(x + 1)(x + 2) > 0

We find the critical values of that which are the solutions to
the equation which is found by replacing the > by =:
                        
                             (x + 1)(x + 2) = 0
                             x + 1 = 0    x + 2 = 0
                                 x = -1;      x = -2

We put those two critical nubers on a number line:

                ----------o--o-------------------
                -5 -4 -3 -2 -1  0  1  2  3  4  5

We test a value for x less that -2, say x = -3

 (x + 1)(x + 2) > 0
(-3 + 1)(-3 + 2 > 0
       (-2)(-1) > 0
              2 > 0

That is true so we shade the number to the left of -2

                <=========o--o-------------------
                -5 -4 -3 -2 -1  0  1  2  3  4  5

Test a value for x between -2 and -1, say x = -1.5

     (x + 1)(x + 2) > 0
(-1.5 + 1)(-1.5 + 2 > 0
        (-0.5)(0.5) > 0
              -0.25 > 0

That is false so we do not shade the number line between -3 and -2.

We test a value for x greater that -1, say x = 0

 (x + 1)(x + 2) > 0
  (0 + 1)(0 + 2 > 0
         (1)(2) > 0
              2 > 0

That is true so we shade the number line to the right of -1

                <=========o--o==================>
                -5 -4 -3 -2 -1  0  1  2  3  4  5

                x < -2 or x > -1   [choice (e)]

The interval notation is an abbreviation for that number line:

                (-∞, -2) U (-1, ∞)


Edwin