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2. Polynomials
medium
Find the value of $k$, if $x -1$ is a factor of $p(x)$ in this case : $p(x)=2 x^{2}+k x+\sqrt{2}$
A
$k =-2-\sqrt{2}$
B
$k =-2+\sqrt{2}$
C
$k =2-\sqrt{2}$
D
$k =2+\sqrt{2}$
Solution
Here $p ( x )=2 x ^{2}+ kx +\sqrt{2}$
For $x-1$ be a factor of $p(x), \,p(1)=0$.
Since, $p (1)=2(1)^{2}+ k (1)+\sqrt{2}=2+ k +\sqrt{2}$
$\because p (1)$ must be equal to $0 .$
$\therefore k+2+\sqrt{2} =0 $ $\Rightarrow k =-2-\sqrt{2}$
or $k=-(2+\sqrt{2}) $ .
Std 9
Mathematics
