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About this Lesson
 Type: Video Tutorial
 Length: 14:30
 Media: Video/mp4
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 Download: MP4 (iPod compatible)
 Size: 157 MB
 Posted: 06/26/2009
This lesson is part of the following series:
College Algebra: Factoring Techniques and Patterns (9 lessons, $16.83)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Intermediate Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/intermediatealgebra. The full course covers real numbers, equations and inequalities, exponents and polynomials, rational expressions, roots and radicals, relations and functions, the straight line, systems of equations, quadratic equations and quadratic inequalities, conic sections, inverse and exponential and logarithmic functions, and a variety of other AP algebra and advanced algebra.
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 UTAustin 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, padic 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.
About this Author
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It's one thing to multiply two binomials together and get some trinomial using the foil method, but what about doing that process in reverse? Suppose that I gave you a trinomiala trinomial just means that I have three terms of this form, some number, let me call it "a", with the variable x^2 + some other number, "b", with the variable x + c. I know this looks sort of complicated. Let me immediately write down an example so you can see what I'm talking about, like x^2 + 7x + 12 for example. These are just some fixed numbers and the x is the unknown. It's called a trinomial because I have three terms and there's a constant term. There's a term that's something times x and there's a term, which is something times x^2. If I think about this as like a backwards thing, this is the answer that we got by foiling something out. The question is what were the two things that we had to foil in order to get this? What I'm really trying to do here is to factor trinomials and I want you to think about having these two terms and I want to lift them, separate them, and therefore when I put them together with the foiling method, I'm going to see a trinomial. It's to go from the trinomial and lift and separate and that should give us the factorization that we want.
Let's see how to do that. Now, the way to do that is to really think about this thing and work backwards. I'm going to have two terms here. At the beginning here we're going to try some careful thinking and then look for patterns. The secret here is to look for patterns. This term here is an x^2 so that clues to me that I'll have to have an x here and an x here in order to produce that x^2 term when I take the first times the first. Let me just keep this up here to remind you that the end game is undoing this. So the first times the first piece, so I'm going to put an x here and an x here. Then I'm going to have some number here and some number here so that when I do the outsides, the insides and combine them, I get 7x and then when I do the last I'm going to get the 12. In fact if you think about that, what are two numbers that multiply to give 12? Well, there are a whole bunch of them. There's 1(12). There's 2(6). There's 3(4) and then of course, there are these numbers written in the reverse order, 4(3), 6(2), and 12(1), and basically each of those possibilities might fit in right there. I could put a 1 here and a 12 here, a 2 here and a 6 here, and so on and with any one of those particular pairs, you'll notice that the last term, last times the last, will give me 12. The product of any of these two produces the 12. Which one of those, if any, would produce the factorization? What I've got to do is ask what about the outside plus the inside terms? Will any of these add up to give me the 7? Let's take a look and see. For example, if I put the 1 and the 12 here, let's just try that as a little test. Let's put the + 1 and let's put the + 12 here. By the way, with factoring like this, you might try something and it might not work. In mathematics, the power of mathematical thinking is to learn from mistakes. If you're not making mistakes in math, you're doing something wrong. What you've got to do is make lots of mistakes and then learn from those mistakes, those false steps. So if you don't get it right the first time that means you're doing it perfectly. If you get it right the first time, something's not right. If we try this, the question is, are the outside terms, which give me a 12x + the inside term of 1x, does that give me 7? Well, is 12x + 1x = 7x? No, it's not. It's 13x. Now which one, if any of these, would work? Let's see. What if I put the 2 and the 6 in there? Well, that would give me a 6x and a 2x. That's not 7. What about the 3 and the 4? If I put the 3 and the 4 in there, look what happens. I would get a 4x + 3x. That is 7x. This works. If I put in + 3 and + 4 I think we've got it. Let's check by foiling. First times the first, x^2, the outside, 4x + the inside, 3x. 4x + 3x = 7x. Last times the last is 12. We just factored this. The way we did it was trying to look at this and factor that into different types of numbers and see which of those will produce inside, outside just like that. It's a process that requires many, many steps, but once you get these steps down and begin to practice them it becomes a little easier to do.
Let's try another example together. Here's one. How about x^2  9x + 14? Here, again, I'm thinking it's going to be two terms. I see this x^2 here which means that I'm going to have an x here and an x here. Now, this seems a little difficult. I've got to factor 14. What are all the ways of factoring 14? Well, there is 1 and 14, 2 and 7, 7 and 2, and 14 and 1. No matter how I put those in here there's no way for me to combine them to get this negative sign. This negative sign is holding me up. How can I make these two numbers negative and yet have their product be positive 14? Well, this is where I remember a negative times a negative is a positive. What must be happening here, if this is really going to be able to be factored, both the numbers must be negative because a negative times a negative will still give me a positive 14, but now I'll be able to actually get a negative term in here and now I try the possibilities. If I put a  1 and a  14 in here, the outside term gives me a 14x. The inside term gives me x. That's 15. That's no good. What about 7 and 2? Well, 7 and 2 work, right? Because 7  (2) gives me a 9. In fact, here I see a solution right out here and what I see is a 2 and a 7 and we can check it now by foiling. First times the first, x^2, outside terms, 7x, inside terms, 2x, that's a 9x, last times the last, negative times a negative is a positive, 14. There you have it.
