Talk:Ruffini's rule/Archive

Horner's rule
A quick Google reveals that Ruffini's rule is basically equivalent to Horner's rule. (IANAM; er, I am not a mathematician.) Should we merge the info from this page over there and replace this page with a redirect? --Diberri | Talk 04:20, Jun 30, 2004 (UTC)


 * Although I've never heard of Horner's rule before, reading the articles it's clear that they (the articles, at least) deal with distinct things. (Albeit related.) So no. But probably somebody who's more familiar with the rules should add a sentence or two to each article with a link to the other. Doops 22:18, 30 Jun 2004 (UTC)


 * As far as I know they are the same. I have never heard the term Ruffini's rule before. I would call them Horner algorithm, Horner scheme and complete Horner scheme. I did some work on Horner scheme recently and I will try to merge the pages soon (If nobody objects).MathMartin 17:26, 15 Aug 2004 (UTC)


 * Um, am I missing something? I have no external knowledge of Ruffini or Horner or any of these things; but reading these pages it's clear that they deal with two rather different things:
 * The page on Ruffini's rule gives a shorthand algortithm for polynomial long division. It only works, however, when dividing by linear factors in the form (x+a) or (-x+a).
 * The page on Horner's rule gives a way of quickly evaluating polynomials by repeatedly dividing out x. (This appears to be more efficient and accurate for floating-point processors, or something like that.) It's clear how Ruffini's rule could be useful in this process: we could use it to divide out the linear factor (x+0). (Hence it's an even more special case of polynomial long division.)
 * So it appears that 1) Horner's rule is ONE possible application of Ruffini's rule (there are many others). 2) But Ruffini's rule, on the other hand, isn't entirely necessary for Horner's rule; we could just use regular polynomial long division if we preferred. Hence I would conclude that the two are, properly, two separate topics to be dealt with separately. Now, maybe the pages are missnamed; for all I know they should be called "Fred's algorithm" and "Drusilla's process"; but, under whatever name, they do seem like two separate things to me. So, again: am I missing something? Doops 21:50, 17 Aug 2004 (UTC)

The important part about Horners rule is not the rearrangement of the polynomial into the form
 * $$p(x) = a_0 + x(a_1 + x(a_2 + \cdots x(a_{n-1} + a_n x)))$$

We just do this to make the recursive evaluation of the polynomial clearer.

The important part is this definition

\begin{matrix} b_n & := & a_{n}            \\ b_{n-1} & := & a_{n-1} + b_{n} x \\ & \vdots &             \\ b_{0} & := & a_{0} + b_{1} x \end{matrix} $$

We can write this recursive definition in a graphic scheme (called the Horner scheme).

|       an        an-1        ... a1        a0     | x |                 bn-1x       ... b1x       b0x |-    |        an     an-1+bn-1x   ... a1+b1x      a0+b0x |    |      = bn-1     = bn-2       ... = b0       = s

Lets take the example from the Ruffini article. Here we are dividing
 * $$P(x)=2x^3+3x^2-4\,\!$$

by


 * $$Q(x)=x+1\,\!$$

resulting in the follwing table.

|    2     3     0     -4     |  -1 |          -2    -1      1 |     |     2     1    -1     -3     |{result coefficients}{remainder}

The result is


 * $$P(x)=Q(x)*R(x)+s\,\!$$, where


 * $$R(x) = 2x^2+x-1\,\!$$ and $$s=-3\,\!$$.

But this is just one interpretation of the table data. We could also forget about the result coefficients and just look at the remainder and say we used Horners rule to evaluate P(-1) as


 * $$P(-1)=3$$

See ? Same algorithm, different interpretation of result. Sorry for the bad formatting and the bad explanation, I am tired :) MathMartin 23:42, 17 Aug 2004 (UTC)

