Question

find the value of k if x-1 is a factor of p(x) = kx2 - √2x + 1​

Answer 1

Given:

• Remainder is given x-1.
• Factor is also given p(x)=kx^2-√2x+1

To Find:

• Value of k

Solution:

Firstly we have to solve remainder theorem

$\: \: \sf \: g(x) = x - 1 = 0 \\ \\ \: \sf = x = 1$

Now find the value of k and put the value of x in p(x)

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

Hence, the value of k is √2 -1.

Answer 2

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