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According to Descartes' Rule of Signs, can the polynomial function have exactly $2$ positive real zeros, including any repeated zeros? Choose your answer based on the rule only. $\newline$ $h(x) = x^6 - 6x^3 - 4x^2 - 6x + 1$ $\newline$ 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

2

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

\newline

h(x) = x^6 - 6x^3 - 4x^2 - 6x + 1

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the coefficients of $h(x) = x^6 - 6x^3 - 4x^2 - 6x + 1$ . Coefficients: $1, 0, 0, -6, -4, -6, 1$ . Sign changes: $0$ to $-6$ ( $1$ change), $-6$ to $1$ ( $1$ change). Total sign changes: $2$ .
Descartes' Rule of Signs: According to Descartes' Rule of Signs, the number of positive real zeros is either equal to the number of sign changes or less than that by an even number. $\newline$ With $2$ sign changes, $h(x)$ can have $2$ or $0$ positive real zeros.
Positive Real Zeros: Since $h(x)$ can have $2$ positive real zeros, the answer to the question prompt is yes.

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