If x = 1 is a root of equation
$x^{4} + a x^{3} + b x^{2} + c x - 1 = 0$ repeated thrice, then (a + b + 2c) is equal to

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Solution

The correct option is C 2
Product of roots $= - 1$
Fourth root $= - 1$
$x^{4} + a x^{3} + b x^{2} + c x - 1 = (x - 1)^{3} (x + 1)$
$\Rightarrow$ $(x^{3} - 1 - 3 x (x - 1)) (x + 1)$
$\Rightarrow$ $(x^{3} - 3 x^{2} + 3 x - 1) (x + 1)$
$\Rightarrow$ $x^{4} - 3 x^{3} + 3 x^{2} - x$ $+ x^{3} - 3 x^{2} + 3 x - 1$
$\Rightarrow$ $x^{4} - 2 x^{3} + 2 x$ $- 1 = 0$
$a = - 2$
$b = 0$
$c = 2$
$\Rightarrow a + b + 2 c = - 2 + 0 + 4 = 2$

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