How do you solve #(x+4)(x-2)(x-6)>0#?

1 Answer
Jan 22, 2017

Answer:

The answer is #=x in ]-4,2 [uu]6,+oo[#

Explanation:

Let #f(x)=(x+4)(x-2)(x-6)#

The domain of #f(x)# is #D-f(x)=RR#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##2##color(white)(aaaaaa)##6##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

#color(white)(aaaa)##x-6##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)>0# when #x in ]-4,2 [uu]6,+oo[#