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The maximum possible number of real roots of equation $x^5-6 x^2-4 x+5=0$ is
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The correct answer is:
$3$
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 .
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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