How do you solve #(x+8)(x+2)(x-3)>=0# using a sign chart?

1 Answer
Dec 19, 2016

Answer:

The answer is #x in [-8,-2] uu [3, +oo[ #

Explanation:

Let #f(x)=(x+8)(x+2)(x-3)#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-8##color(white)(aaaa)##-2##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+8##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

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

Therefore,

#f(x)>=0#, when #x in [-8,-2] uu [3, +oo[ #

graph{(x+8)(x+2)(x-3) [-154.2, 146, -60.4, 89.8]}