SOLUTION: Please help me solve this equation: (x^2- 4x) (x^2- 4) < 0

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Question 737117: Please help me solve this equation:
(x^2- 4x) (x^2- 4) < 0
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%5E2-+4x%29+%28x%5E2-+4%29+%3C+0+

That's factorable. Look for all the zeros, and those define your intervals. You then pick each interval and check the sign for every binomial factor and determine the sign of the evaluated expression. Either it is less than zero or it is not less than zero.

%28x%5E2-+4x%29+%28x%5E2-+4%29=x%28x-4%29%28x-2%29%28x%2B2%29

At each specific value for x of 0, or +2, +4 or -2, the whole expression is zero, therefore not less than zero, therefore false. The intervals to check are to be:
x%3C-2
-2%3Cx%3C0
0%3Cx%3C2
2%3Cx%3C4
4%3Cx
Pick any value in each interval and check the resulting sign for x%28x-4%29%28x-2%29%28x%2B2%29 and decide if less than zero or not.

Test the interval +x%3C-2:
choose x=-3.
x%28x-4%29%28x-2%29%28x%2B2%29
(-)(-)*(-)(-) becomes (-)(-)(-)(-), which is (+), indicating "positive".
Meaning x<-2 IS NOT PART OF THE SOLUTION...
Test the other intervals similarly in order to finish and get the full answer solution set.