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

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Solution

If $(x - 1)$ is a factor of $p (x)$
Then $p (1) = 0$
$p (x) = 2 x^{2} + k x + \sqrt{2}$
$p (1) = 2 + k + \sqrt{2}$
$0 = (2 + \sqrt{2}) + k$
So $k = - (2 + \sqrt{2})$ .

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Find the value of k, if (x −1) is a factor of p(x) in each of the following cases:
$(i) p (x) = x^{2} + x + k$
$(i i) p (x) = {2 x}^{2} + k x + \sqrt{2}$ [4 MARKS]

Q. Find the value of

k

, if

x - 1

is a factor of

p (x)

in each of the following cases:
(i)

p (x) = x^{2} + x + k

(ii)

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

(iii)

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

(iv)

p (x) = k x^{2} - 3 x + k

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

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

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

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Find the value of k, if x-1 is a factor of p(x) in each of the following cases p(x)=2x2+kx+√2

If (x−1) is a factor of p(x)Then p(1)=0p(x)=2x2+kx+√2p(1)=2+k+√20=(2+√2)+kSo k=−(2+√2).

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

If $(x - 1)$ is a factor of $p (x)$
Then $p (1) = 0$
$p (x) = 2 x^{2} + k x + \sqrt{2}$
$p (1) = 2 + k + \sqrt{2}$
$0 = (2 + \sqrt{2}) + k$
So $k = - (2 + \sqrt{2})$ .