Question 470988
Solving {{{(x+2)(x-20)(x+1)>0}}}



First, find the zeros of {{{y=(x+2)(x-20)(x+1)}}}.



{{{(x+2)(x-20)(x+1)=0}}}.



{{{x+2=0}}} or {{{x-20=0}}} or {{{x+1=0}}}.



{{{x=-2}}} or {{{x=20}}} or {{{x=-1}}}.



In ascending order, the zeros are: -2, -1, 20



So this means we have the following intervals to check: *[Tex \LARGE \left(-\infty,-2\right)], *[Tex \LARGE \left(-2,-1\right)], *[Tex \LARGE \left(-1,20\right)], *[Tex \LARGE \left(20,\infty\right)]



Now in the interval *[Tex \LARGE \left(-\infty,-2\right)], the expresssion {{{(x+2)(x-20)(x+1)}}} is negative. Simply plug in any number less than -2 to see this (eg: plug in x=-3 to get (-3+2)*(-3-20)*(-3+1)=-46)



In the interval *[Tex \LARGE \left(-2,-1\right)], the expresssion {{{(x+2)(x-20)(x+1)}}} is positive. Simply plug in any number between -2 and -1 to see this (eg: plug in x=-1.5 to get (-1.5+2)*(-1.5-20)*(-1.5+1)=5.375)



In the interval *[Tex \LARGE \left(-1,20\right)], the expresssion {{{(x+2)(x-20)(x+1)}}} is negative. Simply plug in any number between -1 and 20 to see this (eg: plug in x=10 to get (10+2)*(10-20)*(10+1)=-1320)



Finally, in the interval *[Tex \LARGE \left(20,\infty\right)], the expresssion {{{(x+2)(x-20)(x+1)}}} is positive. Simply plug in any number greater than 20 to see this (eg: plug in x=21 to get (21+2)*(21-20)*(21+1)=506)



So the following intervals yield positive outputs: *[Tex \LARGE \left(-2,-1\right)] and *[Tex \LARGE \left(20,\infty\right)]



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Answer:



The solution set is *[Tex \LARGE \left(-2,-1\right)\cup\left(20,\infty\right)]



Here's a graph to visually confirm this



{{{ graph(700,700,-7,25,-1475.56091173297,85.3757265477785, (x+2)(x-20)(x+1)) }}}