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According to Descartes' Rule of Signs, can the polynomial function have exactly $0$ positive real zeros? Choose your answer based on the rule only. $\newline$ $h(x) = x^6 + 2x^5 + 3x^4 + 9x^3 + 7x^2 + 7$ $\newline$ Choices: $\newline$ (A) yes $\newline$ (B) no

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Descartes' Rule of Signs

Full solution

Q. According to Descartes' Rule of Signs, can the polynomial function have exactly

0

positive real zeros? Choose your answer based on the rule only.

\newline

h(x) = x^6 + 2x^5 + 3x^4 + 9x^3 + 7x^2 + 7

\newline

Choices:

\newline

(A) yes

\newline

(B) no

Count Sign Changes: Count the number of sign changes in the coefficients of $h(x) = x^6 + 2x^5 + 3x^4 + 9x^3 + 7x^2 + 7$ . Coefficients: $1, 2, 3, 9, 7, 7$ . Sign changes: $0$ .
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. $\newline$ Since there are $0$ sign changes, $h(x)$ can have $0$ or $2$ or $4$ or $6$ positive real zeros.
Possible Positive Zeros: Can $h(x)$ have exactly $0$ positive real zeros? Yes, because there are $0$ sign changes, so it's possible for $h(x)$ to have $0$ positive real zeros.

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