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I'm trying to understand why polynomial long division works and I've hit a wall when trying to understand why we multiply all terms of the divisor by the partial quotient. Consider:

$$\frac{x^2 + 3x + 2}{x + 2}$$

During the first step we divide ${x^2}$ by ${x}$ giving us a partial quotient of ${x}$. Next, we multiply the partial quotient by the first term of the divisor, giving us ${x^2}$. So far so good, but this next step is what I don't understand: why do we then multiply the next term of the divisor by the partial quotient (i.e. ${x}$ * ${2}$)? It seems as though we're testing to see if ${x}$ can be divided into the first term of the dividend, and if it can, then we distribute the result over the whole divisor which is then subtracted. I don't understand how we can do that when we're only testing the divisibility of those first terms.

I tried a slightly different example to see what happens which highlights what I mean: $$\frac{x^2}{x + 1}$$

First, we test the divisibilty of ${x^2}$ by ${x}$. Obviously it goes ${x}$ times, but then I wanted to see if I could carry on and multiply ${x}$ * ${1}$. I carried on and the quotient becomes ${x + \frac{1}{x + 1} - 1}$. Checking $({x + 1})({x + \frac{1}{x + 1} - 1})$ does give the original dividend of ${x^2}$. Having worked through that problem I just can't see why we're able to do the multiplication of the second term of the divisor, subtract it and get everything to hold true. Essentially, if I subtract ${x^2}$ from ${x^2}$ I end up with nothing. So the second multiplication, to me at least, seems unclear as to why it works and what the purpose is.

I know I'm missing something simple here but I can't seem to make the connection. Could someone explain this to me please?

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2 Answers 2

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Think about polynomial long division in much the same way that you think about division of real numbers.

For instance, say you want to divide 48820 by 28. In a sense, doing division in this way is "greedy" -- you divide by the largest possible parts first, then work down from there.

Now, , when you long divide by 28, you first see if 28 goes into the leading term, 4. It does not, so then you check if it goes into 48 (which it does exactly once). Then, when you "bring down" the next term, you multiply the "partial quotient" -- 1 -- by 28. Right?

Now think of 28 as 2*10+8. Same operations hold.

Now, replace 10 with x. In order to keep the algorithm the same, you have to multiply by the whole denominator.

Edit: To expand a little...

Consider the previous example. 48820/28 = 1743+16/28.

Now, let's compute $\frac{4x^4+8x^3+8x^2+2x+0}{2x+8}$.

By doing polynomial long division, we obtain: $2x^3-4x^2+20x-79+\frac{632}{2x+8}$. Now you might say, "wait a minute, that has a leading coefficient of 2, but obviously 28 goes into 48 only once!" Yes, this is true, but notice that we have minus signs here. That is key.

Now, let's set $x = 10$ and see what we compute: $2000-400+200-79+\frac{632}{28}$. Doing the arithmetic, we see this is exactly our desired result.

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  • $\begingroup$ Thank you for your answer. Sorry for my delayed reply but I've been snowed under since I asked my question. Whilst your answer has helped me, I've found this part confusing: Now think of 28 as 2*10+8. Same operations hold. I tried something similar so I feel like I'm on the right track, but it's not clicking for me. To keep it simple, I tried 48 / 28, written as (4 * 10 + 8) / (2 * 10 + 8). Working that problem gives me 2 - (8 / 2 * 10 + 8). I can see that it works, but I can also see that we've divided by more than what we originally had. $\endgroup$
    – John H
    Jul 22, 2012 at 15:44
  • $\begingroup$ I can see that it rebalances, but I'd like to understand how someone developed this method, assuming this was created for real numbers first, and realising it would work like this algebraically. The result is the same but the rebalancing by having a negative remainder seems very different to the method for real numbers (at least to me). $\endgroup$
    – John H
    Jul 22, 2012 at 15:46
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    $\begingroup$ Don't think about it as a negative correction, think of it as a natural operation. It just happens that when dividing numbers, we do it in such a way that the terms are always positive. We could, instead, do it in such a way that the terms are always negative. As far as why the operations are analogous, it is sufficient to note that $\Bbb R$ is isomorphic to the field of polynomials with real coefficients, $\Bbb R\left[x\right]$. $\endgroup$
    – Emily
    Jul 22, 2012 at 20:33
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    $\begingroup$ @JohnH to expand a bit on my earlier comment (now that I am no longer on my phone), the way we learn long division of numbers in school is quite "unnatural" in the sense that the absence of negative terms is by design, not a mathematical consequence of the operation. Indeed, when we divide 48 by 28, we ask "how many 28s go into 48 without going over?" We could instead frame long division in such a manner as to say, "how many 28s are necessary to completely cover 48?" This would lead to negative terms. The algorithm would be complex and probably too difficult to teach young students, however. $\endgroup$
    – Emily
    Jul 23, 2012 at 1:33
  • $\begingroup$ Ah, now that makes complete sense. I think that was probably what I couldn't understand all along. I went back over the 48 / 28 problem and realised that I'd get 1 + (20 / 28). Writing that as (4 * 10 + 8) / (2 * 10 + 8) gives 2 - (8 / 28). Then I realised they are the same answer, expressed differently. I can't thank you enough for your help. $\endgroup$
    – John H
    Jul 23, 2012 at 13:52
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You start with $$\dfrac{x^2}{x + 1}.$$

You want the leading term of the denominator to be the same as the leading term of the numerator so write this as $$x \cdot \dfrac{x^2}{x^2 + x} = x \cdot \dfrac{x^2+x - x}{x^2 + x} =x \cdot \dfrac{x^2+x}{x^2 + x} + \dfrac{-x}{x + 1} =x + \dfrac{-x}{x + 1}.$$

Then do it again for the next term $$x -1\cdot \dfrac{-x}{-x -1} =x -1\cdot \dfrac{-x-1+1}{-x -1} =x -1\cdot \dfrac{-x-1}{-x -1} + \dfrac{1}{x + 1} =x -1 + \dfrac{1}{x + 1} . $$

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  • $\begingroup$ Thank you, too, for your answer and again, I'm sorry it's taken such a long time for me to reply. I like that you've given me a completely different way to view the problem. $\endgroup$
    – John H
    Jul 22, 2012 at 15:47

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