if f(x)=x(x-1)(x-4)^2 give all values for x where f(x)>0?

Question

using interval notation

Answer 1

The function will only be negative when either exactly one or all three of the factors are negative. Since (x-4)^2 can never be negative, it will only be negative when x or x-1, but not both, are negative.

If x>1, then both factors will be positive, and if x<0, then they will both be negative. The only values of x that makes one factor negative and not the other are 0
Therefore, f(x)>0 when x<0 or x>1, written in interval notation as
(-∞,0)U(1,∞)

Answer 2

(x-4)^2 is always >0 whatever the value of x. Square of a negative number is always positive.

So, either +*+=+ or -*-=+
x(x-1) > 0 => x>0 & x-1>0 => x>0 & x>1 => x>1 OR
x(x-1) > 0 => x<0 & x-1< 0 => x< 0 & x<1=> x<0

Result X>1 or X<0

Answer 3

If x>1 then f(x)>0
If x<0 the f(x)>0

Answer 4

if f(x)=x(x-1)(x-4)^2 give all values for x where f(x)>0.

zeros at which f(x) changes sign: x = 0,1
f(x)>0 on x<0 or x>1