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According to Descartes' Rule of Signs, can the polynomial function have exactly $4$ positive real zeros, including any repeated zeros? Choose your answer based on the rule only. $\newline$ $g(x) = x^7 + 5x^6 - 8x^5 - 4x^4 + 9x^3 - 5x - 4$ $\newline$ 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

4

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

\newline

g(x) = x^7 + 5x^6 - 8x^5 - 4x^4 + 9x^3 - 5x - 4

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the coefficients of $g(x) = x^7 + 5x^6 - 8x^5 - 4x^4 + 9x^3 - 5x - 4$ . Coefficients: $1, 5, -8, -4, 9, -5, -4$ Sign changes: $1$ (from $5$ to $-8$ ), $2$ (from $-4$ to $9$ ), $3$ (from $9$ to $1, 5, -8, -4, 9, -5, -4$ $0$ ).
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. $\newline$ So, $g(x)$ can have $3$ , $1$ , or $0$ positive real zeros.
Number of Positive Zeros: Since $g(x)$ can have at most $3$ positive real zeros and not $4$ , the answer is no.

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