Question

O fire
the value of k if (x-1) is a factore
polynomial
Kx2-root2x
+ 1
​

Answer 1

Answer:

sorry I can't help you

Answer 2

Answer:

The answer is k= root 2 - 1

Put divisor = 0

X-1 = 0

X = 1

P(x) = kx^2 - root 2x + 1

P(1) = k(1)^2 - root 2(1) + 1

= k- root 2 + 1

P(1) = k - root 2 +1

K - root 2 +1 = 0

= k= root 2 - 1

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