A digression on thinking vs. algorithms
Just a quick POV comment: I strongly urge against beginners learning and using this technique, since it involves abdicating your role as a thinking human being. Unless you see what's going on, step by step, when applying Ruffini's rule -- and this article doesn't focus on helping you see it -- you're just pushing numbers around on paper. So do polynomial long-division the real way! Respect yourself! Doops 22:18, 30 Jun 2004 (UTC)
 * Well, the whole POINT of easy-to-use algorithms is to "abdicate your role as a thinking human being". Do you really want to think through WHY Gaussian elimination works and write out systems of equations in all variables? I don't. Of course, I DO understand why G-J works, and can justify when pressed, but once I learned why, I forgot it. This seems to be a similar situation. Asking that people understand why an algorithm works before using it is helpful and instructive...demanding that they SHOULDN'T use it unless they understand it is ridiculous. If that were true, I'd have to throw away my TI-85, since I'm not familiar with many of the internal algorithms programmed into the computer. Revolver 23:21, 30 Jun 2004 (UTC)
 * I might hasten to add that polynomial division ITSELF is a notational algorithm, so by your logic, we should stop doing that, too. (I would guess many people don't understand why polynomial long division works.) Revolver 23:23, 30 Jun 2004 (UTC)
 * Sorry; I was just being grumpy. :) You're right that they're both algorithms; but the reason why I like polynomial division better is that is more transparent -- that is to say, if you look at it while you're doing it you can see what's going on. (Maybe that would be true with synthetic division too if I practiced it a lot.) To be perfectly honest, my math career's been mostly theoretical and I've never really been in the situation of needing to use any of these algorithms much; so I should get off my high horse. Doops 00:13, 1 Jul 2004 (UTC)
 * My math career is pretty theoretical, too (algebraic number theory), but I still don't want to give up MAPLE or MuPad! :-) Revolver 00:20, 1 Jul 2004 (UTC)
 * I think this is simply long division applied. I added that as a remark to the article Eef (A) 12:17, 1 Jul 2004 (UTC)

Errors
As far as I can see, there are a number of errors, oversights, or at least imprecisions in this article right now. But since I'm not entirely sure what the author meant to say, and since I really have no experience using this algorithm, I haven't tried to fix them. Please, please, somebody who can figure out what's going on-- help out!
 * in the "Uses of the rule" section on Polynomial root-finding, the author says he/she will find rational roots for a monic polynomial. S/he then suggests that the possible candidates are all the divisors of the linear constant. As an example, s/he gives {+1, -1, +2, -2} as the divisors of -2. But what about fractions? Isn't 1/2 a divisor of 2? (Yup.) So I almost changed the sentence to say "all integer divisors" which I suspect is what the author meant. (I believe the author meant integer divisors. Revolver) But here's the problem: if those braces are supposed to include all candidate roots, we can't limit ourselves to integers; for example, the factors of P(x)=x2-2.5x+1 are (x-2) and (x-1/2); 1/2 is a root. So maybe the method only finds the integer roots of a monic polynomial? (Not the rational roots as promised.)
 * Okay, here's what's going on. If you have a polynomial $$f(x) = a_n x^n + \dots a_1 x + a_0$$ (not necc monic) with integer coefficients, then every rational root of f(x) is of the form p/q, where p is an integer divisor of a0 and q is an integer divisor of an. If f(x) is monic, then q = 1, and this is what the contributor was saying. In the example you give, the problem is that 2.5 is not an integer, you have to multiply through by 2 to clear the denominator (no need to compensate...we're solving f(x) = 0). Then, you have 2x2 + 5x + 2, and the rational roots show up. This method will not give you any way of determined irrational or non-real roots. Revolver
 * Ah. See, I was assuming you could take a polynomial and make it nomic forcibly (i.e. by dividing through), which could very easily lead to non-integer coefficients. So: it's clear what to do if your polynomial is monic w/ integer coefficients. But if not -- if your polynomial is non-nomic or has non-integer coefficients (which comes to the same thing) -- is there a way worth including in this article to find all roots of that polynomial? Doops 01:27, 1 Jul 2004 (UTC)
 * Not really. Welcome to wild world of numerical analysis. Revolver
 * Okey-dokey. Well, I'll make the article clearer on this point. Doops 16:02, 1 Jul 2004 (UTC)


 * Later in the root-finding example, I cleaned up some of the language about the two "methods"; but only on blind faith. What exactly is the status of "repeated roots"? I suspect that the first method does in fact pick them up; it just doesn't tell you that they are repeated. Is that true?
 * It should pick them up, yes, you're correct. You just have to count them and keep track. Revolver


