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

Solution:

If $x$ - 1 is a factor of polynomial $p(x)$, then $p(1)$ must be 0 .

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

$p(1)=0$

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

$\Rightarrow 2+k=0$

$\Rightarrow k=-2$

Therefore, the value of $k$ is $-2$.

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

$p(1)=0$

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

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

$\Rightarrow k=-2-\sqrt{2}=-(2+\sqrt{2})$

Therefore, the value of $k$ is $-(2+\sqrt{2})$.

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

$p(1)=0$

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

$\Rightarrow k-\sqrt{2}+1=0$

$\Rightarrow k=\sqrt{2}-1$

Therefore, the value of $k$ is $\sqrt{2}-1$.

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

$\Rightarrow p(1)=0$

$\Rightarrow k(1)^{2}-3(1)+k=0$

$\Rightarrow k-3+k=0$

$\Rightarrow 2 k-3=0$

$\Rightarrow k=\frac{3}{2}$

Therefore, the value of $k$ is $\frac{3}{2}$.