Question

find the value of k if x-1 is a factor of p(x) = 2x²+kx+ underoot 2

Please give me answer urgent​

Answer 1

Question:

find the value of k if x-1 is a factor of

$\sf{p(x) ={2x}^{2}+kx+\sqrt{2}}$

Given:

• x-1 is a factor of
• $\sf{p(x) ={2x}^{2}+kx+\sqrt{ 2}}$

To Find:

• Value of k.

Solution:

By Reminder theorem,

$\rightarrow\mathsf{x-1=0}$

$\rightarrow\mathsf{x=1}$

Now,

$\dashrightarrow\mathsf{p(x)={2x}^{2}+kx+\sqrt{2}}$

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

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

We also know,

${p(x)=0}$

So,

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

$\dashrightarrow\mathsf{-2=k+\sqrt{2}}$

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

Hence, Value of k is $\sf{-2-\sqrt{2}}$.

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