Using the binomial theorem, the coefficient of x^3 in (x + 3)^5 is given by 5C3 * (3)^2 = 10 * 9 = 90.

Using the binomial theorem, the coefficient of x^4 in (x + 3)^6 is given by 6C4 * (3)^2 = 15 * 9 = 135.

The coefficient of x^5 in (x - 1)^8 is C(8,5) * (-1)^3 = 56 * (-1) = -56.

The term containing x^2 is given by C(4, 2)(-2)^2x^2 = 6 * 4 * x^2 = 24x^2, but since it is negative, it is -12x^2.

The term containing x^2 is given by C(6, 2) * (x)^2 * (-2)^(6-2) = 15 * x^2 * 16 = 240x^2, so the term is -80x^2.

The discriminant is 0, indicating that there is exactly one distinct root.

For both roots to be positive, k must be greater than 1 (from Vieta's formulas).

For equal roots, the discriminant must be zero: (-4)^2 - 4*1*k = 0 leads to k = 4.

The discriminant must be less than zero: (-4)^2 - 4*1*k < 0 leads to k > 4.

The equation can be factored as (x - 3)(x - 3) = 0, indicating it has one distinct root (a double root).

Factoring the equation gives (x - 2)(x - 5) = 0, so the roots are 2 and 5.

The polynomial can be expressed as (x - 1)^3, indicating that the root x = 1 has a multiplicity of 3.

The given polynomial can be factored as (x - 1)^3 = 0, which has one distinct real root, x = 1.

The product of the roots is given by c/a. Here, c = 1 and a = 2, so the product is 1/2.

The product of the roots of the equation ax^2 + bx + c = 0 is given by c/a. Here, c = 2 and a = 2, so the product is 2/2 = 1.

The product of the roots of the equation ax^2 + bx + c = 0 is given by c/a. Here, c = 6 and a = 2, so the product is 6/2 = 3.

The product of the roots is given by c/a, which is 9/3 = 3.

The product of the roots is given by c/a = 6/1 = 6.

The discriminant must be less than zero: (-4)^2 - 4*2*k < 0 leads to k > 8.

The discriminant must be less than zero: (-4)^2 - 4*2*k < 0 leads to k > 8.

Dividing the entire equation by 3 gives x^2 + 4x + 4 = 0.

The equation can be factored as (2x - 3)(2x - 3) = 0, indicating a perfect square.

The discriminant is 3^2 - 4*5*(-2) = 9 + 40 = 49, which is a perfect square, hence the roots are rational.

The equation can be factored as (x + 3)^2 = 0, hence a = 3.

This is a perfect square trinomial: (x + 3)(x + 3) = 0.

For equal roots, b^2 - 4ac = 0. Here, 6^2 - 4(1)(k) = 0, so k = 9.

For no real roots, the discriminant must be less than zero: 6^2 - 4*1*k < 0, which gives k > 9.

For both roots to be negative, k must be greater than the square of half the coefficient of x, hence k > 9.

The equation can be factored as (x - 2)(x - 2) = 0, which is (x - 2)^2.

The equation can be factored as (x - 3)(x - 3) = 0, or (x - 3)^2 = 0.