The maximum possible number of real roots of | vedclass

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Question

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.For positive real roots,we check the sign changes in $f(x) = x^5 - 6x^2 - 4x + 5$. The signs are $(+, -, -, +)$.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.For positive real roots,we check the sign changes in $f(x) = x^5 - 6x^2 - 4x + 5$. The signs are $(+, -, -, +)$. There are $2$ sign changes.For negative real roots,we check the sign changes in $f(-x) = (-x)^5 - 6(-x)^2 - 4(-x) + 5 = -x^5 - 6x^2 + 4x + 5$. The signs are $(-, -, +, +)$. There is $1$ sign change.By Descartes' Rule of Signs,the maximum number of positive real roots is $2$ and the maximum number of negative real roots is $1$.Therefore,the maximum possible number of real roots is $2 + 1 = 3$.

The maximum possible number of real roots of the equation $x^5 - 6x^2 - 4x + 5 = 0$ is

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