Question 43596
So you are right!


{{{f(x)=x(x-1)(x-4)^2}}}
For all real values of 'x', {{{(x-4)^2 > 0}}} except when x = 4.
So, for f(x) to be greater than 0, {{{x(x-1)>0}}}.
This is possible if


1) Case I
both x > 0 and (x - 1) > 0 which implies x > 1 [as x > 1 automatically satisfy both the conditions]


2) Case II
both x < 0 and (x - 1) < 0 which implies x < 0 [as x < 0 automatically satisfy both the conditions]


So for f(x) > 0, either x > 1 [from Case I] or x < 0 [from Case II]; also x not equal to 4.
Hence, the answer is (-{{{infinity}}},0)U(1,4)U(4,{{{infinity}}}).