Now look for a pattern here. One pattern that emerges is the following. If I see a plus sign in the last termso if c is positive as in these two examplesthen I know the signs are going to be the same. They're either going to be both positive, or they're going to be both negative because the product has to be positive. If I see a positive constant here, like +12 or +14, I know the signs of the two factors are going to be the same and that same sign will be dictated by the sign of the middle term, the value of b. If it's positive then I know these are both positive. If it's negative, I know these will both be negative. There's a pattern emerging here. The pattern is if the number c is positive, I know the signs will be the same and that sign will be the same as the sign of b, the middle term there. That helps us figuring out this factorization.
Let's try now one that ups the ante a little bit and challenges us a little bit more. How about x^2 + x  6? Let's try to factor that. How would that look? Well, I'm going to have two terms, hopefully, if this can be factored. I'm doing foil. x, xbut now I see a negative so this times this has to be 6. How can I have two numbers multiply to give me negative? Can they both be positive? No, positive times positive is a positive. Could they both be negative? No, because a negative times a negative is a positive. The only way a product of two numbers will be negative is if one is positive and the other one is negative. When I see a negative sign here that tells me that I'm going to have different signs here. I don't know what those different signs are yet. I've got to factor 6. Well, I can factor 6 as 1 and 6, 2 and 3, 3 and 2, and 6 and 1 and let's see if any of these work. It's a little bit tricky. We've got to be careful. When I put in a 1 and a 6 here, the outside term is a 6x + 1x is 5x. That's not good. In fact, since I see this is positive, it tells me that the positive number here should be bigger than this number so when I combine it the net gain will be positive. In fact, while we're here, why don't we just try this then? Let's put the 6 there and the 1 there and jump right to the end. Then I've got a 6x  1x. That's +5x. That's positive but too positive. I need to have these numbers closer together. Let's try the 2 and the 3 option and see if that works. Notice there's a lot of thinking involved here but once you get in the habit of doing this kind of thinking it's not so bad. 2 and 3now my outside term is 3x, inside term is +2x. That's a gain of x, not quite right but so close. Then I remember this has to be the bigger number so I should switch these people and if I switch them look what happens. I have a 2x + 3x. That's a gain of +x. This works. What I see here is that I put a 3 here and a 2 here and I factor this and you can check by just saying, "Well, first times the first is x^2, outside terms 2x, inside term 3x, that's a gain of +x, last times last, 6."
How about one last one? Let me show you one that's really tricky because now I'm going to have stuff in front of the x^2. Now these actually take a lot of time to think about because now I not only have to factor the second stuff, I've got to actually factor the first stuff, too. I know there's going to be an x and an x but now there's going to be stuff in front of the x. Well, how do you break 3 down? Well, 3 can only be broken down as 3(1), since 3 is a prime. Let me try putting a 3 here and a 1 here. If this were a 4, for example, I'd have to try 4 and 1 and maybe the possibility of 2 and 2, but this is a 3. This negative sign tells me that I'm going to have opposite signs here somehow and I don't know how they're going to come out yet. I have to factor 8. Let's see if we can factor 8. Well, we could factor 8 as 1 and 8, 2 and 4, 4 and 2, and 8 and 1. Those are all different ways of factoring 8. I know the signs are going to be different and so let's make all these pluses and make all these minuses and then we have to take it to the possibility of flipping these things around. Let's see what happens here. If I put in a 1 here and a 8 here, will that work? You've got to be really careful. See it's 8(3), which is 24, and then I add 1. That's way away from 10. I need something that's a little bit closer together than that. Let's try the 2 and the 4, 4(3) = 12x and then 2x, it's a positive 2x, 12x + 2x gives me a 10x. I actually hit it right there. If I hadn't hit it there I'd keep going down the list but I see that a +2 and a 4 should work. Let's check it and see. First times the first is 3x^2check. Outside term, 12x, inside term +2x, that's a gain of 10x, and then 2(4) = 8.
When you have stuff in front you've got to be careful because remember, when you're doing the outsides and the insides you might have to multiply it by whatever's in front there. It requires a lot of checking and careful analysis to make sure that you've got it in the right position. But, armed with the idea that if I see a negative sign in the end, I know that these are going to be differing signs, that allows us to start placing things in a reasonable way and reducing the possible choices and options for how to factor.
Undoing foiling, factoring is big business and it's hard business. The bottom line is factoring polynomials in general is really hard. Mathematicians have trouble with it. Don't feel badly if this takes you a while to get, but the trinomial factorization, if it can be factored, can be factored like this. Think about it, have some fun and enjoy the challenge. Feel the burn. See you soon.
Exponents and Polynomials
Techniques for Factoring
Factoring Trinomials: The Guess and Check Method Page [1 of 2]
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