How do solve #2x^3+x^2>6x# and write the answer as a inequality and interval notation?

1 Answer
Feb 23, 2017

Answer:

The solutions are #-2 < x < 0# and #x > 3/2#
Or in interval notation, #x in ]-2,0[uu]3/2,+oo[#

Explanation:

Let's rewrite and factorise the inequality

#2x^3+x^2>6x#

#2x^3+x^2-6x>0#

#x(2x^2+x-6)>0#

#x(2x-3)(x+2)>0#

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

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaa)##0##color(white)(aaaaa)##3/2##color(white)(aaaaaa)##+oo#

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

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

#color(white)(aaaa)##2x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##+#

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

Therefore,

#f(x)>0# when #-2 < x < 0# and #x > 3/2#

Or in interval notation, #x in ]-2,0[uu]3/2,+oo[#