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112. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:

p(x) = kx² - $square root of 2 to the power of straight x end root plus 1$

p(x) = kx² - $square root of 2 x plus 1$
If x - 1 is a factor of p(x), then p(1) = 0
| By Factor Theorem
$rightwards double arrow$ k(1)² - $square root of 2 to the power of open parentheses space 1 space close parentheses end exponent end root$ + 1 = 0
$rightwards double arrow space space space space space space space space space space space space straight k minus square root of 2 plus 1 equals 0 rightwards double arrow space space space space space space space space space space space space space straight k space equals space square root of 2 minus 1$