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

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

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Answer 1 · 2018-07-17T20:47:31

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

#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#.