My understanding of division in general is this:

dividend / divisor = quotient

The quotient X divisor will give you the dividend.

So if we had an equation:

$(x - 2)\left){\vphantom{1{2{x^3} - 11x + 6}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - 11x + 6}}}$

If we just address the $2{x^3}$ expression: How is it that $2{x^3} \div x$ would give you the same quotient as $2{x^3} \div (x - 2)$, i.e.:

$\eqalign{ & 2{x^3} \div (x - 2) = 2{x^2} \cr & 2{x^3} \div (x) = 2{x^2} \cr} $

Surely the (x-2) answer has to be different to the answer where you are just dividing by x? I know I havent unravelled some big fallacy in division, I'd just like to understand it and I'd love for someone to explain where I'm going wrong, thank you.

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    $\begingroup$ First of all, division in general has a remainder. Second, what makes you think $2x^3$ divided by $x-2$ is, or should be, $2x^2$? $\endgroup$ – Gerry Myerson Sep 23 '13 at 13:13
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    $\begingroup$ $2{x^3} \div (x - 2) = 2{x^2}$ remainder $4x^2$. i.e. $2{x^3} \div (x - 2) = 2{x^2}+\frac{4x^2}{x-2}$. This can be divided further to give $2x^2+4x+8$ remainder $\frac{16}{x-2}$. $\endgroup$ – Mufasa Sep 23 '13 at 13:15

As Gerry Myerson says, division in general has a remainder. You don't have $$\text{dividend} \div \text{divisor} = \text{quotient}\tag1$$ unless the remainder is zero. In general, what you have is $$\text{dividend} = (\text{quotient}\times\text{divisor}) + \text{remainder}$$ and when the remainder is zero, it simplifies to (1). For example, we can't say $$100 \div 7 = 14$$ which is wrong, but we can say that $$100 = (14\cdot 7) + 2.\tag2$$ The 2 here is the remainder.

There's a theorem, often just called the division theorem, that holds in a great many situations, and says that for any $a$ and $b\ne 0$, there are a unique quotient $q$ and remainder $r$ such that $$a = qb+r$$ where (this is important) $r$ is in some sense "smaller" than $b$. When $a$ and $b$ are integers, "smaller" can be understood to mean that $|r| < |b|$; in (2) above we had $r=2$ and $b=7$ so we did indeed have $|r|<|b|$. When $a$ and $b$ are polynomials, "smaller" is a little different: it means that the degree of $r$ is less than the degree of $b$.

Now what's $2x^3 \div (x-2)$? Here the division theorem tells us that we can find quotient $q$ and remainder $r$ such that $$2x^3 = q(x-2) + r$$ and the degree of $r$ is less than the degree of $x-2$. That is, $r$ is a constant. In this case the solution has

$$2x^3 = (2x^2+4x+8)(x-2) + 16$$

The remainder is 16, not zero, because $x-2$ does not divide evenly into $2x^3$; this is analogous to the way that $7$ does not divide evenly into $100$ and leaves a remainder of 2. On the other hand $2x^3$ does divide evenly by $x$ because the remainder is zero; we have

$$2x^3 = 2x^2\cdot x + 0$$

and again the degree of the remainder, 0, is less than the degree of the divisor, $x$. In this case we also have the opposite equation

$$2x^2\div x = 2x^2.$$

When we work in a context called a "field", such as the rational numbers, then everything always divides, and the remainder guaranteed by the division theorem is always zero. Going back to our $100\div 7$ example, if we are working in the rationals instead of in the integers, then the quotient and remainder guaranteed by the division theorem are $\frac{100}{7}$ and $0$, and we have $$100 = \frac{100}{7}\cdot 7 + 0.$$ Since the remainder is $0$, we can also say $$100\div 7 = \frac{100}7$$.

  • $\begingroup$ I understand it better now, thank you! $\endgroup$ – seeker Sep 23 '13 at 14:05
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    $\begingroup$ Polynomial long division. It is completely analogous to integer long division: you start by estimating what monomial $M_2$ will make $M_2\cdot(x-2) \approx 2x^3$. The answer is clearly $M_2 = 2x^2$. Then you subtract $M_2\cdot(x-2) = 2x^3 - 4x^2$ from the divisor $2x^3$, obtaining a remainder $R_2 =4x^2$, and repeat the process, again finding a monomial $M_1$ that makes $M_1\cdot(x-2)\approx 4x^2$, and so on. The remainder decreases at each stage. At the end, the quotient is $M_2+M_1+M_0$ and the final remainder $R_0$ is the remainder. $\endgroup$ – MJD Sep 23 '13 at 18:22

The same way multiplication is addition many times, division is subtraction many times. When you do long division you are trying to subtract the divisor from the dividend as many times as you can, until you cannot do it any further.

For example: If we try to divide $47$ by $2$. In long division we take the $4$ and divide it by $2$, i.e. we know that we can subtract $2$, two times from $4$. We know that that $4$ is actually representing a $40$ in that $47$, when we use decimal notation. This is how decimal notation works: $47=4\cdot 10+7$. Then we really subtract $2\cdot 10$ times $2$ from $47$. We get $7$, and then it is easy to see we can still subtract $2$ another extra $3$ times. In total we subtracted $2$, $2\cdot10+3$ times. So $47=2\cdot 23+1$.

Polynomials are pretty much the same as decimal notation, just replace the $10$ by an $x$: as if we write $4x+7$ instead of $4\cdot 10 +7$. Notice how the procedure we did above also gives you the division of $4x+7$ by $2$.

For numbers of more digits (and polynomials of higher degree) it works similarly. You try to subtract as much as possible (you do it as much as possible to be as efficient as possible. This is of course not the only way to divide number or polynomials. There are also other ways to try to be efficient). That is why you try to subtract as much as possible the terms of higher degree (as with numbers you try to subtract the digit that is multiplied by the higher power of $10$ in decimal notation).

  • $\begingroup$ So, when you multiply the polynomial $x$ by the polynomial $x$, you are adding $x$ to itself $x$ times? I don't see any way to make sense of that. $\endgroup$ – Gerry Myerson Sep 25 '13 at 13:38
  • $\begingroup$ Don't worry. Not everybody is good at generalizing ideas. $\endgroup$ – OR. Sep 25 '13 at 14:16
  • $\begingroup$ Ways to make sense of that: No1: Polynomials are determined by their values at the integers (actually you only need as many values as their degree). So you can make sense of it as '$x$ times' meaning actually times, where you replace $x$ by integers in large enough set, or perhaps all integers. $\endgroup$ – OR. Sep 25 '13 at 14:21
  • $\begingroup$ No2: Purely symbolic, as a direct generalization of the procedure of Euclidean division, to that of polynomials. This way you get the notion of Euclidean domain in general. Replacing the role of the absolute value in the integers by the degree function in the polynomials. $\endgroup$ – OR. Sep 25 '13 at 14:23
  • $\begingroup$ I can do without the condescension. How do you interpret $\sqrt2\times\sqrt3$ as "addition many times"? You may also want to have a look at Keith Devlin's many columns on this topic, beginning with maa.org/external_archive/devlin/devlin_01_10.html and working your way backwards. But I suppose you don't think Devlin is much good at generalizing ideas. $\endgroup$ – Gerry Myerson Sep 26 '13 at 0:09

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