Factoring a Multivariate Polynomial: Strategies Beyond GroupingDate: 11/11/2008 at 09:33:32 From: Greg Subject: Factoring a Multivariate Polynomial Hi, Dr. Math. Could you help me factor this polynomial? x^3 + y^3 + z^3 - 3xyz Also, could you tell me the general techniques for factoring multivariate polynomials? (I'm looking for a technique other than "grouping," which doesn't work here.) And how can I tell when it's "fully factored?" Would I have a product of binomials? I simply have no idea how to factor multivariate polynomials like this. I tried a few things and they didn't work. The only way I know how to factor multivariate polynomials is to put the terms in "groups" and get a common factor out of each "group." (This is similar to factoring quadratic equations after splitting the middle term.) I tried that and I could not get any common factors. Also, I looked at the expression as a cubic equation and tried to find possible roots of x using the rational root theorem; it got really confusing and I couldn't find the roots. I don't know what else to try. Date: 11/15/2008 at 19:33:16 From: Doctor Jerry Subject: Re: Factoring a Multivariate Polynomial Hello Greg, Thanks for writing to Dr. Math. I noticed your question but didn't reply because I felt that, besides the factorization, I had nothing useful to say. The factorization is x^3 + y^3 + z^3 - 3*x*y*z = (x + y + z) (x^2 - x y + y^2 - x z - y z + z^2) My thoughts were not clear at all. I noticed that the given polynomial was symmetric. One expression that came to mind was x + y + z, and I tried that. I have no background in factoring polynomials in several unknowns. If you have questions about this, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 11/15/2008 at 20:09:39 From: Greg Subject: Thank you (Factoring a Multivariate Polynomial) Hello! Thank you for replying. I checked my book and the answer is (1/2)(x + y + z)[(x - y)^2 + (y - z)^2 + (x - z)^2] I don't understand what you wrote. What is a symmetric polynomial, and why would that imply a factor of x + y + z? Date: 11/15/2008 at 22:25:39 From: Doctor Jerry Subject: Re: Thank you (Factoring a Multivariate Polynomial) Hello Greg, Thanks for writing to Dr. Math. A symmetric polynomial is one for which interchanging x and y, or x and z, or y and z does not change the polynomial. One might guess that the factors will also be symmetric. That's all I meant. If you have questions about this, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 11/15/2008 at 22:39:41 From: Doctor Pete Subject: Re: Thank you (Factoring a Multivariate Polynomial) Hi Greg, This is another Math Doctor responding to your question. First, I should point out that the factorization Doctor Jerry provided is equivalent to the one in your book, although my preference is for Doctor Jerry's form. The reason is that representing the second-degree factor as the sum of squares is not necessary, and confers no additional information about the factors themselves -- it is not a "more factored" expression. Also, the writing of the second-degree factor in such a way then puts a constant factor of 1/2 out in the front, which again is unnecessary. However, they are both legitimate ways of rendering a factorization of the original polynomial. Second, a polynomial is symmetric if it is invariant under any permutation of its variables. For example, in your case, we have F[x,y,z] = x^3 + y^3 + z^3 - 3xyz. There are six distinct permutations of the variables x, y, z: {x,y,z} {x,z,y} {y,x,z} {y,z,x} {z,x,y} {z,y,x}. If we applied the function to each of these permutations, then we would still have the same polynomial; i.e., F[y,z,x] = F[x,y,z] = F[z,x,y] = .... This is what we mean by "invariant under any permutation." Other examples of symmetric polynomials include x + y, x^2 y + y^2 z + z^2 x, and xyz + yzw + zwx + wxy, where the number of variables is 2, 3, and 4, respectively. Doctor Jerry mentioned symmetry because -- as a general heuristic approach to factorization -- if a symmetric polynomial can be factored, its factors are, themselves, likely to be symmetric. For example, x^2 + 2xy + y^2 = (x + y)^2 is symmetric, and its factors are symmetric. Also, x^3 + y^3 = (x + y)(x^2 - xy + y^2) is a symmetric polynomial with symmetric factors. When we look at the structure of F[x,y,z] = x^3 + y^3 + z^3 - 3xyz, we might try to factor it by comparing it to other expressions that have easy factorizations that we already know. For example, (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2 y + y^2 z + z^2 x) + 3(x y^2 + y z^2 + z x^2) + 6xyz. This implies F[x,y,z] = (x + y + z)^3 - 3(x^2 y + y^2 z + z^2 x) - 3(x y ^2 + y z^2 + z x^2) - 9xyz. We can then try to pull out a factor of x + y + z from the remaining terms, because the first term has such a factor. We observe that (x + y + z)(xy + yz + zx) = x^2 y + y^2 z + z^2 x + x y^2 + y z^2 + z x^2 + 3xyz, so we now have F[x,y,z] = (x + y + z)^3 - 3(x + y + z)(xy + yz + zx) = (x + y + z)[(x + y + z)^2 - 3(xy + yz + zx)]. At this point, we have successfully pulled out one factor. The remaining term, (x + y + z)^2 - 3(xy + yz + zx), simplifies to x^2 + y^2 + z^2 - (xy + yz + zx), which gives the result that Doctor Jerry provided. The lesson is that factorization is something of a guessing game. There do exist algorithms that computers use to factor polynomials. These are generally difficult for humans to apply to arbitrary cases, however; so you have to try different approaches to see if they solve the puzzle -- and there's no guarantee that any of them will work. