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

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

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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.
$x + 3$
$- - - - - - - -$
$x^{3} - 3 x^{2} + x - 3 ∣ ∣ ∣ x^{4} + 0 x^{3} + 6 x^{2} + 5 x^{4} - 3 x^{3} + x^{2} - 3 x - + - + - --------------- - 3 x^{3} + 5 x^{2} + 3 x + 5 3 x^{3} - 9 x^{2} + 3 x - 9 - + - + - -------------- - 14 x^{2} + 14 = 14 (x^{2} + 1)$
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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