3 Tutor System
Starting just at 265/hour

# 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 $$p(x) = kx^2 - 3x + k$$,
$$\therefore p(1) = 0$$
$$\Rightarrow 0 = k(1)^2 - 3(1) + k$$
$$\Rightarrow 0 = k - 3 + k$$
$$\Rightarrow 3 = 2k$$
$$\Rightarrow k = \frac{3}{2}$$