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According to Descartes' Rule of Signs, can the polynomial function have exactly $0$ positive real zeros? Choose your answer based on the rule only. $\newline$ $f(x) = x^5 + 2x^4 + 8x^3 - 4x^2 + 5$ $\newline$ 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

0

positive real zeros? Choose your answer based on the rule only.

\newline

f(x) = x^5 + 2x^4 + 8x^3 - 4x^2 + 5

\newline

Choices:

\newline

(A)yes

\newline

(B)no

Count Sign Changes: Count the number of sign changes in the polynomial $f(x) = x^5 + 2x^4 + 8x^3 - 4x^2 + 5$ . Coefficients: $1, 2, 8, -4, 5$ . Sign changes: $1$ (from $8$ to $-4$ ).
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 we have $1$ sign change, the possible number of positive real zeros is $1$ or $1-2= -1$ , which doesn't make sense, so we stick with $1$ .
Number of Positive Zeros: Since the polynomial can have $1$ positive real zero, it cannot have exactly $0$ positive real zeros.

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