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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. $\newline$ $g(x) = x^6 + 2x^5 + 6x^4 + x - 8$ $\newline$ 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 + 2x^5 + 6x^4 + x - 8

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the coefficients of $g(x) = x^6 + 2x^5 + 6x^4 + x - 8$ . Coefficients: $1, 2, 6, 0, 0, 1, -8$ . Sign changes: $0$ to $1$ (from $1$ to $-8$ ).
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 only $1$ sign change, $g(x)$ can have at most $1$ positive real zero.
Number of Positive Zeros: Since $g(x)$ can have at most $1$ positive real zero and we're asked if it can have exactly $1$ , the answer is yes, it can have exactly $1$ positive real zero.

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