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

k = - 2

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k = 2

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k = - 2 - \sqrt{2}

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k = - 2 + \sqrt{2}

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Solution

The correct option is C $k = - 2 - \sqrt{2}$
$p (x) = {2 x}^{2} + k x + \sqrt{2}$
$p (1) = 0$ .

$p (1) = {2 (1)}^{2} + k (1) + \sqrt{2} = 0$

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

$k = - 2 - \sqrt{2}$

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Similar questions

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

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

Q. 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}

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

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 the following case: p(x)=2x2+kx+√2

The correct option is C k=−2−√2p(x)=2x2+kx+√2 p(1)=0. p(1)=2(1)2+k(1)+√2=0 2+k+√2=0 k=−2−√2

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

The correct option is C $k = - 2 - \sqrt{2}$
$p (x) = {2 x}^{2} + k x + \sqrt{2}$
$p (1) = 0$ .

$p (1) = {2 (1)}^{2} + k (1) + \sqrt{2} = 0$

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

$k = - 2 - \sqrt{2}$