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

The zero of x - 1 is 1.


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


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


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


iv)So, as x - 1 is a factor of \(p(x) = kx^2 - 3x + k\),
\(\therefore p(1) = 0\)
\(\Rightarrow 0 = k(1)^2 - 3(1) + k\)
\(\Rightarrow 0 = k - 3 + k\)
\( \Rightarrow 3 = 2k\)
\(\Rightarrow k = \frac{3}{2} \)