Question 469683
First, find the zeros of {{{y=(x+18)(x-12)(x+3)}}}. 



(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 *[Tex \LARGE \left(-\infty,-18\right)], (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 *[Tex \LARGE \left(-18,-3\right)], (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 *[Tex \LARGE \left(-3,12\right)], (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 *[Tex \LARGE \left(12,\infty\right)], (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: *[Tex \LARGE \left(-18,-3\right)] and *[Tex \LARGE \left(12,\infty\right)]




So combine them with a union symbol to get the final answer: *[Tex \LARGE \left(-18,-3\right)\cup\left(12,\infty\right)]



Here's a graph to visually confirm this



{{{ graph( 500, 500, -20, 20, -1500, 1500, (x+18)(x-12)(x+3)) }}}