 * The polynomial factorization description is clearly wrong. A factored polynomial is by definition factored; it can't have the format R(x) + S(x)! Are we not "factoring" but "partially factoring," or something like that? Oh -- no wait! Won't S(x) always be, by definition, zero? So maybe we don't want P(x) mod R(x), we actually want P(x) / R(x); and then we want to multiply S(x) by (rather than add it to) R(x)? Example: consider x3-2x2+x-2. By the def'n given, R={2}, and R(x)=x-2. Using long division, we find that, as expected, P(x) mod R(x) is zero; but P(x) divided by R(x) is x2+1 -- and then, in fact, this quantity times x-2 (aka R(x)) is our original P(x). So is that what was meant? (I'll betcha it is; but I won't change until an expert confirms it.
 * I believe the author is trying to perform the division algorithm (long division). Your observation just confirms by definition that P = 0 (mod R), by definition this means P = RT, for some T. What they did is to confirm that each of the 3 numbers was a root, then the product of the linear factors must divide the polynomial. (The whole point of this whole article is that with polynomials over a field, (x &minus; a) divides F(x) if and only if F(a) = 0. This should be mentioned somewhere. It doesn't happen over ordinary rings, e.g.) It could certainly be cleaned up and made easier to follow. Revolver
 * What's written in the article right now is definitely wrong. It says P(x)=R(x)+S(x), which by its definition of those polynomials, in the case of my example above, gives x3-2x2+x-2 = (x-2) + 0. This is false, QED.
 * No, R(x) = product of all linear (x - root), not just the (x - 2).


 * But let's also think about the goal of this section.
 * 1. It could be "to factor any polynomial with only first-order factors." In that case, the thing is trivial and we can lose all the S(x) nonsense altogether. [The problem is, you have no way of knowing this ahead of time, unless, of course, you use calculus. Revolver]
 * 2. It could be "to factor any polynomial whatsoever. In that case, remember that we've already found all linear factors of it... [No, we've only found all linear factors (x - a), where a is rational. Revolver] ...so we multiply all the linear factors (including duplicates) together, and divide (via polynomial long division) the polynomial by the result; this gives us all the higher-order stuff which won't go away. Now if we write out that "what's left" next to all the linear factors we can say we've factored the original polynomial nearly completely. (We actually can't be sure, though, that we've factored it utterly completely -- what if "what's left" is a fourth-order polynomial which can be factored into two quadratics? So this procedure is of limited utility and doesn't really apply to "any polynomail whatsoever.")
 * Again, welcome to wild, wild world of factorisation of polynomials over Z (integers). There is no general algorithm, and the problem is even a current research topic. In certain areas of math, we know far less than it seems we should. As to your earlier idea, "nearly completely"...every polynomial over the integers can be factored uniquely as product of linear factors and irreducible quadratics, each over the reals. This is a particular case of a general result in field theory. If you allow complex linear factors, of course every polynomial splits (the famous fundamental theorem of algebra. Other than that, I don't know much else to say. Revolver 09:27, 1 Jul 2004 (UTC)
 * 3. It could be something else which hasn't occurred to me.
 * At any rate, if option 1 above, as I noted, we can lose the S(x) nonsense altogether. If option 2, here's my hypothesis of what the author actually meant: not P(x)=R(x)+S(x) but P(x)=R(x) * S(x), where S(x) isn't the remainder (as currently stated, which is by def'n always zero) but the quotient. And if we're actually dealing with option 3 -- well, just let me know what option 3 is first! :) Doops 01:27, 1 Jul 2004 (UTC)
 * I'm almost certain the author meant P = R + S, (which is actually correct), not R*S, here S is the remainder, it just happens to be zero. Option 1 is possible but unrealistic, because you don't know if all factors over the integers are linear. Remember, it's always important WHAT YOU'RE FACTORING OVER, the integers?, the rationals?, the reals?, the complexes?, modulo an integer?, modulo a polynomial ideal?, etc., etc. Option 2 is really not well-defined problem, and even if it were, it would be hopelessly impossible to solve completely. Option 3...I don't know, come up with something, maybe you can contribute to the problem. In any case, I hope this example has sufficiently disproven anyone's ideas that mathematics has "solved" all simply stated, elementary important problems. Our knowledge, after 2,000+ years, is still pretty dim. Revolver 09:27, 1 Jul 2004 (UTC)

As you can see, I represent the triumph of innocence over experience. Doops 00:13, 1 Jul 2004 (UTC)

