The order of the multiple root of $- 1$ of the equation $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = 0$ is

1

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2

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3

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4

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Solution

The correct option is A $4$
$x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = 0$
Let $f (x) = x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1$
$f (- 1) = 0$
$\Rightarrow f^{'} (x) = 4 x^{3} + 12 x^{2} + 12 x + 4$
$f^{'} (- 1) = 0$
$f^{''} (x) = 12 x^{2} + 24 x + 12$
$f^{''} (- 1) = 0$
$f^{''} (x) = 24 x + 24$
$f^{'''} (- 1) = 0$
$f^{''''} (x) = 24$
$f^{''''} (- 1) \neq 0$
Hence, the order of multiple root $- 1$ is $4$ .

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