How do you solve #(x-4)(x+1)>=0# using a sign chart?

1 Answer
Dec 1, 2016

Answer:

The answer is #x in ] -oo,-1 [ uu [4, +oo[#

Explanation:

Let #f(x)=(x-4)(x+1)#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##4##color(white)(aaaa)##+oo#

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

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

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

So, #f(x)>=0# when #x in ] -oo,-1 [ uu [4, +oo[#

graph{(x-4)(x+1) [-12.66, 12.65, -6.33, 6.33]}