Am I doing something wrong in making this inequality or do algebra tools just usually not factor? So, I thought of something nerdy...I was making an algebra problem to disguise it so that to see the joke, you'd have to a bit of work, but I think that I might be doing something wrong, because neither Wolfram|Alpha nor MyAlgebra can solve it...I fixed a few things that I had wrong initially, but I don't see anything wrong now...you're supposed to solve for i.
$$\begin{array}{rcl}
    (5u)^2-(i+2u)^2 &\gt& 20u^2+4u(i+2u)\\
    (5u+(i+2u))(5u-(i+2u))&\gt& 4u(5u+(i+2u))\\
    5u-(i+2u)&\gt& 4u\\
    3u-i&\gt& 0\\ 
    i&\lt& 3u
\end{array}
$$
I figured that you could factor the left side and then divide (5u+(i+2u)) out of both sides, but maybe not?
 A: Your mistake lies in trying to cancel $5u +(i+2u)$ without being careful about whether it is positive, negative, or $0$. Added: There is also a miscalculation in the third line.
Remember that "cancelling factors" is really multiplying by the inverse. But when you multiply an inequality by a number, you have to be careful about whether the number is positive or negative: if you have $a\gt b$ and $c\gt 0$, then $ac\gt bc$; but if $c\lt 0$, then you have $ac\lt bc$ (multiplying by negative numbers reverses the inequality).
When you try cancelling $5u+(i+2u) = 7u+i$, by not reversing the inequality you are implicitly assuming that $7u+i\gt 0$; that is, that $-7u\lt i$. So in fact, your conclusion is that if $-7u\lt i$ and the inequality holds, then $i\lt 3u$. But what if $7u+i\lt 0$? Then when you divide both sides by $7u+i$ in order to cancel the factor, you must reverse the inequality, so you actually get $5u -(i+2u) \lt 4u$.
Added: At this point you made a slight mistake with the signs. The left hand side is $5u - (i+2u) = 3u-i$; you had $5u - i + 2u$ (missing the parenthesis) so you incorrectly got $7u - i$. 
So, if $7u+i \gt 0$, then you get $3u-i\gt 4u$, or $i\lt -u$. Thus, if $i\gt -7u$, then you must also have $i\lt -u$, or $-7u\lt i \lt -u$. Note that this can only occur if $u\gt 0$.
If $7u+i \lt 0$, then when you cancel that factor you get $3u-i \lt 4u$, or $i\gt -u$; so you have that both $i\lt -7u$ and $i\gt -u$ must hold. That is, $-u\lt i\lt -7u$. Note that this can only occur if $u\lt 0$. 
So in fact you have two parts of the solution: you either have $u\gt 0$ and $-7u \lt i\lt -u$, or else you have $u\lt 0$ and $-u\lt i\lt-7u$. 
