Final answer:
Using Descartes' Rule of Signs on the polynomial f(x) = 2x⁷ - 3x⁶ + 5x⁴ + x³ - 6x² + 1, we find four sign changes suggesting up to four positive real zeros. Subtracting even numbers from four gives us possible counts of four, two, or zero positive real zeros. The best choice from the options given is Option C) 4.
Step-by-step explanation:
To determine one of the best possible numbers of positive real zeros for the polynomial function f(x) = 2x⁷ - 3x⁶ + 5x⁴ + x³ - 6x² + 1, we use Descartes' Rule of Signs. This rule states that the number of positive real zeros of a polynomial function is either equal to the number of sign changes between consecutive coefficients or it is less than that number by an even number. By examining the given function, we can observe the sign changes: +2 to -3 (one change), -3 to +5 (second change), +5 to +1 (no change), +1 to -6 (third change), and -6 to +1 (fourth change). Therefore, there are four sign changes, which suggests that the maximum number of positive real zeros is four.
However, since the Rule of Signs indicates that the actual number of positive real zeros can be less than the number of sign changes by an even number, we subtract even numbers from four to find the possible number of positive real zeros. Subtracting two from four yields two, and subtracting four from four yields zero. Thus, the best possible numbers of positive real zeros are four, two, or zero. Among the provided options, the best is Option C) 4.