Revolver, can you explain what S is in case P(x) = x^5 - 1 ? This can be factorized over Z in (x - 1) * (x^4 + x^3 + x^2 + x + 1) The latter of which is irreducible over Z. This last polynomial can't be S, because the two parts sum up to a polynomial of degree 4. Thanks. Eef (A) 12:34, 1 Jul 2004 (UTC)


 * You're right. I was thinking over C, as usual. If R is just the linear factors corresponding to rational roots, then there is a quotient, in this case the irreducible cyclotomic, yes. "S" is a bad choice of notation, IMO, "Q" for quotient seems more natural. Revolver 20:03, 1 Jul 2004 (UTC)

I'm sorry: I was thinking in the clouds when I wrote the polynomial factorization part: if the result has to be factored, it can't contain an addition! I've corrected it with a somewhat-tested way to factor polynomials. The major problem is that we have to handle 3 different cases: when S(x)!=1, when S(x)=1 but an=1 and when both are =1. I've handled that through a case equation. Habbit 14:36, 1 Jul 2004 (UTC)

Doops tries again
Revolver: Hi. It may be that Habbit's changes have obviated the issue, but I'm still curious to know exactly what you meant. To end the hierarchical-outline-format (which was getting out of hand), I'll try below to summarize the point we've reached on the Polynomial-factoring discussion.

First of all, I have to apologize for being very sloppy and imprecise in my last post. When I wrote "we've found all linear factors" of course I should have specified "we've found all rational linear factors"; when I wrote "factor completely" of course I should have written "factor completely over the rationals." And thanks too for pointing out that I was cart-before-horsing in asking whether a polynomial was factorable or not prima facie. That was very sloppy of me. That said, let's remember what this polynomial section was about: it's about applications of Ruffini's rule. Using that rule we've found the integer roots of a polynomial. Where do we go from there, factorization-wise? Remember, I'm not at all invested in Ruffini's rule -- I don't really care whether its powerful or useless, personally! I just stumbled on the artice, read it, was confused, and tried to sort things out.

So, please help me out: explain my good old polynomial of x3-2x2+x-2. Now, I can factor this completely over the reals: it's (x-2)(x2+1). (The quadratic is irreducible over the reals.) Now: the discussion of P and R and S came in an article discussing how to use Ruffini's rule to find real rational roots of polynomials; so in this case, as I stated earlier, R is simply {2}. The complex roots don't show up in R since the article never told us how to find them! (Indeed, neither would real irrational roots.) So, using the description as given, can't you agree that the article was wrong?

As I see it, if we limit our membership of R to the real rational roots (as everything else in the article does), then in many cases P(x) / R(x) will not = 1; it will equal some higher-order polynomial with no real rational linear factors (such as x2+1). Alternatively, if we expand our membership of R to all roots, irrational and complex as they want to be, then surely P(x) / R(x) will always = 1. But in neither case will P(x) mod R(x) not = 0! I really don't understand how saying P(x)=R(x)+S(x) makes any sense at all! First of all, and most importantly, how can we say something of the format (....)*(....)*(....)+(...) is factored??!! and secondly; isn't S(x) always zero and therefore pointless?

I think Habbit's edits have taken care of the problems; and I will go back through the section of the article to make clearer both the limitations of the rule and the domain being factored over ; but I am still very curious about this whole P(x) = R(x) + S(x) thingummy. Can you help me? Thanks Doops 16:40, 1 Jul 2004 (UTC)


