How do you solve #x^3+2x^2-4x-8>=0#?

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Answer 1 · 2016-11-23T09:12:26

The answer is #x>=2# or #x in [2,+oo[ #

Let #f(x)=x^3+2x^2-4x-8#

Then #f(2)=8+8-8-8=0#
Therefore, #(x-2)# is a factor of #f(x)#

#f(-2)=-8+8+8-8=0#
Therefore, #(x+2)# is a factor of #f(x)#

So, #(x+2)(x-2)=x^2-4# is a factor of #f(x)#

To find the last factor, let's do a long division

#color(white)(aaaa)##x^3+2x^2-4x-8##color(white)(aaaa)##∣##x^2-4#

#color(white)(aaaa)##x^3##color(white)(aaaaaaa)##-4x##color(white)(aaaaaaa)##∣##x+2#

#color(white)(aaaa)##0+2x^2##color(white)(aaaaa)##0-8#

#color(white)(aaaaaa)##+2x^2##color(white)(aaaaaaa)##-8#

#color(white)(aaaaaaaa)##0##color(white)(aaaaaaaaaa)##0#

So, #f(x)=(x+2)^2(x-2)#

As #(x+2)^2>0#

Therefore, the sign of #f(x)# will depend on the sign of #(x-2)#

When #x<2#, #f(x)<0#

and when #x>=2#, #f(x)>=0#

graph{x^3+2x^2-4x-8 [-20.28, 20.27, -10.14, 10.14]}