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)=2x^(2)+kx+sqrt(2)`
(iii) `p(x)=kx^(2)-sqrt(2)x+1`
(iv) `p(x)=kx^(2)-3x+k`

Answer

To find the value of \( k \) such that \( x - 1 \) is a factor of \( p(x) \) in each case, we will use the fact that if \( x - 1 \) is a factor of \( p(x) \), then \( p(1) = 0 \).

### (i) \( p(x) = x^2 + x + k \)

1. Substitute \( x = 1 \) into \( p(x) \):
\[ p(1) = 1^2 + 1 + k = 0 \]
Simplifying this gives:
\[ 1 + 1 + k = 0 \implies 2 + k = 0 \]

2. Solve for \( k \):
\[ k = -2 \]

### (ii) \( p(x) = 2x^2 + kx + \sqrt{2} \)

1. Substitute \( x = 1 \) into \( p(x) \):
\[ p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0 \]
Simplifying this gives:
\[ 2 + k + \sqrt{2} = 0 \implies k + \sqrt{2} = -2 \]

2. Solve for \( k \):
\[ k = -2 - \sqrt{2} \]

### (iii) \( p(x) = kx^2 - \sqrt{2}x + 1 \)

1. Substitute \( x = 1 \) into \( p(x) \):
\[ p(1) = k(1)^2 - \sqrt{2}(1) + 1 = 0 \]
Simplifying this gives:
\[ k - \sqrt{2} + 1 = 0 \implies k + 1 = \sqrt{2} \]

2. Solve for \( k \):
\[ k = \sqrt{2} - 1 \]

### (iv) \( p(x) = kx^2 - 3x + k \)

1. Substitute \( x = 1 \) into \( p(x) \):
\[ p(1) = k(1)^2 - 3(1) + k = 0 \]
Simplifying this gives:
\[ k - 3 + k = 0 \implies 2k - 3 = 0 \]

2. Solve for \( k \):
\[ 2k = 3 \implies k = \frac{3}{2} \]

### Summary of Results
- (i) \( k = -2 \)
- (ii) \( k = -2 - \sqrt{2} \)
- (iii) \( k = \sqrt{2} - 1 \)
- (iv) \( k = \frac{3}{2} \)