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

1 Answer
Jul 23, 2018

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)xcolor(white)(aaaa)-oocolor(white)(aaaaaaa)-8color(white)(aaaaaa)-7color(white)(aaaaa)-5color(white)(aaaa)+oo

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

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

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

color(white)(aaaa)f(x)color(white)(aaaaaaa)-color(white)(aaaa)0color(white)(aaa)-color(white)(aa)0color(white)(aaaa)-color(white)(a)0color(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]}