What is the HCF of $(x^{4} - x^{2} - 6)$ and $(x^{4} - 4 x^{2} + 3)$ ?

x^{2} - 3

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x + 2

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x + 3

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x^{2} + 3

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Solution

The correct option is A $x^{2} - 3$
$x^{4} - x^{2} - 6$ can be factorized as
$= x^{4} - 3 x^{2} + 2 x^{2} - 6$
$= x^{2} (x^{2} - 3) + 2 (x^{2} - 3)$
$= (x^{2} - 3) (x^{2} + 2)$
and $x^{4} - 4 x^{2} + 3$ can be factorized as
$= x^{4} - 3 x^{2} - x^{2} + 3$
$= x^{2} (x^{2} - 3) - 1 (x^{2} - 3)$
$= (x^{2} - 3) (x^{2} - 1)$
Therefore, H.C.F is $(x^{2} - 3)$ .

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