How do solve #x^3+2x^2<=8x# and write the answer as a inequality and interval notation?

1 Answer
Oct 24, 2016

Answer:

The answer is #-oo<=x<=-4# and #0<=x<=2#

Explanation:

Let us factorise the inequality and make a sign chart

#x^3+2x^2-8x<=0#

#x(x^2+2x-8)<=0#

#x(x-2)(x+4)<=0#

So the values are #x=0#, #x=2#and #x=-4#

#x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##-4##color(white)(aaaaa)##0##color(white)(aaaaa)##2##color(white)(aaaaa)##+oo#
#x##color(white)(aaaaa)##color(white)(aaaa)##-##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#x-2##color(white)(aaaa)##color(white)(aa)##-##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#x+4##color(white)(aaaa)##color(white)(aa)##-##color(white)(aaaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#
#X##color(white)(aaaaaa)##color(white)(aa)##-##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
So the answer is #-oo<=x<=-4# and #0<=x<=2#