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3.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 :

The zero of x - 1 is 1.
i)So, as x - 1 is a factor of \(p(x) = x^2 + x + k\), therefore, \(p(1) = 0\)
i.e. \(0 = (1)^2 + 1 + k\)
i.e.\(0 = 1 + 1 + k\)
i.e.\(0 = 2 + k\)
i.e.\(k = -2\)

ii)So, as x - 1 is a factor of \(p(x) = 2x^2 + kx + \sqrt{2}\), therefore, \(p(1) = 0\)
i.e. \(0 = 2(1)^2 + k(1) + \sqrt{2}\)
i.e.\(0 = 2 + k + \sqrt{2}\)
i.e.\(k = -2 - \sqrt{2}\)

iii)So, as x - 1 is a factor of \(p(x) = kx^2 - \sqrt{2}x + 1\), therefore, \(p(1) = 0\)
i.e. \(0 = k(1)^2 - ( \sqrt{2})1 + 1\)
i.e.\(0 = k - \sqrt{2} + 1\)
i.e.\(k = \sqrt{2} - 1\)

iv)So, as x - 1 is a factor of \(p(x) = kx^2 - 3x + k\), therefore, \(p(1) = 0\)
i.e. \(0 = k(1)^2 - 3(1) + k\)
i.e.\(0 = k - 3 + k\)
i.e.\(3 = 2k\)
i.e.\(k = 3/2\)