Factoring Polynomials over Real and Complex Numbers
Date: 07/17/2006 at 16:43:40 From: Anil Subject: Factoring Polynomials that are irreducible in Q, R, and C I am having difficulties factoring polynomials into a product of polynomials that are irreducible in Q[x], R[x], and C[x]. For example, x^4 - 15x^2 - 75. I'm not sure where to start. Using the rational root test didn't bring any results trying +/- 1,3,5,15,25,75. I looked at the graph on my calculator and it clearly goes through the x-axis two times. My conclusion is that it is irreducible in Q[x], but how do I take this into R[x] and C[x]?
Date: 07/18/2006 at 08:58:45 From: Doctor Vogler Subject: Re: Factoring Polynomials that are irreducible in Q, R, and C Hi Anil, Thanks for writing to Dr. Math. The three problems (1) factoring over the complex numbers, (2) factoring over the real numbers, (3) factoring over the rational numbers (or the integers), are three very different problems, but I'll talk a little about each. In case (1), this is a matter of finding all of the complex roots, since every polynomial factors into a product of linear factors over the complex numbers. So the polynomial factors as a(x - r1)(x - r2)...(x - rn) where n (the number of roots or factors) is the degree of the polynomial, a is the leading coefficient (coefficient of x^n), and r1 through rn are the n complex roots. Since you generally can't represent the roots exactly except by saying something like "the third root of the polynomial ____," one usually makes do with a decimal approximation, which means you use numerical methods to find them, such as Newton's Method (or a graphing calculator). In case (2), the easiest way is to find all of the complex roots and then pair up the non-real complex factors and put them into quadratic factors. That is, you change (x - (r + si))(x - (r - si)) into x^2 - (2r)x + (r^2 + s^2) Every polynomial with real coefficients factors into a product of linear factors and quadratic factors over the real numbers. As in case (1), you generally can't find the roots exactly, so you use numeric methods (e.g. Newton's method) to get as accurate values as you'd like. Case (3) is more interesting, for the reason that it's more challenging. Polynomials might factor in many different ways over the rational numbers. You do have one theorem: A polynomial with integer coefficients that factors over the rationals will also factor over the integers. This means that factoring over rationals and factoring over integers are really the same thing, at least if you start with a polynomial with integer coefficients. The easy way to factor a polynomial over the integers is to have a math program do it for you. For example, the very nice program GNU Pari can be downloaded for free at http://pari.math.u-bordeaux.fr/ and you can just ask it for factor(x^5+x+1) You can also use more expensive math programs, like Mathematica, which you can tell Factor[x^5+x+1] These programs use sophisticated factoring algorithms. One such is the Berlekamp algorithm, which you can read about here Berlekamp-Zassenhaus Algorithm http://mathworld.wolfram.com/Berlekamp-ZassenhausAlgorithm.html or here Polynomial Factorization http://math.berkeley.edu/~berlek/poly.html or by searching Google for something like berlekamp algorithm or just polynomial factorization This kind of thing is more suited to computers and programming than it is to pencil and paper, though. If you have a small polynomial that you want to factor by hand, then the easiest way is perhaps the naive approach: You consider how the polynomial might factor. You only have to consider two factors, because if it factors into more than two factors, then it also factors into two factors. For example, a fifth- degree polynomial that factors will have one of the following forms: (degree 1) * (degree 4), or (degree 2) * (degree 3). The first form is easy to check. You use the Rational Roots Theorem: Any rational root has a numerator that divides the constant term of your polynomial and a denominator that divides the leading coefficient. A degree-1 factor means you have a rational root, and the Rational Roots Theorem tells you what those roots could be. So you try all the possibilities and see if any work. If none work, then your polynomial can't factor into a degree-1 times a degree-4. For the other form, you suppose that your polynomial equals (x^2 + ax + b)*(x^3 + cx^2 + dx + e). For example, if you wanted to factor x^5 + x + 1, then you set x^5 + x + 1 = (x^2 + ax + b)*(x^3 + cx^2 + dx + e), and then you multiply out the right side of the equation and set the coefficients equal to one another. That gives you five equations in five variables. Sometimes they can be hard to solve, but sometimes it's easier. And you can use the fact that all of the variables have to be integers. For example, be = 1 implies that either b = e = +1, or b = e = -1. That's the only way to factor the number 1 into two integers. So try solving these equations, and see what you get. You can see another example worked out at Factoring Quartics http://mathforum.org/library/drmath/view/56403.html 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/
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