Understanding simplification of an algebra problem I have the problem $x^3 - 2x^2$. My book tells me that this problem is simplified to $x^3 (1 -(\frac{2}{x}))$. How does that work?
This step of my book I am in about the "end behavior" of trying to graph a polynomial function. Apparently once you get $x^3 (1 -(\frac{2}{x}))$, its equivalent is just $x^3$.
 A: As we take $|x|\to\infty$, we indeed find that $$\left|\frac{-2}{x}\right|=\frac2{|x|}\to 0,$$ so $$\frac{-2}{x}\to 0,$$ so $$\frac{x^3-2x^2}{x^3}=1-\frac{2}x\to 1.$$ Thus, since the functions $x^3$ and $x^3-2x^3$ have only finitely-many zeros, we say that they are asymptotically equal (as $|x|\to\infty$).
Intuitively speaking, not only is their end behavior the same, but we can make them get as close to each other as we like, if we pick $x$ values sufficiently far from $0$.
A: This is only valid when $x \neq 0$.  It follows from the distributive rule.  If $x \neq 0$, then
\begin{align*}
x^3 \left(1 - \frac{2}{x} \right) &= x^3 \cdot 1 - x^3 \cdot \frac{2}{x} \quad \text{(by distributive rule)}\\
&= x^3 - 2x^2.
\end{align*}
As Cameron Buie explained nicely, $x^3$ and $x^3(1 - \frac{2}{x})$ are asymptotically equal as $|x| \to \infty$.
A: Seeing that $x^3 - 2x^2$ and $x^3(1 - \frac{2}{x})$ is (nearly) the same thing is straight forward. Just expand $x^3(1 - \frac{2}{x})$, and observe that you get $x^3 - 2x^2$. That that those two expressions are not completely identical - $x^3 - 2x^2$ is defined for all $x$, while $x^3(1 - \frac{2}{x})$ is defined only for $x \neq 0$.
Regarding your second question - observe that happens if you put a very large number $x$ into $x^3(1 - \frac{2}{x})$. What can you say about $\frac{2}{x}$? And what thus about the term $(1 - \frac{2}{x})$? 
