Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x^3$.

share|cite|improve this question
$x^3 - 2x^2 = x^3(1 - \frac{2}{x})$ 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
up vote 2 down vote accepted

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 $|x| \to \infty$.

share|cite|improve this answer
$x^3$ 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 $0$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.