Question 1126217
set f(x) = 0 and you get (x-2) * (x+4)^2 = 0
the roots are x = 2 and x = -4.
f(x) = 0 when x = 2 and x = -4
the intervals between these zero points are:
(-infinity,-4) which means x is smaller than -4.
(-4,2) which means x is greater than -4 and less than 2.
(2,+infinity) which means x is greater than 2.


since the function is continuous, if x is smaller than 0 in an interval, it will stay smaller than x for the whole interval and, if x is greater than 0 in an interval, it will stay greater than 0 for the hole interval.


pick any value in each interval to test whether the function is positive or negative in that interval.


for the interval where x is less than -4, i chose -5.
f(x) = (x-2) * (x+4)^2 becomes f(x) = -7 * 1 which becomes f(x) = -7 which is negative. 


for the interval where x is greater than -4 but less than 2, i chose 0.
f(x) = (x-2) * (x+4)^2 becomes f(x) = -2 * 16 which becomes f(x) = -32 which is negative.


for the interval where x is greater than 2, i chose 5.
f(x) = (x-2) * (x+4)^2 becomes f(x) = 3 * 81 which becomes f(x) = 243 which is positive.


the function is negative when x < -4 and when -4 < x < 2.
the function is positive when x > 2.


i interval notation, this results in:


the function is negative during the interval (-infinity, -4) union (-4, 2).
the function is positive during the interval (2, +infinity).


the graph of the function confirms this analysis is true, as shown below.


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