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 $\frac{3}{2}$.
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 $\frac{3}{2}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.