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

1 Answer
Apr 17, 2017

Answer:

The solution is #x in (-oo,2]#

Explanation:

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

#f(2)=8+8-8-8=0#

Therefore,

#(x-2)# is a factor of #f(x)#

To find the other factors, perform a long division

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

#color(white)(aaaa)##x^3-2x^2##color(white)(aaaaaaaaaaaa)##|##x^2+4x+4#

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

#color(white)(aaaaaaa)##+4x^2-8x#

#color(white)(aaaaaaaaaa)##0+4x-8#

#color(white)(aaaaaaaaaaaa)##+4x-8#

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

So,

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

We build a sign chart

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

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

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

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

Therefore,

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

graph{x^3+2x^2-4x-8 [-21.2, 19.34, -12.25, 8.03]}