Find the value of $k$ , if $x - 1$ is a factor of $p (x)$ in the following case
$p (x) = {k x}^{2} - \sqrt{2} x + 1$

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Solution

If $(x - 1)$ is a factor of $p (x)$ then $p (1) = 0$

$p (x) = k x^{2} - \sqrt{2 x} + 1$

$p (1) = k - \sqrt{2} + 1$

$0 = k - \sqrt{2} + 1$

So $k = \sqrt{2} - 1$ .

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Find the value of k, if x − 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x² + x + k (ii)

(iii) (iv) p(x) = kx² − 3x + k

Find the value of k, if x - 1 is a factor p(x) in each of the following cases:
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(iii) $p (x) = k x^{2} - \sqrt{2} x + 1$
(iv) $p (x) = k x^{2} - 3 x + k$

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Find the value of k, if x−1 is a factor of p(x) in the following casep(x)=kx2−√2x+1

If (x−1) is a factor of p(x) then p(1)=0p(x)=kx2−√2x+1p(1)=k−√2+10=k−√2+1So k=√2−1.

Find the value of $k$ , if $x - 1$ is a factor of $p (x)$ in the following case
$p (x) = {k x}^{2} - \sqrt{2} x + 1$

If $(x - 1)$ is a factor of $p (x)$ then $p (1) = 0$

$p (x) = k x^{2} - \sqrt{2 x} + 1$

$p (1) = k - \sqrt{2} + 1$

$0 = k - \sqrt{2} + 1$

So $k = \sqrt{2} - 1$ .