If the HCF of $p (x) = x^{2} + 3 x - 10$ , $q (x) = 2 x^{2} - k x - 4$ is $h (x) = x - k$ , find the value of $k$ .

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Solution

Since $x - k$ is the HCF of polunomials $p (x)$ and $q (x)$ , it is common factor of polynomials $p (x)$ and $q (x)$ . Hence $p (k) = 0$ and $q (k) = 0$ . Putting $x = k$ , we get
$0 p (k) = k^{2} + 3 k - 10$ , $0 = q (k) = 2 k^{2} - k^{2} - 4$ .
The second relation gives $k^{2} = 4$ or $k = \pm 2$ . If $k = - 2$ , we get
$p (- 2) = {(- 2)}^{2} + 3 (- 2) - 10 = - 12$ .
Since $p (k) = 0$ , $k \neq - 2$ . Taking $k = 2$ , we may verify that $p (2) = 2^{2} - 3 (2) - 10 = 0$ .
Thus $k = 2$ .

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If the HCF of p(x)=x2+3x−10, q(x)=2x2−kx−4 is h(x)=x−k, find the value of k.

If the HCF of $p (x) = x^{2} + 3 x - 10$ , $q (x) = 2 x^{2} - k x - 4$ is $h (x) = x - k$ , find the value of $k$ .