How do you solve #x^3+2x^2<=8x# using a sign chart?

1 Answer
Feb 21, 2017

Answer:

The solution is #x in ]-oo,-4]uu[0, 2]#

Explanation:

Let's rewrite and factorise the inequality

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

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

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

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

Let #f(x)=x(x-2)(x+4)#

Now, we can build the sign chart

#color(white)(aaaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##0##color(white)(aaaaa)##2##color(white)(aaaaa)##+oo#

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

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aaaaa)##-##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)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0#, when #x in ]-oo,-4]uu[0, 2]#