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About this Lesson
 Type: Video Tutorial
 Length: 15:22
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 165 MB
 Posted: 12/02/2008
This lesson is part of the following series:
College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Basics & Prerequisites (37 lessons, $52.47)
Beginning Algebra Review (19 lessons, $37.62)
College Algebra: Factoring Techniques and Patterns (9 lessons, $16.83)
In this lesson, you will learn how to factor trinomials using a reverseFOIL trialanderror method. You will start by simplifying the trinomial as much as possible, by removing any common factors or grouping any possible combinations. Then, try the reverseFOIL by first breaking up the squared term. He also gives you a hint that when the last factor of the trinomial is negative, you know that the last terms of the binomials have to be opposites. You will walk through this process with a onevariable trinomial, a twovariable trinomial, and a trinomial with a positive last term.
Learn the FOIL method: http://www.mindbites.com/lesson/954begalgebrathefoilmethod
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Beginning Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/beginningalgebra. The full course covers linear equations, inequalities, polynomials, rational expressions, relations and functions, roots and radicals, quadratic equations and systems of equations.
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
 Thinkwell
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11/13/2008
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FACTORING TRINOMIALS COMPLETELY
The most common things that we are going to factor are actually trinomials. And that’s going to be things that basically have the shape––we’re going to have something squared in it, we’re going to have something sort of alone, and then we’ll just have a constant, and we have to sort of factor it into two factors. And actually this is where you sort of undo all the foiling stuff that we do when we actually multiply out two binomials. Let me sort of give you a sense of the theme of this and some pointers and let me just do a whole bunch of examples so you can just sort of remind yourself of how all this goes.
Suppose that we take a look at the following: we want to factor 3x2  x  4. Now the first thing that you might want to do, which once you sort of get in the habit of it can be really quick, is to go through all the easy things. First of all, is there anything that I can pull out of all of them? Is there any kind of common factor in all these monomials? The answer is no, because this only has a factor of 2 and 4, this has only x, this has a 3 and an x, I can’t pull anything out. Any grouping possibilities? I can group out an x here, but then I’m still left with that 4 dangling in the wind, so this doesn’t look too good.
So let’s just see if we can try to factor this and say, well, what would it look like to factor this? Well, it would have two pieces if it were to be factorable. Let me actually try to write in those two pieces here, and these are both going to be binomials, there are going to be two pieces to this, and now what are we going to do? I’m going to want those two pieces to have the property that when I foil it all out it’s going to equal this. So let’s just think through how that’s going to play out and the possibilities.
Well, the first two terms here I want them to be 3x2. Now there are a lot of ways of breaking this up. I could break it up like, for example, a 3 here and an x2 here, that would work. I could put in a 3x and an x here, I could put in an x and a 3x here, and so forth. I could do a lot of possibilities here, in fact there’s even more exotic ones, but I won’t even think about those right now. Let me try some easy ones. Actually putting in a 3 and an x2 here is not a great idea only because then later on in life when I’m doing the inside stuff and the outside stuff, I’m going to get some x2 somewhere in here. And that actually might not play out too well since this is the only x2 I have. So what I traditionally try to do is to break up the x2 by putting some of the x here and the other part of the x here. So let me put a 3x here, and an x here. And by the way, I am not giving any kind of warranty or guarantee that this is always going to work, but this is the way most of these things tend to go. So I’m trying to show you a general way of starting to think about it and seeing if it works. If this doesn’t work you will have to try something else, so don’t panic.
Anyway, notice this has the pleasant property that when I multiply these two things out I do get that first term. Okay, but now I have to get everything else to work out. Now how am I going to get everything else to work out? Well, let’s see, these two terms have to multiply to hopefully give me the 4. Now if you think about that for a second, the product of two numbers here is going to be negative. That tells me that the signs of these numbers have to be opposite. Do you see that? Because if they are both positive then that product would be positive, and if they were both negative that product would be positive. So, in fact, what I have to do is have opposite signs. And it has to be two numbers whose product is 4. Well, there are a lot of possibilities, there’s 4 and 1, there’s 2 and 2, there’s all sorts of possibilities. So what we have to do now is think about which possibility is going to give us everything else.
