If $p (x) = x^{2} - 2 x + 1$ , $q (x) = x^{4} - 1$ and $r (x) = x^{3} - 2 x^{2} - 5 x + 6$ , find their HCF.

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Solution

We factorise all of them:
$p (x) = x^{2} - 2 x + 1 = {(x - 1)}^{2}$ ,
$q (x) = x^{4} - 1 = (x^{2} - 1) (x^{2} + 1) = (x - 1) (x + 1) (x^{2} + 1)$ ,
$r (x) = x^{3} - 2 x^{2} - 5 x + 6 = x^{3} - x^{2} + x - 6 x + 6$
$= x^{2} (x - 1) - x (x - 1) - 6 (x - 1)$
$= (x - 1) (x^{2} - x - 6) = (x - 1) (x - 3) (x + 2)$ .
Observe that $H C F (p (x), q (x)) = x - 1$ and $H C F (x - 1, r (x)) = x - 1$ . Hence the HCF of $p (x), q (x), r (x)$ is $x - 1$ .

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