How do you solve and write the following in interval notation: #x(x-2)(x+1)>0#?

1 Answer
Feb 12, 2017

Answer:

The answer is #x in ]-1, 0[uu]2, +oo[#

Explanation:

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

Now, we build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##0##color(white)(aaaaaa)##2##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x##color(white)(aaaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

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