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

1 Answer
Jan 17, 2017

Answer:

The answer is #x in [-2,-1] uu [1,+ oo[ #

Explanation:

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

#f(1)=1+2-1-2=0#

So, #(x-1)# is a factor of #f(x)#

To find the other factors, we do a long division

#color(white)(aaaa)##x^3+2x^2-x-2##color(white)(aaaa)##∣##x-1#

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

#color(white)(aaaa)##0+3x^2-x#

#color(white)(aaaaaa)##+3x^2-3x#

#color(white)(aaaaaaaa)##+0+2x-2#

#color(white)(aaaaaaaaaaaa)##+2x-2#

#color(white)(aaaaaaaaaaaaa)##+0-0#

Therefore,

#(x^3+2x^2-x-2)/(x-1)=x^2+3x+2=(x+1)(x+2)#

So,

#f(x)=(x+2)(x+1)(x-1)#

Now, we can construct the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##-1##color(white)(aaaa)##1##color(white)(aaaaaa)##+oo#

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

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

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

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

Therefore,

#f(x)>=0#, when #x in [-2,-1] uu [1,+ oo[ #