How do you solve #((x+3)(x+5))/(x+2)>=0#?

1 Answer
Mar 17, 2017

Answer:

The solution is # x in [-5,-3] uu]-2, +oo[#

Explanation:

We solve this inequality with a sign chart

Let #f(x)=((x+3)(x+5))/(x+2)#

The domain of #f(x)# is #D_f(x)=RR-{-2}#

Let build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##-3##color(white)(aaaaaa)##-2##color(white)(aaaaaaa)##+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+3##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,-3] uu]-2, +oo[#