Question 469683: I need to know how to solve this equation. (x+18)(x-12)(x+3)>0. If anyone can help, I would appreciate it, thanks in advance.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! First, find the zeros of  .
(x+18)(x-12)(x+3)=0
x+18=0 or x-12=0 or x+3=0
x=-18, x=12, or x=-3
In ascending order, the zeros are: -18,-3,12
Now in the interval ) , (x+18)(x-12)(x+3) is negative. Simply plug in any negative number less than -18 to see this (eg: plug in x=-20 to get (-20+18)(-20-12)(-20+3)=-1088)
In the interval ) , (x+18)(x-12)(x+3) is positive. Simply plug in any number in the interval to see this (eg: plug in x=-5 to get (-5+18)(-5-12)(-5+3)=442)
In the interval ) , (x+18)(x-12)(x+3) is negative. Simply plug in any number in the interval to see this (eg: plug in x=0 to get (0+18)(0-12)(0+3)=-648)
Finally, in the interval ) , (x+18)(x-12)(x+3) is positive. Simply plug in any number greater than 12 to see this (eg: plug in x=50 to get (50+18)(50-12)(50+3)=136952)
So the following intervals make (x+18)(x-12)(x+3) positive: and
So combine them with a union symbol to get the final answer: \cup\left(12,\infty\right))
Here's a graph to visually confirm this
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