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According to Descartes' Rule of Signs, can the polynomial function have exactly $1$ positive real zero, including any repeated zeros? Choose your answer based on the rule only. $\newline$ $g(x) = x^3 + 6x^2 - x - 9$ $\newline$ 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

1

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

\newline

g(x) = x^3 + 6x^2 - x - 9

\newline

Choices:

\newline

(A) yes

\newline

(B) no

Count Sign Changes: Count the number of sign changes in the coefficients of $g(x) = x^3 + 6x^2 - x - 9$ . $\newline$ Coefficients: $1, 6, -1, -9$ . $\newline$ Sign changes: $1$ (from $6$ to $-1$ ).
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's $1$ sign change, $g(x)$ can have $1$ or $0$ positive real zeros.
Number of Positive Zeros: Since $g(x)$ can have $1$ positive real zero, the answer to the question is yes.

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