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

Since, $x - 1$ is factor of $p (x) = k x^{2} - \sqrt{2} x + 1$
$\Rightarrow p (1) = 0$
Now,
$p (1) = k (1)^{2} - \sqrt{2} (1) + 1$
$= k - \sqrt{2} + 1$
So, $k - \sqrt{2} + 1 = 0$
$\Rightarrow k = \sqrt{2} - 1$

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

Since, x−1 is factor of p(x)=kx2−√2x+1 ⇒p(1)=0 Now, p(1)=k(1)2−√2(1)+1 =k−√2+1 So, k−√2+1=0 ⇒k=√2−1

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

Since, $x - 1$ is factor of $p (x) = k x^{2} - \sqrt{2} x + 1$
$\Rightarrow p (1) = 0$
Now,
$p (1) = k (1)^{2} - \sqrt{2} (1) + 1$
$= k - \sqrt{2} + 1$
So, $k - \sqrt{2} + 1 = 0$
$\Rightarrow k = \sqrt{2} - 1$