Let’s think of the 2 and 2 possibility for a second. Imagine I write a 2 here and a 2 here, well, what happens? I would get the 4 over here, but this would give me a 2x and this would give me a 6x, and the question is, is there any way to combine a 6x and a 2x and make it into just an x by itself, or a negative x? And if you have a 6x and if you added the 2x or if you subtracted from 2x, I’m not going to get it, so that 2, 2 factor doesn’t look too good.
So it is probably going to be a 1 and a 4 factor that’s the next thing I’m going to try––it may not be, by the way, I want to point out it may not, this may not be factorable. But if it is going to be factorable in an easy nice way, then probably we are going to have a 1 and a 4 thing, but I have to consider all the possible signs. For example, 1  4, 1, 4 and then putting the 4 here maybe, 4  1 and then 4, 1. A lot of things to check and let me show you how I would actually check those things.
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Let me write the problem back here, this problem is 3x2  x  4, and let me write down all those cases I have to consider for that bottom thing, because here is where we are right now, we have this. We know it’s going to look something like that and now what I want to do is figure out what’s going to go in these holes. Well, what are the possibilities? Well, I could put in a +1, but then I have to put a 4 here; I could put a 1 and then put a 4 here; I could put a 4 here and a 1 here; and then I could put in a 4 and then a + 1 here. Those are all the four possibilities of putting the two different signs in different places. You see, they are all different possibilities. We know the 2, 2 thing can’t happen, because I can’t get that middle term.
And now what I’m going to do is just of go by trial error and see which one of these actually might give us the answer, maybe none of them will, but let’s try. So the first one, if I do this, how would that look? Well, let’s see, the first terms give me the first term here, the inside term gives me a +x, the outside term gives me a 12x so that combines to give a 11x. Already I can stop the game because I know that’s not right. So, let’s move on.
What about here? Well, in this case the first two terms of course are good, the inside term I get a x, the outside term now I get a +12x and that gives me a +11x––hold the presses, we’re done, it doesn’t work.
Let’s go there, this might not be factorable folks, who knows. Here what I see is the first term is okay. Then I have a 4x, and then I have a 3x, so 4x  3x is actually +x. So close, do you see it? It’s still plus and I need minus, so even though I don’t want to I have to throw it away and hope for the best down here.
Let’s see what happens here. Here I see a 3x2, here I see a 4x, here I see a +3x, and a 4x + 3x is a x, and the last two terms give me the 4. This is actually a correct factorization, great. So, in fact, this is a way that one can think about trying to factor a trinomial into two binomials. And the idea is to break up that term with the square in it in a reasonable way. And then try to fiddle with the constant term thinking about the signs and seeing how the signs would play out to make this thing work, and then trying those possible cases. Now this is quite an exorbitant way of doing this. In reality you might actually be able to try them pretty quickly once you get into the habit of it.
Okay, let’s try another example together and practice a couple more of these. Let’s consider the following: 6y2  48y  120. Okay, a lot of big numbers there, at least big for me, and so I’m getting a little bit nervous about this. But notice something hey, wait a minute, I can factor something out of all these terms and that will actually make my work a lot easier. Now what can I factor out? Well, I can factor out a 6 from here, I can factor out a 6 from here. Can I factor a 6 from here out? Well, let’s see, I can certainly factor out a 2, can I factor out a 3? Yes, I should be able to factor out a 6 from here. So first we’ll pull out the greatest common factor here. See how I am trying to do that to save myself some effort? If I pull out this, a 6 from here, I think I’m left with just an 8, and if I pull out a 6 from here, I think I’m just left with 20. Okay, now we’re talking my language, because this is something I can actually deal with.
So let’s see what this equals, and let’s see what this equals, so I keep that 6 in front though and now I’m going to hope that this may be factored into two pieces; the y2––I’m going to put a y and a y here. And now what about the last term? Well, let’s see. The last term is negative so I know I’m going to have opposite signs, but notice that since these are the exact same things, y and y––remember before it was 3x and x––since these are the same it doesn’t make a difference how I put them down. I just have to make sure that they are opposite, so I will put a plus here and a minus here. Okay, and now what I need to have is two numbers that multiply to give me 20, but then when they combine in this fashion, insideoutside, they give me 8. So let’s try some possibilities, for example, there is 2 and 10. So if I put in 10 here and 2 here, notice that product is 20, but inside I have +10 and outside I have 2, that gives me a +8. So close. That closeness tells me maybe I should do this: flip them. So let’s actually put the 2 here and the 10 here and notice the outside term is 10y and the inside term is +2y and I get the 8y, the last two terms, 20, great. Here’s the correct factorization with the 6 there.
