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According to Descartes' Rule of Signs, can the polynomial function have exactly $4$ positive real zeros, including any repeated zeros? Choose your answer based on the rule only. $\newline$ $h(x) = x^4 + 5x^3 + 5x^2 + 4x + 1$ $\newline$ Choices: $\newline$ (A)yes $\newline$ (B)no

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Descartes' Rule of Signs

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

4

positive real zeros, including any repeated zeros? Choose your answer based on the rule only.

\newline

h(x) = x^4 + 5x^3 + 5x^2 + 4x + 1

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the coefficients of $h(x) = x^4 + 5x^3 + 5x^2 + 4x + 1$ . $\newline$ Coefficients: $1, 5, 5, 4, 1$ . $\newline$ There are no sign changes.
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. Since there are no sign changes, $h(x)$ can have $0$ positive real zeros.
Answer Explanation: Since $h(x)$ can have $0$ positive real zeros and we're asked if it can have exactly $4$ , the answer is no.

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