x^2 - 5x + 6 can be factored as (x-2)(x-3) since -2 and -3 add to -5 and multiply to 6.

f(-1) = 4(-1)^2 + 3(-1) - 5 = 4 - 3 - 5 = -4.

g(1) = 1^3 - 2(1^2) + 1 - 2 = 1 - 2 + 1 - 2 = -2.

p(2) = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0.

The product of the roots of ax^2 + bx + c is given by c/a. Here, 6/2 = 3.

The sum of the roots of ax^2 + bx + c is given by -b/a. Here, -6/3 = -2.

For a double root, the discriminant must be zero: k^2 - 4(1)(9) = 0. Thus, k^2 = 36, giving k = ±6.

Using the quadratic formula x = [-b ± √(b² - 4ac)] / 2a, we have a = 2, b = -8, c = 6. The discriminant is (-8)² - 4(2)(6) = 64 - 48 = 16. Thus, x = [8 ± 4] / 4, giving x = 3 or x = 1.

The polynomial x^2 - 4 factors to (x-2)(x+2), so the roots are 2 and -2.