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

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To do:

We have to find the value of $k$, if $x-1$ is a factor of $p(x)$ in each of the given cases.

Solution :

Factor Theorem:

The factor theorem states that if $p(x)$ is a polynomial of degree $n >$ or equal to $1$ and $a$ is any real number, then $x-a$ is a factor of $p(x)$, if $p(a)=0$.

Therefore,

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

$x-1$ is a factor of $p(x) =x^{2}+x+k$.

$p(1) = (1)^2+1+k = 0$

$1+1+k = 0$

$k+2=0$

$k = -2$

The value of $k$ is $-2$.

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

$x-1$ is a factor of $p(x) =2 x^{2}+k x+\sqrt{2}$.

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

$2(1)+k+\sqrt2 = 0$

$k+2+\sqrt2=0$

$k = -(2+\sqrt2)$

The value of $k$ is $-(2+\sqrt2)$.

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

$x-1$ is a factor of $p(x) =k x^{2}-\sqrt{2} x+1$.

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

$k(1)-\sqrt2+1 = 0$

$k+1-\sqrt2=0$

$k = \sqrt2-1$

The value of $k$ is $\sqrt2-1$.

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

$x-1$ is a factor of $p(x) =k x^{2}-3 x+k$.

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

$k(1)-3+k = 0$

$2k-3=0$

$2k = 3$

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

The value of $k$ is $\frac{3}{2}$.

Updated on 10-Oct-2022 13:39:07