Let’s try a little more exotic one. How about this: 5r4  7r2s  6s2. Now this one, there is something to the fourth, so it looks like maybe we can’t even do this trick. But the thing to notice here is that there is something to the fourth, but then there is also that same thing squared. So in our minds we could think about it this way, this is sort of a blob thing here, and this is sort of the blob squared, so if I think about it that way then I do have all squares there even though it looks like a fourth. Let me actually write that out so you can see that this is just a camouflage square. You see, it’s
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something squared and then minus that same something unsquared. You see, so this is just now something, let’s just treat it as one big variable, one big blob, and let’s try to factor. Any common factors here? I don’t see any, so I’m going to hope that this can factor. I see a 5 something squared, so I’m going to try 5 something and then just the something. Notice that the something is whatever is in here. You see how I just brought that down, because the product is 5 something squared, so five r squared squared, so that’s the r4. Okay, I see a negative sign again so that means I’m going to switch signs, but now I have to be careful where to put the signs because these aren’t the same. So it could be plus minus, or it could be minus plus, I don’t know, by the end of the day these terms have to give me this, 6s2, so I hope that I should have an s here at least. And now I have to figure out factors of 6 that combine with this 5––it’s now getting pretty complicated––to give me this 7.
Well, how should that play out? Well, let’s see, some factors for 6 would be for example, 3 and 2; there’s also 6 and 1, so we could try 6 and 1 here and here; and then 3 and 2 here and here and so forth. Let’s see what happens if we put in the 2 here and a 3 here. So visualize a 2 right here and a 3 right here, maybe I have some of these things, let me just check here in my plethora of tricks. I’ve got a 5, but not very useful, and do I have a 3? I’ve got a little teeny 3, look at this, a little teeny 3, and let’s see, I’ve got a little teeny 2, well just for fun let’s just do that. Suppose we put this in here, now remember we have to put in opposite signs so if I put in a plus here and a minus here let’s see what happens. At the end of the day we do get the 6s2. Now what is the inside term? Remember, we have a plus here now and a minus here, this would be a +2sr2, and then here I have a 15sr2, 15 + 2 does not equal 7, so that doesn’t look too good.
Let’s try to reverse these roles. So I still have the plus sign here and the minus here, let’s see what happens here. Here I have a 3sr2 and here I have a 10sr2, so 10sr2 + 3sr2 is, yeah, 7sr2. So this is great, so in fact this is the right factorization. I could take off these little pathetic 3’s and 2’s and replace them by their larger font counterpoint. And now I’ve actually factored this correctly. And a great thing to do by the way, a really great thing to do is to check your answer.
Let’s quickly check, really fast. This times this, foiling, is 5r4, perfect. The inside term gives me a +3r2s, the outside term gives me a 10r2s. They combine to give 7r2s, and the last times the last is a 6s2 as we hoped. So, in fact, this is factored correctly, neat.
Now, let me just say one last thing about this and that is let’s look at something that is a lot easier than this. Let’s look at the following: how about x2  5x + 6. I want to factor this real fast because I want to now show you what you do when there is a positive sign here. When there is a positive sign here everything is the same except you can actually say a little bit more. The x2 I’m going to break up as x and x, but now I have the positive sign. But there’s two possibilities, because both these numbers can either both be positive, in which case they’ll provide us a positive, or they can both be negative. But here is a great little device to help you, I can actually tell you what the answer is by looking at this next term here. Because see, the terms have to combine to give this value, right? They have to add up to give this value. Well, if they were both positive here, if I have a positive thing plus a positive thing I can’t make it negative. So, in fact, if I see a positive here then I know the signs will have to be the same and that sign will actually be whatever is here. So since this is a negative I know these are the same and they are both negative, that’s a great trick. If this were a positive here by the way, then I would know these would both be positive. So if you see a positive out here that means same sign over here and the sign will be whatever this is. If you see a negative you have to look at all the possible cases of plus and minus and you don’t know what’s going on, no matter what this is. And here I can put in a 2 and a 3 very quickly and you’ll see that happily I have an x2,  2x  3x =  5x, and 2 times 3 is happily a +6––this factors. So when this is positive you know the signs will be the same and that same sign will equal whatever the sign this is. When they are negative, you are going to have staggering signs and they can go any which way you can.
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