The polynomial can be factored as (x - 2)(x + 2). Setting each factor to zero gives us x - 2 = 0 or x + 2 = 0, so the roots are x = -2 and x = 2.

We can factor the polynomial as f(x) = (x - 3)(x + 3). Setting each factor to zero gives us the roots x = -3 and x = 3.

The polynomial can be factored as (x + 3)(x + 3) = 0, giving the root x = -3.

Factoring gives (x - 2)(x - 2) = 0, so the repeated root is x = 2.

The polynomial can be factored as (x - 3)(x - 3) = 0, giving a repeated root of x = 3.

The polynomial can be factored as (x + 3)(x + 3) or (x + 3)^2. Therefore, the roots are x = -3 and x = -3.

x^2 - 4 can be factored as (x - 2)(x + 2) = 0. Therefore, the roots are -2 and 2.

To factor, we look for two numbers that multiply to 24 and add to -10. The numbers -4 and -6 work, so the factored form is (x - 4)(x - 6).

The expression x^2 - 4 is a difference of squares and can be factored as (x - 2)(x + 2).

The polynomial factors to (x - 2)(x - 3) since the roots are 2 and 3.

First, add 3 to both sides: 2x < 8. Then divide by 2: x < 4.

Solving the inequality: 2x < 4, thus x < 2.

Using Vieta's formulas, the sum of the roots (4 + (-4)) = 0, so k = 0.

Using the fact that the sum of the roots is -k, we have 4 + 4 = -k, so k = -8.

Using Vieta's formulas, the sum of the roots (3 + (-3)) = -k, so k = 0. The product of the roots (3 * -3) = 9, which is consistent. Thus, k = 0.

Using Vieta's formulas, the sum of the roots is -k = 4 + (-4) = 0, so k = 0.

Using the sum of roots, 3 + 4 = k, we find k = 7.

The polynomial can be factored as (x - 2)(x - 5). Setting each factor to zero gives us x = 2 and x = 5, so 2 is a root.

Using the Rational Root Theorem, we can test x = 1, 2, and 3. Testing x = 3 gives 3^3 - 6(3^2) + 11(3) - 6 = 0, so x = 3 is a root.

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

The polynomial can be factored as (x - 3)(x - 3) or (x - 3)^2.

First, simplify the equation: 2x^2 = 8, so x^2 = 4. The roots are x = ±2.

The polynomial x^2 + 2x + 1 can be factored as (x + 1)(x + 1), indicating that -1 is a root.