# 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 $x^3$.

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$x^3 - 2x^2 = x^3(1 - \frac{2}{x})$ if you multiply the right hand side out you will end up with the left hand side. – Sam Jones Nov 6 '12 at 9:55
I don't think the later is simpler than the former. – ᴊ ᴀ s ᴏ ɴ Nov 6 '12 at 9:58
Your initial sentence is unclear. You say "I have the problem $x^3-2x^2$." But that doesn't identify any particular problem. There must be some words asking some question about $x^3-2x^2$, or asking you to do something with $x^3-2x^2$. But you don't tell us what those words say. – Michael Hardy Nov 6 '12 at 10:24
@MichaelHardy the third sentence states my question about the problem. Instead of criticizing my wording, can you give me an example of a more "proper" way of stating my question? – Tyler Zika Nov 6 '12 at 18:30
@TylerZika : You could say: "I am trying to find the end behavior of $x^3-2x^2$. My book tells me that this expression is simplified to $x^3\left(1−\left(\frac2x\right)\right)$." etc.... – Michael Hardy Nov 6 '12 at 19:17

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 $|x| \to \infty$.

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$x^3$ is equivalent to x^3(1-(2/x)) in graphing when you are using large x values, as x approaches -infinity, in addition when x approaches infinity. That is what my book says : / – Tyler Zika Nov 6 '12 at 10:01
Ah, I see. I just edited my reply accordingly. – littleO Nov 6 '12 at 10:12
I like how you stated the first part of your response though. :) – Tyler Zika Nov 6 '12 at 10:14

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 $0$.

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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 $(1 - \frac{2}{x})$?

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