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According to Descartes' Rule of Signs, can the polynomial function have exactly 11 positive real zero, including any repeated zeros? Choose your answer based on the rule only.\newlineg(x)=x3+6x2−x−9g(x) = x^3 + 6x^2 - x - 9\newlineChoices:\newline(A) yes\newline(B) no

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Full solution

Q. According to Descartes' Rule of Signs, can the polynomial function have exactly 11 positive real zero, including any repeated zeros? Choose your answer based on the rule only.\newlineg(x)=x3+6x2−x−9g(x) = x^3 + 6x^2 - x - 9\newlineChoices:\newline(A) yes\newline(B) no
  1. Count Sign Changes: Count the number of sign changes in the coefficients of g(x)=x3+6x2−x−9g(x) = x^3 + 6x^2 - x - 9.\newlineCoefficients: 1,6,−1,−91, 6, -1, -9.\newlineSign changes: 11 (from 66 to −1-1).
  2. Descartes' Rule of Signs: According to Descartes' Rule of Signs, the number of positive real zeros is equal to the number of sign changes or less by an even number.\newlineSince there's 11 sign change, g(x)g(x) can have 11 or 00 positive real zeros.
  3. Number of Positive Zeros: Since g(x)g(x) can have 11 positive real zero, the answer to the question is yes.

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