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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. $\newline$ $g(x) = x^6 + 6x^3 + 8x^2 + x - 6$ $\newline$ 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^6 + 6x^3 + 8x^2 + x - 6

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the coefficients of $g(x) = x^6 + 6x^3 + 8x^2 + x - 6$ . Coefficients: $1, 0, 0, 6, 8, 1, -6$ Sign changes: $1$ (from $1$ to $-6$ )
Apply Descartes' Rule: 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 we have $1$ sign change, $g(x)$ can have $1$ or $0$ positive real zeros.
Determine Real Zeros: Since $g(x)$ can have $1$ positive real zero, the answer to the question is yes, $g(x)$ can have exactly $1$ positive real zero.

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