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

1 Answer
Feb 4, 2017

Answer:

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

Explanation:

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

Before, we need to find the factors of #f(x)#

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

Therefore, #(x+1)# is a factor

To find the other factors, we perform a long division

#color(white)(aaaa)##x^3+x^2+4x+4##color(white)(aaaa)##|##x+1#

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

#color(white)(aaaaa)##0+0+4x+4#

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

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

Therefore,

#f(x)=(x+1)(x^2+4)#

#AA x in RR,(x^2+4)>0 #

So, we can build the sign chart

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

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

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

Therefore,

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