How do you solve #(x^2+3)(x^2+3x-4)^3(x-1)^3 > 0#?

My answer: #(-4;-3)uu(3;+oo) #

1 Answer
Jul 17, 2018

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

Explanation:

#x^2+2>0# holds for all real #x#. So we have only consider the factors #x^2+3x-4# and #x-1#

Note that #x^2+3x-4=(x-1)(x+4)# so both factors are positive if #x>1#

Both factors are also positive if both factors are negative.This is the case for #-4 < x <1#.