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According to Descartes' Rule of Signs, can the polynomial function have exactly 00 positive real zeros? Choose your answer based on the rule only.\newlineh(x)=x3−3x2+8x+6h(x) = x^3 - 3x^2 + 8x + 6\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 00 positive real zeros? Choose your answer based on the rule only.\newlineh(x)=x3−3x2+8x+6h(x) = x^3 - 3x^2 + 8x + 6\newlineChoices:\newline(A)yes\newline(B)no
  1. Count Sign Changes: Count the number of sign changes in the coefficients of h(x)=x3−3x2+8x+6h(x) = x^3 - 3x^2 + 8x + 6.\newlineCoefficients: 1,−3,8,61, -3, 8, 6.\newlineSign changes: 11 (from 11 to −3-3), 22 (from −3-3 to 88).
  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 than that by an even number.\newlineSo, h(x)h(x) can have 22 or 00 positive real zeros.
  3. Number of Positive Zeros: Since h(x)h(x) can have 00 positive real zeros, the answer to the question is yes, h(x)h(x) can have exactly 00 positive real zeros.

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