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

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$. By 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 are: $(1$ to $-6)$ and $(-4$ to $5)$. Thus,there are $2$ sign changes,so there are 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 change is: $(-6$ to $4)$. Thus,there is $1$ sign change,so there is 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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