According to Descartes' Rule of Signs, can the polynomial function have exactly 5 positive real zeros, including any repeated zeros? Choose your answer based on the rule only.f(x)=x7+8x5+x4+x3+8x2−7x−5Choices:[A]yes[B]no
Q. According to Descartes' Rule of Signs, can the polynomial function have exactly 5 positive real zeros, including any repeated zeros? Choose your answer based on the rule only.f(x)=x7+8x5+x4+x3+8x2−7x−5Choices:[A]yes[B]no
Apply Descartes' Rule of Signs: Let's apply Descartes' Rule of Signs to the polynomial function f(x)=x7+8x5+x4+x3+8x2−7x−5 to determine the number of possible positive real zeros. Descartes' Rule of Signs states that the number of positive real zeros of a polynomial function is equal to the number of sign changes between consecutive non-zero coefficients, or less than that by an even number.
Count sign changes in coefficients: We will count the number of sign changes in the coefficients of the polynomial. The coefficients are: 1 (for x7), 8 (for x5), 1 (for x4), 1 (for x3), 8 (for x2), x70 (for x71), and x72 (for the constant term). There is only one sign change, which occurs between the coefficient 8 (for x2) and x70 (for x71).
Determine possible positive real zeros: According to Descartes' Rule of Signs, the polynomial f(x) can have at most 1 positive real zero, since there is only one sign change. It cannot have exactly 5 positive real zeros because the number of positive real zeros must be equal to the number of sign changes or less than that by an even number. Since we have only one sign change, the possible numbers of positive real zeros are 1 or 0 (1 minus an even number, which in this case can only be 0).
Conclusion based on Descartes' Rule of Signs: Therefore, based on Descartes' Rule of Signs, the polynomial function f(x) cannot have exactly 5 positive real zeros.