x³ + 2x² - 25x - 50 = 0

Factor the first two terms on the left side by taking out GCF x²

        x²(x + 2) - 25x - 50 = 0

Now factor the last two terms on the left side by taking out GCF -25

       x²(x + 2) - 25(x + 2) = 0

Now factor out the GCF (x + 2) leaving x² and -25 in parentheses:

            (x + 2)(x² - 25) = 0

Factor the second parentheses as the difference of squares:

       (x + 2)(x - 5)(x + 5) = 0

Use the zero-factor principle by setting each factor = 0:

x + 2 = 0;  x - 5 = 0;  x + 5 =  0
    x =-2;      x = 5;      x = -5

The real-number solutions are -5, -2, and 5.

Edwin