 * "Ruffini's rule" isn't really the way of finding integer roots. That comes from the result I gave you earlier about p/q, where p divides the constant term, q divides the leading coeff. Ruffini's rule is really just a special case of synthetic division when the divisor is linear. But, if I understand what "Ruffini's rule" is supposed to mean, the algorithm should work whatever you're working over -- ring, field, alg. closed or not, whatever. It's just a shorthand notation for polynomial division, and there is nothing special about the integers then. You could use Ruffini's rule to factor over the reals R (try this with x^3 - x^2 - x, see what you get). Of course, you have to recognize x = 0 is a root and already know one of the other roots (use quad formula, then use the rule, it should give you the other linear factor).
 * What we have here is putting the 2 together -- first, use the p/q result to get all possible integer roots, then use synthetic division to factor out the linear factors and get a quotient. Beyond this, over the integers, you can try some ways to test for irreducibility -- obviously, every quadratic is irreducible, or it would be the product of 2 linear factors. For other factors, there's a limited choice of weapons, the most popular are reduction modulo a prime (e.g. if you can show f(x) is irreducible over the field Z/pZ, i.e. it can't be reduced doing arithmetic mod p, then it's irreducible over Z, because if it factored over Z, then it would factor over Z/pZ), this works well for cubic, because you just check for roots mod p. Fourth-degree and higher -- good luck. The other main tool is Eisenstein's criterion. I'm sure there are more, I'm not familiar with them.
 * It's not neccessary the restriction the polynomial always be monic for this article. For example, consider 3x^3 - x^2 - x - 1. Clearly, x = 1 is an integer root. You get this by the general p/q result I mentioned earlier. Of course, there may not always be roots, the point is, the polynomial doesn't have to be monic to have integer roots, and in fact, checking all divisors of the constant term should find all integer roots for ANY polynomial with integer coefficients (if q is not = 1, then p = qn for some n, since p divides constant term, n divides constant term, and we could have originally chosen p' = n, q' = 1 and gotten p/q = n that way).
 * So, using the description as given, can't you agree that the article was wrong? Yes, certainly!
 * As I see it, if we limit our membership of R to the real rational roots (as everything else in the article does), then in many cases P(x) / R(x) will not = 1; it will equal some higher-order polynomial with no real rational linear factors (such as x2+1). This is exactly right! Most polynomials don't factor well. Finding nice linear factors with integer coeffs is a pleasant surprise...in practice, in most examples you choose at random, even "nice" examples, there will be many higher-order factors you don't know what to do with.
 * Alternatively, if we expand our membership of R to all roots, irrational and complex as they want to be, then surely P(x) / R(x) will always = 1. Exactly right, again! What you've said is precisely the statement of the fundamental theorem of algebra.
 * isn't S(x) always zero and therefore pointless? Yes, you're right. I was thinking about things over C for some reason, and thinking S(x) had to be 0, as you say. I would say, make it *, but change S to something like Q, so it reminds one of a quotient.

Revolver 20:03, 1 Jul 2004 (UTC)
 * A comment over the last edition made: the last author said:

"&quot;this method will not necessarily even factor our polynomial completely over the rationals.Consider the polynomial x5-3x4+3x3-9x2+2x-6.Using Ruffini's method we will find only one root (x=-3); factoring it out gives us P(x)=(x4+3x2+2)(x-3). But the fourth-degree polynomial isn't fully factored; it is actually factorable into two irreducible quadratics (x2+1)(x2+2). In short, Ruffini's rule only helps us (with a limited class of polynomials) to find rational linear bynomials; finding any others remains a tricky proposition.&quot;"
 * I think this is FALSE, just because the two irreducible quadratics do not solve to RATIONAL roots:
 * $$x^2+1=0\,\!$$
 * $$x^2=-1\,\!$$
 * $$x=\sqrt{-1}=i\,\!$$
 * And that is not a rational number. The same happens to $$x^2+2$$, so I've deleted the comment from the article. Think before you post, or you'll regret it as I did with my first post about polynomial factoring.
 * Habbit 20:31, 1 Jul 2004 (UTC)


 * Sorry, Habbit, I'll clarify what I meant. [By the way, first of all, I think you mean "real," not "rational."] I quite realized that those quadratics have no real roots; that's why I said they were irreducible! All I was trying to say is: in colloquial usage, a polynomial is factored "completely" when there is no more factoring to do. This varies from circumstance to circumstance; as Revolver points out, if we're allowed to enter the realm of complex numbers, all polynomials are completely factorable to a product of linear terms. But assuming we're limited to real numbers, you can't (as you noted) factor x2+1 further into linear terms. You can, however, factor (x4+3x2+2) further if you want to: it's (as I noted) (x2+1)(x2+2). So here's all I'm trying to say: if we want to factor something "completely", we can't stop at (x4+3x2+2)(x-3); we have to continue to (x2+1)(x2+2)(x-3), and that's the step which Ruffini's rule doesn't help with. Of course, unless we're prepared to enter the realm of complexity, we can't go any further; so (x2+1)(x2+2)(x-3) can be called a "complete factorization" over the reals in a way that (x4+3x2+2)(x-2) can't. See what I mean? Doops 20:57, 1 Jul 2004 (UTC)


