How do you solve and write the following in interval notation: # (b + 4)(b - 1)(b - 3) <0#?

1 Answer
Mar 31, 2017

Answer:

The solution is #b in ]-oo,-4[uu]1,3[#

Explanation:

Let #f(b)=(b+4)(b-1)(b-3)#

Let's build a sign chart

#color(white)(aaaa)##b##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##1##color(white)(aaaa)##3##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##b+4##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

#color(white)(aaaa)##b-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(b)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore

#f(b)<# when #b in ]-oo,-4[uu]1,3[#