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Taking common factor from trinomial

Break down the process of taking common factors from trinomials. Learn how to identify the greatest common factor of a trinomial expression and use it to simplify the expression. Follow along as Sal factors 4x⁴y-8x³y-2x² as 2x²(2x²y-4xy-1) by taking the greatest common factor. Created by Sal Khan and Monterey Institute for Technology and Education.

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Video transcript

We're told to factor 4x to the fourth y, minus 8x to the third y, minus 2x squared. So to factor this, we need to figure out what the greatest common factor of each of these terms are. So let me rewrite it. So we have 4x to the fourth y, and we have minus 8x to the third y, and then we have minus 2x squared. So in the other videos, we looked at it in terms of breaking it down to its simplest parts, but I think we have enough practice now to be able to do a little bit more of it in our heads. So what is the largest number that divides into all of these? When I say number I'm talking about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet. And we say, well, the largest, of, the largest common factor of 2, 8 and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor. So let's write that down. So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them. 2x squared. And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't, so there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as the product of the 2x squared and something else. And to figure that something else we can literally undistribute the 2x squared, say this is the same thing, or even before we undistribute the 2x squared, we could say look, 4x to the fourth y is the same thing as 2x squared, times 4x to the fourth y, over 2x squared. Right? If you just multiply this out, you get 4x to the fourth y. Similarly, you could say that 8x to the third y-- I'll put the negative out front-- is the same thing as 2x squared, our greatest common factor, times 8x to the third y, over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared-- so we have that negative sign out front-- if we factor out 2x squared, it's the same thing as 2x squared, times 2x squared, over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1. That simplifies to 1, maybe I should write it below. But what do these simplify to? So this first term over here, this simplifies to 2x squared times-- now you get 4 divided by 2 is 2, x to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y, and then you have minus 2x squared times, 8 divided by 2 is 4. x to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared time-- this right here simplifies to 1-- times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared times this term, minus this term, minus this term. Right, if you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y, minus 4xy, and then you have minus 1, minus 1, and we're done. We've factored the problem. Now, it looks like we did a lot of steps. And the reason why I kind of of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head. You might say OK, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2, so let me put that 2, let me factor 2 out. Well, all of these are divisible by x squared. That's the largest degree of x. Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. So let's see, it's going to be 2x squared times-- and what's this guy divided by 2x squared? Well 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1, there's no other y degree that we factored out, so it's just going to be a y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And then you have y divided by say, 1, is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared 1, so 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head, but I really want you to understand what we did here. There is no magic. And to realize that there's no magic, you could just use the distributive property to multiply this out again, to multiply it out again, and you're going to see that you get exactly this.