SOLUTION: could you please help me solve {{{ 2x(x+3)(x-3)less than or equal to 0 }}} I got x=0, x=-3, x=3 for the critical points. I just dont understand how to getr the rest of the answer

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: could you please help me solve {{{ 2x(x+3)(x-3)less than or equal to 0 }}} I got x=0, x=-3, x=3 for the critical points. I just dont understand how to getr the rest of the answer       Log On


   

Question 304646: could you please help me solve +2x%28x%2B3%29%28x-3%29less+than+or+equal+to+0+
I got x=0, x=-3, x=3 for the critical points. I just dont understand how to getr the rest of the answer sicne x=0 for one of the critical points.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Break up the number line into four regions.
(-infinity,-3)
(-3,0)
(0,3)
(3,infinity)
Then for each region pick a point (not the endpoints).
Test the inequality.
If it solves it, that region is part of the solution.
Then move on to the next region.
Region 1:
(-infinity,-3)
Let x=-4
2x%28x%2B3%29%28x-3%29%3C0
2%28-4%29%28-4%2B3%29%28-4-3%29%3C0
2%28-4%29%28-1%29%28-7%29%3C0
-56%3C0
True, first region is part of the solution.
Move on.
.
.
.
Region 2:
(-3,0)
Let x=-1
2x%28x%2B3%29%28x-3%29%3C0
2%28-1%29%28-1%2B3%29%28-1-3%29%3C0
2%28-1%29%282%29%28-4%29%3C0
16%3C0
False, second region is not part of the solution.
.
.
.
Continue to the third and fourth regions to complete the solution.
Re-post if you need more help.