How do you solve #(x+5)(x-2)(x-1)(x+1)<0# using a sign chart?

1 Answer
Jan 23, 2017

Answer:

The answer is #x in ]-5,-1 [uu ]1 ,2[#

Explanation:

Let #f(x)=(x+5)(x-2)(x-1)(x+1)#

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)##-5##color(white)(aaaa)##-1##color(white)(aaaa)##1##color(white)(aaaa)##2##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

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

Therefore,

#f(x)<0# when #x in ]-5,-1 [uu ]1 ,2[#