Question

Determine one of the best possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f(x) = 2x^7 - 3x^6 + 5x^4 + x^3 - 6x^2 + 1 by using Descartes' Rule of Signs.**

A) 1, 1, 5
B) 2, 2, 3
C) 2, 3, 2
D) 1, 3, 3

Answer 1

Using Descartes' Rule of Signs for the function f(x), there can be either 3 or 1 positive real zeros, and either 2 or 0 negative real zeros. The total degree of the polynomial is 7, which means there must be 7 zeros in total. The option that fits this analysis is Option A, with 1 positive real zero, 1 negative real zero, and 5 imaginary zeros.

Step-by-step explanation:

To determine the number of positive real zeros, negative real zeros, and imaginary zeros for the function f(x) = 2x7 - 3x6 + 5x4 + x3 - 6x2 + 1, we use Descartes' Rule of Signs. First, we count the number of sign changes in the original polynomial to find the maximum number of positive real zeros. The polynomial has 3 sign changes, suggesting up to 3 positive real zeros, but by Descartes' Rule, this number could also be lower by an even integer, hence 3 or 1.

Next, we evaluate f(-x) to find the number of negative real zeros. f(-x) is -2x7 - 3x6 - 5x4 + x3 + 6x2 + 1, which has 2 sign changes, indicating there may be 2 or 0 negative real zeros.

Since the polynomial is of degree 7, the total number of zeros (real and imaginary) must be 7. If we assume the maximum number of positive real zeros (3) and the maximum number of negative real zeros (2), we would have 5 real zeros and thus 2 imaginary zeros (as complex zeros always come in pairs). Keeping the rule in mind that complex zeros come in pairs, the possible scenarios could be (3 positive real, 0 negative real, 4 imaginary) or (1 positive real, 2 negative real, 4 imaginary).

The best possible number of positive real zeros is either 3 or 1, the number of negative real zeros is either 2 or 0, and the number of imaginary zeros would accordingly be 2 or 4 to make up the total of 7 zeros. Looking at the options provided, the answer that fits this analysis is Option A: 3 or 1 positive real zeros, 2 or 0 negative real zeros, and 4 or 2 imaginary zeros. Therefore, Option A: 1 positive real zero, 1 negative real zero, and 5 imaginary zeros is the best fit according to the scenario with the least number of real zeros, which allows for more imaginary zeros.