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

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Solution

The correct option is A 3
Let $f (x) = x^{5} - 6 x^{2} - 4 x + 5 = 0$ Then the number of change of sign in $f (x)$ is $2$ therefore $f (x)$ can have at most two positive real roots . Now, $f (- x) = - x^{5} - 6 x^{4} + 4 x + 5 = 0$ Then the number of change of sign is 1.
Hence $f (x)$ can have at most one negative real root. So that total possible number of real roots is 3.

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