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enter image description hereIf we were to use long divison to divide $x^2 - y^2$ by $x + y$, the long division easily gives $x-y$. But if I were to use long division to divide $x^2 - y^2$ by $y + x$ instead, the quotient becomes lengthy and fruitless. Forgive me if I'm fundamentally flawed. I don't seem to understand this at all.

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    Would you mind showing us what you mean? Doing long division in MathJax is *extremely hard*, so a picture would suffice here – 2017-02-21
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    More importantly - why are you using long division here? Just use the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ – 2017-02-21
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    @brevan ellefsen I am well aware of the identity. I have difficulty with a more complex expression but I'd rather have a simple example for such a basic question. – 2017-02-21
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    @SiddharthJossy see my updated answer, if it helps please upvote and mark as the correct answer! :) – 2017-02-21

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Everything will be turned around, if you are dividing by $y+x$ then just turn the equation around to read $-y^2+x^2$ and it will come out nicely as you say. However, it does not matter the order. $$\frac{x^2-y^2}{y+x}=-y+\frac{x^2+yx}{y+x}=-y+x$$ Solution

As you can see, you just have to be careful!:)

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    my question is why does the order matter so much....it never matters in multiplication. :) – 2017-02-21
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    Oh I do not think it matters, you just have to remember which term it is you are focusing on, ill update my answer for you. – 2017-02-21
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    @SiddharthJossy sorry quick teacher meeting! I am updating with example now! – 2017-02-21
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    Always! Just please mark my answer correct with the green check:) @SiddharthJossy – 2017-02-21
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The order doesnt matter as long as you maintain the signs whenever you move the variables around in the equation. I.E. $$y-x=-x+y$$ There will be no difference in the quotient.

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Dividing $x^2-y^2$ by $y+x$ gives:$$\frac{x^2}{y}-y-\frac{x^3}{y^2}+x+\frac{x^4}{y^3}-\frac{x^2}{y}-\frac{x^5}{y^4}+\frac{x^3}{y^2}+\frac{x^6}{y^5}-\frac{x^4}{y^3}-\frac{x^7}{y^6}+\frac{x^5}{y^4}+\frac{x^8}{y^7}-\frac{x^6}{y^5}-\cdots$$ The pattern of signs is + - - +, and every fractional term cancels with the fifth term following, leaving $-y + x$. So going at it this way is lengthy but not fruitless. It seems the order does not affect the result, only the ease of attaining it, at least in this example.

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    `polynomial division can yield an infinite series--or not,--depending on how the terms are ordered` Don't know what you mean by that, and it's plain wrong at face value. Anyway, remember that the question was about [polynomial ***long*** division](https://en.wikipedia.org/wiki/Polynomial_long_division) which by definition can *never* yield infinite series. – 2017-02-21
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    @dxiv--OP asks why the result seems different when dividing by y+x instead of x+y. I meant to say above that the quotient is the same, even though it goes on and on when you divide by y+x. Is it mistaken to call such a quotient an infinite series? – 2017-02-22
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    You seem to be working with the series expansion: $$ \frac{x^2-y^2}{x+y}=\frac{\frac{x^2}{y}-y}{1+\frac{x}{y}}=\left(\frac{x^2}{y}-y\right)\left(1-\frac{x}{y}+\frac{x^2}{y^2}-\cdots\right) $$ This *not* what polynomial long division is about, see the link in my previous comment. The polynomial long (or Euclidian) division calculates the quotient and remainder of the polynomial division, which are both polynomials, and are both uniquely defined. – 2017-02-22
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    @dxiv--I wasn't working with the series expansion, but just doing what I thought was long division of x^2-y^2 by y+x and got, after a full page, the result given above, which I probably should have offered as a comment rather than an answer. I've edited my "answer" to make it more modest. If there's still something amiss in it, I'm listening. What do you get for a result? What would you say to the OP? I didn't know if by "messy" he meant the calculation broke down, or just that it was long. Thanks for the link, I'll try to get straight on the definition of polynomial long division. – 2017-02-22
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    Polynomial long division refers to univariate polynomials. In the second example, the OP swapped the "main" variable in the divisor but not in the dividend, so what followed is no longer a legitimate long division, as pointed out in Nick Pavini's answer. No matter which variable you choose as your primary variable, the result of a properly done long division should be the same $x-y$. – 2017-02-23
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    @dxiv--ok, must choose primary variable ("which term you are focusing on", in Nick Pavini's comment). Thanks. – 2017-02-23