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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. $\newline$ $g(x) = x^7 - 5x^6 - 7x^5 - 3x^2 + 2x - 2$ $\newline$ 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^7 - 5x^6 - 7x^5 - 3x^2 + 2x - 2

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the coefficients of $g(x) = x^7 - 5x^6 - 7x^5 - 3x^2 + 2x - 2$ . Coefficients: $1, -5, -7, 0, 0, -3, 2, -2$ . Sign changes: $1$ to $-5$ (no change), $-5$ to $-7$ (no change), $-7$ to $0$ (no change), $0$ to $-3$ (no change), $-3$ to $1, -5, -7, 0, 0, -3, 2, -2$ $1$ (change), $1, -5, -7, 0, 0, -3, 2, -2$ $1$ to $1, -5, -7, 0, 0, -3, 2, -2$ $3$ (change). Total sign changes: $1, -5, -7, 0, 0, -3, 2, -2$ $1$ .
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 than that by an even number. $\newline$ So, $g(x)$ can have $2$ or $0$ positive real zeros.
Number of Positive Zeros: Since $g(x)$ can have $2$ or $0$ positive real zeros, it cannot have exactly $1$ positive real zero.

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