How do you solve #(x-2)(x+1)(x+2)>0# using a sign chart?

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Answer 1 · 2016-12-12T12:36:47

The answer is #x in ] -2,-1 [ uu] 2,+oo [ #

Let #f(x)=(x-2)(x+1)(x+2)#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##-1##color(white)(aaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

So, #f(x)>0#, #x in ] -2,-1 [ uu] 2,+oo [ #