 * I would agree if we were factoring over $$\mathbb{C}\,\!$$ (complex numbers), but we are factoring over $$\mathbb{Q}\,\!$$ (rational numbers), and in rationals neither $$(x^2+1)$$ nor $$(x^2+2)$$, (that solve to $$x=\sqrt{-1}$$ and $$x=\sqrt{-2}$$) can be roots of any polynomial. I'll explain: by definition, the root of a polynomial P(x) is the value of x for P(x)=0. In a factorized polynomial, when x is a root, one of the factors equals to 0, and then the product is 0. If we were factoring over $$\mathbb{C}\,\!$$, $$(x^2+1)$$ would be a root, because $$x^2=-1$$, so $$x=\sqrt{-1}$$, and $$P(\sqrt{-1})=0$$. However, we are factoring over $$\mathbb{Q}\,\!$$, and neither $$(x^2+1)$$ nor $$(x^2+2)$$ solve to any rational number, so there is no rational number for which $$P(x)=(x^2+1)(x^2+2)(x-3)=0$$ except for the 3, the rational root found by Ruffini's rule. To factor a polynomial, you need to know its roots, and if you're factoring over rational numbers and the roots are not rational, forget them. Furthermore, you're polynomial $$P(x)=(x^4+3x^2+2)(x+3)$$ would not be properly factored even if we were factoring over$$\mathbb{C}\,\!$$, because you could factor $$P(x)=(x^2+1)(x^2+2)(x-3)$$ into $$P(x)=(x-\sqrt{-1})(x-\sqrt{-2})(x-3)$$.:Habbit 23:14, 1 Jul 2004 (UTC)


 * I think there's a language barrier here. For me the word "factor" means to "unmultiply." So, for example, I can factor 12 into 6 * 2. If I want, I can then factor 6 into 2 * 3. If I really want to, I could even factor 2 into root(2) * root(2), or 3 * 2/3, or anything else. That's all I mean by factor. I'm somewhat intrigued to hear you say "to factor a polynomial, you need to know its roots," since that way of thinking about it is precisely the opposite of mine. I would say "to know a polynomial's roots, you need to factor it." :) For example, to find the roots (the zeros) of x2-3x+2, we would set it equal to zero, factor, and, getting 0=(x-2)(x-1), know what the roots are (aha! they are 2 & 1)! Neither your way nor mine is wrong; but the way we use words can affect the way we look at the world. For me, factoring is a process; that's why it makes sense, in my eyes, to say "I'll factor this as much as I can; then I'll stop." See what I mean? Doops 00:23, 2 Jul 2004 (UTC) P.S. I should add that when I was in high school we practiced factoring as a purely mental exercise long before we started talking about "roots," learned the quadratic formula, etc. Perhaps not all schools around the world do things this way. - D.


 * I agree with Doops. Factoring [as in this case, over Q] means to rewrite a polynomial as a product of all irreducible terms [over Q]. So (x2+1)(x2+2)(x-3) is the correct factorisation over Q (and R in this case). I think it should be mentioned in the article that Ruffini's rule can't be used to find (irreducible) quadratic and higher orders factors. Eef (A) 09:19, 2 Jul 2004 (UTC)


 * I think there isn't any language barrier. I agree with Eef (A) in the definition of factorization: writing something (maybe a number, maybe a polynomial) in the most irreducible terms. But factoring is not unmultiplying. Unmultiplying is dividing: when you factor numbers, you usually test-divide the number by prime numbers until you find a divisor. If we want to factor a number, we must write it in the most reduced set of numbers (in this case, the primes), so 12=6*2 isn't a good factorization. The correct would be 12=22*3. If we want to factor a polynomial, we must write it in the most reduced set of polynomials: monic linear bynomials. The way to factor a polynomial into monic linear bynomials is simple: find its roots however you want and pass the number r from x=r to x-r=0. Then write all the monic linear bynomials created from the roots as a product (x-r1)(x-r2)...(x-rn). If you factored over C, you've finished. If you didn't, find the remainder S(x) and write it as another factor. What I try to mean is that (x2+2) would never be a correct factor, (just as 4 isn't a correct factor when factoring numbers: it's not a prime factor; it's 22) because it's not an irreducible term: those are monic linear bynomials, in this case $$x-\sqrt{-2}$$ and $$x+\sqrt{-2}$$. :Habbit 12:27, 2 Jul 2004 (UTC)


