The third term of $x^{3} + 2 x^{2} + x + 1 = 0$ is eliminated by putting $x = y + h$ . The values of $h$ are

1, \frac{1}{3}

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1, - \frac{1}{3}

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- 1, \frac{1}{3}

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- 1, - \frac{1}{3}

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Solution

The correct option is B $- 1, - \frac{1}{3}$ .
$f (x) = x^{3} + 2 x^{2} + x + 1$
Put $x = y + h$
$f (y + h) = (y + h)^{3} + 2 (y + h)^{2} + (y + h) + 1$
Now making coefficient of $y = 0$ we get,
$3 h^{2} + 4 h + 1 = 0$
This gives $h = - 1, - \frac{1}{3}$

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