The H.C.F. of $x^{3} - 3 x^{2} + x - 3$ and $x^{4} + 6 x^{2} + 5$ is .

x^{4} + 6 x^{2} + 5

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x + 3

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x - 3

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x^{2} + 1

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Solution

The correct option is D $x^{2} + 1$
Here we do not find any common terms, so we simply divide.
$x^{3} - 3 x^{2} + x - 3) x^{4} + 6 x^{2} + 5 (x + 3 x^{4} - 3 x^{3} + x^{2} - 3 x - + - + - --------------- - 3 x^{3} + 5 x^{2} + 3 x + 5 3 x^{3} - 9 x^{2} + 3 x - 9 - + - + - -------------- - 14 x^{2} + 14 = 14 (x^{2} + 1)$
Here the highest power of divisor is 3 and that of remainder is 2.
So now remainder becomes divisor and divisor becomes the dividend.
$x^{2} + 1) x^{3} - 3 x^{2} + x - 3 (x - 3 x^{3} + x - - - ----------- - - 3 x^{2} - 3 - 3 x^{2} - 3 - ------------ - \times$
Now the remainder in zero. So the required H.C.F is the divisor i.e. $x^{2} + 1$

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