x(x − 4)(x + 8) > 0 in interval notation?

Here's another way to get the same answers. Use cases around the zero crossings at 0, 4 and -8. None of those zero crossings is in the set, since 0>0 is false, so all intervals will be open.

Case 1: suppose x>0. Then (x+8)>0 too, and x(x+8)>0 means that you can divide the inequality by x(x-8) without changing the direction of the inequaltiy. That leaves:

x - 4 > 0
x > 4

So, the positive solutions are the interval (4, oo).

Case 2: Suppose x<0. Then x-4<0 too, and x(x-4) is positive You can divide by x(x-4), again without changing the inequality and this time:

x + 8 > 0
x > -8
-8 < x
-8 < x < 0 .... remembering that x is negative in this case

So, the negative solutions are in the interval (-8,0)

The full solution is the union of the positive and negative solutions:

(-8,0) U (4, oo)

The graph of y=x(x-4)(x+8) is a cubic crossing the x-axis at
x=-8,x=0 and x=4. Sketch the graph and it is above the
x-axis (i.e. y>0) for -8<x<0 or x>4. This can be expressed as
{x:(-8<x<0)U(x>4)}

____________-8_________0_________4_______
x_______-ve______-ve____0___+ve______+ve
x - 4____-ve______ -ve_______-ve____0___+ve
x + 8___-ve___0___+ve_______+ve_______+ve
Product__-ve__0___+ve___0___-ve___0___+ve

solution set {- 8 < x < 0 ] U { x > 4 }

x(x - 4)(x + 8) > 0

either x > 0, x - 4 > 0 and x + 8 > 0 => x > 4
or x < 0, x - 4 < 0 and x + 8 > 0 => -8 < x < 0
or x < 0, x - 4 > 0 and x + 8 < 0 -- can't be
or x > 0, x - 4 < 0 and x + 8 < 0 -- can't be

answer is x > 4 or -8 < x < 0

see also https://www.google.com/search?q=y+%3D+x%28x+-+4%29%28x+%2B+8%29&ie=utf-8&oe=utf-8

x(x − 4)(x + 8) > 0
(x + 8)x(x − 4) > 0

When x < −8 :
(x + 8)x(x − 4) = (‒ve) × (‒ve) × (‒ve) < 0

When ‒8 < x < 0 :
(x + 8)x(x − 4) = (+ve) × (‒ve) × (‒ve) > 0

When 0 < x < 4 :
(x + 8)x(x − 4) = (+ve) × (+ve) × (‒ve) < 0

When x > 4 :
(x + 8)x(x − 4) = (+ve) × (+ve) × (+ve) > 0

Range of x : ‒8 < x < 0 or x > 4
in interval notation: (‒8, 0) or (4, ∞)