zeroes or no negative real zeroes.
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Descartes' rule is based on the number of sign changes in the non-zero coefficients when the powers of x are in descending order. There are 3 such changes (-2 to +3, +3 to -6, and -6 to +1).
The number of positive real zeroes is either equal to the number of sign changes, or less than that by a multiple of 2. That means there can either be 3 positive real zeroes or 1 positive real zero.
To get the number of negative real zeroes, you have to reverse the sign on the odd powers of x (or equivalently, look at P(-x) instead of P(x)). x^5 is the only odd power of x with a non-zero coefficient in our polynomial, so we're looking at the coefficients 2 3 -6 1. There are two sign changes, so there can either be 2 negative real zeroes or no negative real zeroes.