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ Date: 11/16/2008 at 09:39:03 From: Greg Subject: Thank you (Factoring a Multivariate Polynomial) Thank you everyone for the help! Date: 11/16/2008 at 18:36:41 From: Doctor Vogler Subject: Re: Thank you (Factoring a Multivariate Polynomial) Hi Greg, This is a third Math Doctor. I was watching your question because I'm interested in the same subject myself, although my knowledge of the matter is less than I'd like. Unlike polynomials in a single variable, multivariate polynomials do not always factor into linear terms, even over the complex numbers. In fact, *most* multivariate polynomials are irreducible. So yours was one of the ones that does not factor. Your polynomial is also special in two other ways. Doctors Jerry and Pete mentioned that it is symmetric, which is true. Not all symmetric polynomials factor into symmetric factors. For example, (x + y)(x + z)(y + z) is symmetric in three variables, but none of the factors is. However, a symmetric polynomial that factors into terms of different degrees will have symmetric factors. The reason is that permuting the variables doesn't have to leave all of the factors unchanged, but it does have to leave the collection of factors unchanged, which means that the most it can do is rearrange the factors. (Look at how swapping x and y, for example, will rearrange the factors in my example polynomial.) Your polynomial is also homogeneous, which means that every term in the polynomial has the same degree. For example, x^3 + 2xy^2 + y^3 is homogeneous of degree 3. A homogeneous polynomial always factors into homogeneous factors (of smaller degree when the polynomial is not irreducible). But homogeneous polynomials are really polynomials in one fewer variable than you see. For example, a homogeneous polynomial f (x,y,z) of degree d satisfies the equation f(rx, ry, rz) = r^d * f(x, y, z), which means that f(x, y, z) = z^d * f(x/z, y/z, 1), so that the equation f(x, y, z) = 0 is the same as f(x/z, y/z, 1) = 0. So we can consider this to be a polynomial in the two variables u = x/ z and v = y/z. Anyway, back to the point of factoring multivariate polynomials: we are left with the question of how do you actually do it. Well, there are a variety of ad-hoc techniques like those Doctors Jerry and Pete have described. Those kinds of things usually work well when you're doing this by hand and you really expect to get a factorization. They are less useful for determining that your polynomial is irreducible. There are other techniques that can determine that your polynomial is irreducible. And there are still others that can do either one. But some of them require a lot of work. I am certain that there are sophisticated algorithms which will work efficiently even for rather large polynomials, although I am unfortunately not familiar with such techniques. Let me describe one which -- though quite messy, involving lots of variables and equations -- is sort of like the last method described (for one-variable polynomials) at http://mathforum.org/library/drmath/view/68494.html If a degree-3 polynomial factors, then it must factor into a product of a degree-1 and a degree-2 polynomial. (There are more cases for higher degrees, but for degrees 2 and 3, there is only one possibility.) So you could argue that if your polynomial is irreducible, then it must equal (rx + sy + tz)(ax^2 + bxy + cy^2 + dxz + eyz + fz^2) for some numbers (r, s, t, a, b, c, d, e, f). In fact, since both of these are homogeneous, you can assume that two of these are 1. For example, you can assume that r = 1 and a = 1 if your polynomial has 1 for the coefficient of x^3 (otherwise, you just divide your polynomial by the coefficient of x^3). Then you multiply out the above product, gather like terms, and for each one you get an equation that says that its coefficient must be equal to the coefficient in your polynomial. Continuing my example above, if you multiply through, you'll get sc for the coefficient of y^3; and since your polynomial has y^3 with coefficient 1, that means that sc = 1. Similarly, you'll get te + sf for the coefficient of yz^2; and since your polynomial has no yz^2 term, that means that te + sf = 0. It turns out that since you have more than one variable, this will give you more equations than variables, which makes things easier to solve. But that's still a lot of equations to work out. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 11/17/2008 at 16:02:07 From: Greg Subject: Thank you (Factoring a Multivariate Polynomial) Thank you! |
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