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

1 Answer
Narad T.
Oct 24, 2016

The expression is #>=0# for #x>2#

Explanation:

let #f(x)=x^3+2x^2-4x-8#
Let us start by factorising the expression
By trial and error, ·#(x-2)# is a factor as #f(2)=0#

let us make the long division
#x^3+2x^2-4x-8##color(white)(aaa)##∣##x-2#
#x^3-2x^2##color(white)(aaaaaaaaaaa)##∣##x^2+4x+4#
#0+4x^2-4x#
#color(white)(aaaa)##4x^2-8x#
#color(white)(aaaaa)##0+4x-8#
#color(white)(aaaaaaaaa)##4x-8#
#color(white)(aaaaaaaaaa)##0+0#

so #f(x)=(x-2)(x^2+4x+4)=(x-2)(x+2)^2#
#f(x)>=0#
Looking at the factorisation
#(x+2)^2>=0# for all values of #x∈RR#
#(x-2)>=0# for #x>=2#
So the answer is #x>=2#

Related questions
See all questions in Polynomial Inequalities
Impact of this question
1654 views around the world
You can reuse this answer
Creative Commons License