 * Habbit, whether you think 12=6*2 is a "good" factorization or not, it is certainly a factorization. Obviously it's not the "complete prime factorization" of 12, since that would be 22*3. So what? In going from 12 to 6 * 2, we are factoring. Not factoring completely, perhaps, but certainly factoring. If we write 12=6*2, then 6 is a factor of 12. Not a fully reduced factor, but a factor nontheless. This article (at least until Revolver adds some abstruse method of using Ruffini's rule over other fields) rests entirely among real numbers, and (in fact) among rationals. So we're not factoring over C. This being so, (x-root2i) is not a valid factor! By contrast (x2+2) is. Yes, that's right, it is a factor; that's how I've always heard the word used. We don't limit the word "factor" to monic linear binomial factors! [Furthermore, I cannot understand why you keep objecting to (x2+2) when you never objected to (x4+3x2+2). They are both in the same boat, according to your use of the word "factor," and the former has the merit (unlike the latter) of being irreducible over the rationals.] Finally, I want to reiterate the most important point of my last post (and one which you didn't address in your reply, except obliquely by objecting to my use of the word "unmultiplying): for many people, factoring is simply a process like multiplication or tying your shoes. You can give us an expression and the instructions "factor"; and we'll do it; we've been doing it since 8th grade, long before we knew what a "root" was, long before we started using the complex plane on a regular basis. You can say "please factor ex+ex+y", and we'll say "OK. That equals ex(1+ey)." The notion of roots never entered our heads! That's why I say that factoring is unmultiplying. Doops 17:07, 2 Jul 2004 (UTC)

Thank you, Eef(A). I've put the warning back in again; and I hope that this time I've worded it unambiguously enough for you, Habbit. Doops 17:50, 2 Jul 2004 (UTC)

Habbit, you are wrong. I'm sorry, you're not understanding the word "factor" here. Doops and Eef are correct. What you are talking about is what's called "factoring into irreducible elements", this is what you guys are calling "completely factored polynomial". Now, an irreducible polynomial is one which cannot be written as the product of smaller-degree polynomials OVER THAT FIELD. So, e.g. x^2 + 2 is irreducible over Q, because there are no monomials with rational coeffs that multiply to give it to you. In this sense, x^2 + 2 is "factored completely over Q" in the sense that this is its factorisation into irreducible elements (and, BTW, since Q is a field, Q[x] is a unique factorisation domain, so such unique factorisation is guaranteed to exist). Now, over any field, you can factor to irreducibles. It doesn't matter how the irreducibles split in an extension field. What you're thinking about, Habbit, is the splitting of a polynomial over its splitting field. Or, to be overkill, over its algebraic closure. In this case, the polynomial is completely factored, but that's because every polynomial splits over its splitting field. That's not what we're talking about. We're not talking about irreducible polynomials over some finite extension of Q, we're talking about irreducible polynomials over Q, and in that case, many polynomials (most, in fact) will have their unique factorisation into irreducible polynomials where not all the irreducibles are "binomial" (really, linear is the common term here, not "binomial"). I hope this settles the argument. I haven't read the article, please change it so it reflects these facts. Revolver 09:14, 4 Jul 2004 (UTC)

OK, Habbit gives up against wiser (and probably more experienced) people
If so many people think I'm wrong and they give that bunch of reasons, that should mind something. And that should mind more when I'm only a 15-year-old student. The experience is a status that must be achieved: in my country, we say that the Devil's knowledge does not come from its demonic nature, but from its old age... OK, I'm changing my mental concept of "factor"... But you haven't beaten me!! MUHAHAHA!! I shall return with new ways to take you out of your minds, just like computer engineering, object-oriented and/or procedural programming, distributed network theory... (and I know a heavy bunch of that, I can bet!) :D
 * Habbit 22:30, 5 Jul 2004 (UTC)


 * Don't feel bad. A lot of mental concepts require "adjusting" when you first take algebra...(zero divisors?!, non-associative operations?!) If you're particularly interested in polynomials, the most readable account is probably chapter 16-17 of Gallian's book (Contemporary Abstract Algebra). Revolver 22:57, 6 Jul 2004 (UTC)