How do you solve #x^3+x^2-16x-16>0#?

2 Answers
Jun 1, 2018

Answer:

#-4 < x <-1# or #x>4#

Explanation:

We have
#x^3+x^2-16x-16=(x-4)(x+1)(x+4)#
For
#x>4#
is this product positive.

The product also positive for

#-4 < x < -1#

Jun 1, 2018

Answer:

The solution is #x in (-4,-1) uu(4,+oo)#

Explanation:

The inequality is

#x^3+x^2-16x-16>0#

Factorising,

#(x^2)(x+1-16(x+1))>0#

#(x^2-16)(x+1)>0#

#(x+4)(x-4)(x+1)>0#

Let #f(x)=(x+4)(x-4)(x+1)#

Build a sign chart

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

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

Therefore,

#f(x)>0# when #x in (-4,-1) uu(4,+oo)#

graph{x^3+x^2-16x-16 [-19.55, 26.05, -4.47, 18.34]}