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

1 Answer
Nov 17, 2016

Answer:

The naswer is #x in [-4,-1] uu [1, +oo] #

Explanation:

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

Then, #f(1)=1+4-1-4=0#

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

To find the other factors, let's do a long division

#color(white)(aaaaa)##x^3+4x^2-x-4##color(white)(aaaaa)##∣##x-1#

#color(white)(aaaaa)##x^3-x^2##color(white)(aaaaaaaaaaaaa)##∣##x^2+5x+4#

#color(white)(aaaaaa)##0+5x^2-x#

#color(white)(aaaaaaaa)##+5x^2-5x#

#color(white)(aaaaaaaaaaaa)##0+4x-4#

#color(white)(aaaaaaaaaaaaaa)##+4x-4#

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

Therefore, #f(x)=(x-1)(x+1)(x+4)>=0#

Let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##-1##color(white)(aaaa)##+1##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

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

Therefore #f(x)>=0# if #x in [-4,-1] uu [1, +oo] #

graph{x^3+4x^2-x-4 [-20.28, 20.27, -10.13, 10.13]}