How do you solve #(x+8)^2(x+5)(x+7)^2>=0# using a sign chart?

1 Answer
Jul 23, 2018

Answer:

The solution is #x in {-8} uu {-7}uu[-5,+oo)#

Explanation:

The inequality is

#(x+8)^2(x+5)(x+7)^2>=0#

Let #f(x)=(x+8)^2(x+5)(x+7)^2#

Let 's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##-8##color(white)(aaaaaa)##-7##color(white)(aaaaa)##-5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##(x+8)^2##color(white)(aaaa)##+##color(white)(aaaa)##0##color(white)(aaa)##+##color(white)(aaaaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##(x+7)^2##color(white)(aaaa)##+##color(white)(aaaa)####color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x+5##color(white)(aaaaaa)##-##color(white)(aaaa)####color(white)(aaaa)##-##color(white)(aa)####color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)##0##color(white)(aaa)##-##color(white)(aa)##0##color(white)(aaaa)##-##color(white)(a)##0##color(white)(aa)##+#

Therefore,

#f(x)>=0# when #x in {-8} uu {-7}uu[-5,+oo)#

graph{(x+8)^2(x+5)(x+7)^2 [-9.692, -3.533, -1.228, 1.85]}