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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)=x6+2x5+3x4+9x3+7x2+7h(x) = x^6 + 2x^5 + 3x^4 + 9x^3 + 7x^2 + 7\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)=x6+2x5+3x4+9x3+7x2+7h(x) = x^6 + 2x^5 + 3x^4 + 9x^3 + 7x^2 + 7\newlineChoices:\newline(A) yes\newline(B) no
  1. Count Sign Changes: Count the number of sign changes in the coefficients of h(x)=x6+2x5+3x4+9x3+7x2+7h(x) = x^6 + 2x^5 + 3x^4 + 9x^3 + 7x^2 + 7. Coefficients: 1,2,3,9,7,71, 2, 3, 9, 7, 7. Sign changes: 00.
  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 are 00 sign changes, h(x)h(x) can have 00 or 22 or 44 or 66 positive real zeros.
  3. Possible Positive Zeros: Can h(x)h(x) have exactly 00 positive real zeros? Yes, because there are 00 sign changes, so it's possible for h(x)h(x) to have 00 positive real zeros.

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