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

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3.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

03.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

Let $f (x) = x^{5} - 6 x^{2} - 4 x + 5$
According to Descartes' rule,
$2$ changes of sign $\Rightarrow$ maximum 2 positive roots.

Now $f (- x) = - x^{5} - 6 x^{2} + 4 x + 5$
According to Descartes' rule,
$1$ change of sign $\Rightarrow$ maximum 1 negative root.

Hence the total maximum number of real roots $= 2 + 1 = 3$

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