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(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.According to Descartes' Rule of Signs,the number of positive real roots is at most the number of sign changes in

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.According to Descartes' Rule of Signs,the number of positive real roots is at most the number of sign changes in the coefficients of $f(x)$.The coefficients are $(1, 0, 0, -6, -4, 5)$.The sign changes occur from $1$ to $-6$ and from $-4$ to $5$. Thus,there are $2$ sign changes,meaning $f(x)$ has at most $2$ positive real roots.Now,consider $f(-x) = (-x)^5 - 6(-x)^2 - 4(-x) + 5 = -x^5 - 6x^2 + 4x + 5$.The coefficients are $(-1, 0, 0, -6, 4, 5)$.The sign changes occur from $-6$ to $4$. Thus,there is $1$ sign change,meaning $f(x)$ has at most $1$ negative real root.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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