 So, what we had was the divisor divided by the dividend is equal to the quotient plus the remainder divided by the divisor, right? Or the denominator. Now, we got D, big D and little D and we're going to get rid of that, right? We're not going to stick with two different types of Ds, right? But let's continue this, okay? What we're going to do right now is, you know, in general, most people hate fractions. I personally love them. One of those freaks that, you know, it's like one of those freaks that likes long division, right? I don't mind fractions. However, they get in the way when you're trying to simplify stuff, right? So, what we're going to do right now is multiply the whole equation by small D, right? By whatever's in the denominator, the common denominator, and it just happens to be D right now, small case D. So, we're going to multiply the whole thing by the small D and we're going to get another statement and that's our division statement. That is what we're looking for because that's the way we're going to express our polynomials. You can express them in this form as well and there are some courses, some, you know, books that I've seen that do express in this form but I personally don't like it. It's ugly and it's not as useful because the remainder gives us a lot of information when we're dealing with polynomials because the remainder, when you're doing long division, just on the side, when you're doing long division with polynomials, the remainder is what your Y is when your X is equal to D, okay? So, or the X that you derive from the D, the small case D. Now, let's multiply this whole thing and that's just an aside and we're going to get there, okay? So, what we're going to do right now is just multiply the whole thing by small D. So, we put the whole equation, whole expression, whole equation in brackets and multiply all the terms with D. Now, big D over little D multiplied by little D is just going to leave us big D over here. It's going to be equal to Q times D and the R over small D, the little D is going to kill the big, the little D that you're multiplying with. So, all you're left in this one is just going to be R. So, your final expression is going to be in the following form. Okay. So, what we're going to have is going to be D, your dividend is going to be equal to your quotient times your divisor plus your remainder. So, basically, if we went back to our fraction, 27 divided by 2, this is the way we would write it. So, we have 27 over 2 is equal to 13 plus a half. So, we're going to take that whole equation and multiply it by 2 and the 2 is going to reduce it down to this expression over here, right? Which your big D is going to be 27, your Q is going to be 13, your little D is going to be 2 and your R is going to be 1. So, what we're going to have is 27 is equal to 13 times 2 plus 1. Now, this is our division statement, which is this one really. Now, you know, we've got two different types of D's in here and I personally don't like the two different types of D's. So, what we're going to do is call the big D a P, which is the product, okay, which represents the product. So, and because we're going to start talking about functions, what we're going to do is call them Px is equal to Qx times Dx plus Rx. So, what we're going to have in our division statement, we start dealing with polynomials, is going to be P of x is equal to Q of x times D of x plus R of x, okay? And all of these guys are going to be functions. Sometimes R of x is going to be 0 and if it is 0, it means the x that we put in for P is a y-intercept, which is really, really, really important, okay? Now, I'm just going to lay this out in the division form as well, so you get a better understanding of what it looks like in that form as well, so you know where everything fits in. And then we're going to get into dividing a polynomial. So, if you wrote it down in the long division form, you would have your D goes there, your P, your product is there, your quotient is up top and your remainder goes down here. And all of these are functions, which means they're P of x, Q of x, D of x and R of x, right? In our basic function terminology. So, what we're going to do right now is go through and do a division, do a long division. And remember, the long division, when they give it to you in the format, it's just a fraction. It's just P divided by D. It's just P divided by D and it would ask you to do this. And you can use long division or synthetic division. And after we go through the long division, we're going to go through synthetic division and you're going to see that it's a much sweeter process. What we're going to do is do a long division with polynomials. And basically, the structure is the same. It's basically our long division format. Now, hopefully I can fit it all in here because long division takes a fair bit of space. It takes a fair bit of computation to do it. And we're definitely going to express it in the division statement where we had P of x is equal to Q of x times D of x plus R of x, which is the product is equal to quotient times the divisor, whatever the denominator is, plus the remainder, right? And the remainder is going to give us a lot of information, which is basically going to be what y is, what the function is equal to whenever we plug in a certain x and our Q and P are factors if R is equal to 0. And we're going to talk a lot more about this when we start breaking this down. And you're going to see where this goes. And it comes in really handy when you're trying to find factors of polynomials, especially large polynomials, right? Or high-order polynomials when we're going above quadratic equations, when we're going beyond quadratics and trinomials and the difference of square. We're going to go into larger functions so we need more powerful factoring techniques. So we're going to do a polynomial and you should recognize what we're writing down because you should have gone through our other factoring videos if you're new to this. And it's basically a polynomial, a quadratic equation that we factored before. So the question that we'd be asking you is the following. Negative 3x squared plus 5x plus 2 divided by x minus 2. Now, if you remember from the previous videos, we've already factored that expression two different ways. We factored it using complex trinomial factoring and we factored it using the quadratic formula. Hopefully you'll remember that x minus 2 is a factor of the above polynomial, right, of the above quadratic equation. And what we're going to do right now is do the long division for this. And later on we'll do the synthetic division, right, just to see so you can do a comparison, right. But we're going to do the long division for this. And what's going to happen is the above polynomial is our product. Okay, the x minus 2 is our divisor and the other factor that we're going to get, the quotient is going to be the other factor and our remainder for this is going to be zero. Because what this really means is we're taking x equal to 2 and we're subbing it in and finding out what y is equal to, okay. Now keep this in mind, I keep on throwing these teasers in there because you'll get a better understanding of what's going on once we go through the whole process, right. This is one of those lessons that, you know, you won't get the whole picture until you get to the end and when you get to the end you're going to look back at this and go, oh, wow, okay. And it should have that moment when you realize that all you're doing is seeing if the bottom guy is a factor of the top guy and the other thing that you're doing is seeing what y is equal to when x is equal to whatever this expression is going to be. And this is x minus 2 equals 0 so you bring the 2 over. There's a skater here so I gotta find out if they're gonna go through the bowl, if they are, I gotta get out of here. So unfortunately we might have to come back for this, okay. How you doing? You gonna go around? Yeah. Okay, you need me out of your way, yeah? Yeah. Yeah, okay. Okay, we gotta zoom out of here. Sorry, this is gonna have to wait.