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

1 Answer
Jan 19, 2017

Answer:

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

Explanation:

Let's rewrite the inequality

#x^3+3x^2-4<0#

We must factorise the LHS

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

The domain of #f(x)# is #D_f(x)=RR#

#f(1)=1+3-4=0#

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

To find the other factors, we do a long division

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

#color(white)(aaaa)##x^3-x^2##color(white)(aaaa)####color(white)(aaaaaaaa)##∣##x^2+4x+4#

#color(white)(aaaa)##0+4x^2##color(white)(aaaa)####color(white)(aaaaaaaa)#

#color(white)(aaaaaa)##+4x^2-4x#

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

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

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

Therefore,

#x^3+3x^2-4=(x-1)(x^2+4x+4)=(x-1)(x+2)^2#

So, we can make the sign chart

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

#color(white)(aaaa)##(x+2)^2##color(white)(aaaa)##+##color(white)(aaa)##0##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

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