If $2 x^{3} + 4 x^{2} + 2 a x + b$ is exactly divisible by $x^{2} - 1$ , then the value of $a$ and $b$ , respectively will be

1, 2

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- 1, 4

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1, - 2

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- 1, - 4

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Solution

The correct option is C $- 1, - 4$
Since $f (x) = 2 x^{3} + 4 x^{2} + 2 a x + b$ is exactly divisible
by $x^{2} - 1 = (x - 1) (x + 1)$
$∴ f (1) = 0$ and $f (- 1) = 0$
These give
$2 + 4 + 2 a + b = 0$
or $2 a + b + 6 = 0$ ..(i)
and $- 2 + 4 - 2 a + b = 0$
or $2 a - b - 2 = 0$ ...(ii)
Solving equations (i) and (ii), we get
$a = - 1$ , $b = - 4$
$d i v i s o r (x - 1) \to R = f (1) = 2 a + b + 6 = 0 d i v i s o r (x + 1) \to R = f (- 1) = b - 2 a + 2 = 0$ .

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