Thanks, everyone. Yes, I understand inequalities and yes, I've graphed it (see below). As

@Jomo pointed out it should have been ```>``` rather than ```>=```. I calculated the ```(4, infty)``` as below on the left. So I get that part and I also see that the ```(-infty, 0)``` is valid because the ```x^2``` will be positive for any x value and ```-4(x)``` will also be positive for any x value that is negative, meaning ```x^2 -4x``` will be positive for any value less than 0. My issue is how to find the ```(-infty, 0)``` part algebraically. I can't seem to figure out how to structure an inequality similar to the ```x > 4``` inequality below that gives the ```x < 0```.

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The problem is that you can't divide both sides of an inequality by x unless you know whether it is positive or negative! Your x > 4 is not a valid conclusion.

This is why we teach the method that several of us have mentioned (#2, #5, and my links), which you claim to know but have not used.

There is an alternative method that works more like what you are trying to do, namely the use of cases:

We can say that if x > 0, then you can divide by x and get x > 4; so part of the solution is "x > 0 and x > 4", which simplifies to "x > 4".

On the other hand, if x < 0, then when you divide by x you reverse the direction of the inequality and get x < 4. So the other part of the solution is "x < 0 and x < 4", which simplifies to "x < 0".

Putting those together, the solution is "x > 4 or x < 0".

I find that considerably harder than the factoring method.