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

Question

1185 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-30T00:39:56

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

# x^3+4x^2 > x+4 #
# x^2(x+4) > x+4 #
# x^2(x+4) - (x+4) > 0 #
# (x^2-1)(x+4) > 0 #

We need to find the critical values (where a sign change can occur), which is given by # (x^2-1)(x+4) = 0 #

Either # x^2-1 = 0 => x =+-1#
or # x+4 = 0 => x=-4 #

So we need to examine the behaviour of (x^2-1)(x+4) in each of the intervals:

enter image source here

Sign of # (x^2-1)(x+4) = { ( (+)(-)<0,x < -4 ),( (+)(+)>0,-4 < x < -1 ),( (-)(+)<0,-1< x < 1 ),( (+)(+)>0,x > 1 ) :}#

So the solution for # (x^2-1)(x+4) > 0 # is
#-4 < x < -1 # or # x > 1 #