Solution by Algebraic Reasoning (Method 2):
Remember a = numerator, b = denominator.

1. This inequality already has 0 on one side. We are all set to go.

2. Since we are dealing with > 0, "either a > 0 and b > 0 OR a < 0 and b < 0."

3. Both positive: x² - 2x - 8 > 0 and x + 4 > 0 (Now it gets a bit messy!) This quadratic is going to yield a product of factors which will also need to be addressed with the rule "either both are > 0 or both are < 0."

Both a (numerator) and b (denominator) need to be positive.
Numerator: x2 - 2x - 8 > 0 → (x - 4)(x + 2) > 0 Both factors are positive: x - 4 > 0 and x + 2 > 0 x > 4 and x > -2 x > 4	Denominator: x + 4 > 0 x > -4
Now, x > 4 and x > -4 must BOTH be TRUE, which gives x > 4 where both are true.
But wait! In this problem, the numerator could also be positive when both factors are negative. We need to look further.
Numerator: x - 4 < 0 and x + 2 < 0 x < 4 and x < -2 x < -2	Denominator: x > -4
Now, x < -2 and x > -4 gives -4 < x < -2 where both are true.
So, we have two situations under which both the numerator and denominator can BOTH be positive: -4 < x < -2   OR    x > 4

Both negative: x² - 2x - 8 < 0 and x + 4 < 0 (Here we go again!) This time the quadratic factors will need the rule "one is positive and the other is negative." But, the "order" of which factor is positive and which is negative can occur in two ways.

Numerator: x2 - 2x - 8 < 0 → (x - 4)(x + 2) < 0 First factor positive, second negative: x - 4 > 0 and x + 2 < 0 x > 4 and x < -2 Can never happen!	Denominator: x + 4 < 0 x < -4
Now, x < -2 and x > 4 and x < -4 are NEVER all true at the same time.
Second factor positive, first negative: x - 4 < 0 and x + 2 > 0 x < 4 and x > -2 -2 < x < 4	x < -4
Now, x < 4 and x > -2 and x < -4 are NEVER all true at the same time.

4. The denominator cannot equal -4, so we adjust to -4 < x < -2.