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College Algebra: Lowest Term Rational Expressions

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About this Lesson

  • Type: Video Tutorial
  • Length: 9:17
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 100 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Basics & Prerequisites (37 lessons, $52.47)
College Algebra: Rational Expressions (3 lessons, $4.95)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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Okay, so if you have a rational function, sometimes they can be overly complicated. And what I mean by that is, remember a rational function is just a polynomial divided by a polynomial. But sometimes you can actually cancel stuff away and make it even simpler without changing the value. So I want to show you how you can actually simplify a rational function if someone hands you one. And what I want to do first is illustrate this with a simple example, which I plan to solve incorrectly. So here we go.
So what I can say here is the following: . What someone might say is, "Well, okay, great. Let's just cancel those 2's off there," and what I'd get is the following, what' I'd see is I'd cancel those away. I would just have 2x. You see that canceling? That's some real fancy canceling. In fact, it's fancy false canceling, because that is not allowed. In fact, this is a classic mistake and if you're tended to do that, if you're tended to cancel something away that's sort of sitting out there being added or subtracted to something else, it's a great mistake to make now and then forever hold your peace, because, in fact, this is my classic algebraic mistake number four. It's not number four. Why did I say number four? That's a classic mistake right there. This is actually classic mistake number three. It's moving up the charts, folks, really fast. It's going to be number one any day now, unless you stop it. Only you can prevent that from being number one. So don't buy into it. It's number three, it's the canceling mistake. And you have to remember that we only cancel factors, we don't cancel individual terms that are being added or subtracted. It's a classic mistake.
So this is wrong, by the way. Here I am having this thing on the Web all that time, but it's wrong. So how do you do this correctly? What we have to do is we have to factor out a common factor and then cancel factors on top and bottom and I'll show you why. So, in this case, for example, what I would see is that I could factor this out and I would take a 2 out of the top. That's a common factor. In fact, it's the greatest common factor, and they cancel like this. Now notice what I have here is a multiple of 2 on the top and a multiple of 2 on the bottom. So if you really want to be pedantic, which you will never do in life, but let me just show this to you, I could write it like this: 2x + 1, and on the bottom here I could write this 2, and on the bottom here I could write the invisible 1 that follows us everywhere. But see, now you can see that, since this is a product, is the number 1. It's still the 1 and if you cancel that away, 1 times anything is the anything. So I can't cancel that away without changing the value. And so what I see is the correct factorization, is this. It's a very simple example, I admit, but it's an important point that you don't want to just cancel individual little things. You want to make sure you have a common factor everywhere, including the bottom, and then cancel away.
Now, on with that, let me show this to in an even more dramatic way. So let's take a look at the same example once again. Okay, so this is the exact same problem. You can see . And the important thing to do is to first factor out that 2. So if you factor it out, now it's a factor and I can take these guys and just throw them away and I'm just left with the answer, which is this. So that's the idea behind it.
Okay, let's try now some more exotic ones. In fact, maybe I should give you a chance to try these things. I don't know, what do you think? . Oh, what the heck, I'll do this one for you. Okay, I look at this thing and I say, "Gee, whiz, this is something is going to happen here, because I've got too many common factors going on here." That's what I'm thinking in my head. So the first thing I do is I can factor out a 10 on the top. So let me actually do that, and that would give me 2x + 1. And then on the bottom I can factor out a 15. And then I see 2x + 1. And so these guys can be canceled, and then even here I can do a little cancellation, so I can do a little cancellation here. I can take the 5 out of here and the 5 out of here, and that would leave me with . So this equals . That's great. There's just a little teeny thing that I just remind you of, and that's the domain of this is anything except where the bottom is zero. And when is the bottom zero? You can see that happens when x = . If I put here, that would produce -15 + 15 = 0. So, in fact, this equals that. This is the same thing as , as long as x . If x = , then this doesn't exist, so it doesn't equal . Just a little point that remind you that, in fact, only could look where the thing is defined. And if it's not defined, don't look.
Hey, now it's your turn. How about this one? . These are sort of fun in a way. Anyway, give it a try and tell me what you think.
Okay, well I see I can pull out a 2 on the top, certainly, and I also can pull out an x. Let me pull out 2x on the top. That would leave me with a 5x here and it would leave me with a 4 here. And on the bottom, well, let me actually pull out a 2x there, just because I see the 2x dangling so nicely here. And so, in fact, I'll just write this as 2x(2x). Well, now I see this factor, 2x factor on top, 2x factor on the bottom. I can cancel away and I'm left with just . So you can see this actually simplifies quite nicely to an identical equivalent thing, but much simpler to deal with, so that's sort of fun. Okay, so I have you got .
Okay, how about one last one? I'll use a here, just to really throw you off. "Can we change the A to x and then do it that way?" Sure, yeah. "I can only do it with x's." Okay, so here's the problem, . Factor away and cancel at will.
Okay, how did you make out? Well, there's a little bit of factoring here. Let's see what we can do on the top. So I put an A here, A here. I see that both signs will be the same and they will both be minus. And I need something that multiplies to give 4 and adds to give 4. Well, that's just 2. I make mistakes all the time. Did I ever say I was perfect? No! Don't look at me that way! So here I'm going to have an A and an A. The negative sign means they're going to be opposite signs and since they're both the same A's here, I can put them anyway I want. It's got to multiple to give 6 and combine somehow to give a 1. 3 and 2 have that property if I subtract them the right way. Where should I put the 3? I want the thing to be positive, so I want the big guy to be by the positive and the little guy to be here. Looks like that's a good factorization. Check it, yeah, good.
So now, what do I see? I can cancel. And if I cancel those common factors, they are factors now, so I can cancel them. I'm not violating my classic cancellation mistake, then I'd have A - 2 on the top and A + 3 on the bottom. But again, I just caution you about the domain issue. The domain of this you can see easily is everything, except when A = -3. But remember the original problem, the domain was everything except -3, and also A 2. So you have to understand that these two things aren't equal if A = 2, even though this is defined. I broke the rule by dividing by zero, so you have to understand that this is only okay if A 2. As long as A 2, you're fine. And let's face it, if you pick a number at random, how often are you going to have 2?
Anyway, that's it for simplifying. Not a big deal, factor top, factor bottom, cancel away. Your worries are over.
Prerequisites
Rational Expressions
Writing Rational Expressions in Lowest Terms Page [2 of 2]

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