Why Factor Polynomials?Date: 02/20/98 at 18:52:05 From: Alexandra Wrigley Subject: Factoring polynomials Why is factoring polynomials important - how do we use it in our everyday lives? Date: 02/21/98 at 13:42:31 From: Doctor Sam Subject: Re: Factoring polynomials Alexandra, Well, factoring ISN'T important to most people in everyday life. I mean shopping and cleaning and cooking and going to the movies. But many occupations use different kinds of mathematics, ranging from accountants to carpenters to scientists and engineers to people who work to protect the environment. Many of them will sometimes need to use factoring, but factoring isn't a goal in itself. Factoring is used to solve different kinds of problems. I think you might want to know why you should learn factoring if you aren't ever going to use it "in real life." One answer to your question is that most of us don't know what we will be doing in real life until it happens to us. Sometimes we plan for it and sometimes it takes us by surprise. But it is a good idea to be prepared. If you don't know ANY mathematics then there are hundreds, maybe thousands of jobs that you won't be able to do. For most of these jobs mathematics isn't the main point of the job, it is just one of the many tools that are used. So if you don't know mathematics you may be losing the opportunity to do something that you would find exciting and worthwhile. And now to the mathematical part of your question: how is factoring used? I can think of two important uses of factoring. One is to make complicated things look simpler. For example, I don't have any idea what this fraction 1848 / 16632 means. The numbers are too large. But if I factor the numerator and denominator I get: (3)(8)(7)(11) / (8)(11)(7)(9) and I can see that the 3, the 8, the 7, and a 5 are factors of both the numerator and denominator. Since 3/3, 8/8, 7/7 and 11/11 are all equal to 1 the fraction reduces to 1/3, which is a lot easier to understand and to compute with. It's the same with complex algebraic fractions. (x^4 - x^2)/(x^3+x^2) looks REALLY complicated, but the numerator factors into x^2(x-1)(x+1) and the denominator factors into x^2(x+1), and since x^2/x^2 and (x+1)/(x+1) are both equal to 1, this fraction simplifies to x-1 . . . a LOT easier to work with. A second important use of factoring is in solving equations. You don't need to factor to solve 2x+3 = 5 ... linear equations use a different method. And you don't need to factor second degree equations because you can use the Quadratic Formula (although factoring is often MUCH easier!). But if you need to solve equations where the degree of the highest term is more than 2 then you really have no choice at all because you don't have formulas for most of them. Here is an example of a hard equation to solve: x^5 - 4x^4 - 12x^3 = 0 But if you factor it completely you get (x^3)(x-6)(x+2) = 0 and now it is easy, because this says that a product of things turns out to be equal to zero. If you multiply, the only way to get zero as an answer would be if you multiplied by zero. So one of the three factors has to be zero. If x^3 = 0 then x = 0 If x-6 = 0 then x = 6 If x+2 = 0 then x = -2 So the solutions to this equation are x = 0 or 6 or -2. I hope that helps. -Doctor Sam, The Math Forum Check out our Web site http://mathforum.org/dr.math/ |
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