 So we're just going to go through the routine right now of, you know, figuring out how to lay down the synthetic division statement, basically, and go ahead and doing the synthetic division with something fairly simple. And, you know, watching out for expressions like this where you're constant in front of the X is not just one. But when you get an expression like this, if they ask you to do synthetic division, this is what you're going to have to end up doing. You take the numbers, the coefficients in front of the X terms and the constant back here. So you take all the numbers. You forget about the X's, right? You forget about the variables. You take all the numbers and you lay them down, okay? And you do a sort of upside down root symbol or upside down division symbol. And what you have to do with this guy is you're going to have to set this equal to zero and move it over so it's X equals whatever the number is. Which is going to be negative one over three. And we're going to take all the coefficients and lay them down here and, you know, set up our synthetic division statement and go ahead and start doing the synthetic division. So what we do is set negative three X plus one, our minus one is equal to zero. You bring the one over and then divide by negative three. So what we have is X is equal to negative one over three. You take this guy and you move it up there. You lay down your, and this is your synthetic division statement. This is the way you lay out your problem, okay? You do sort of an upside down division symbol, just a, you know, semi box or half of a box, right? You take all the numbers in your numerator, right? In what you have in the fraction up top and you put them all down there, okay? Now, keep in mind the sign in front of the numbers always go with the numbers. And if you have any missing X terms, you have to put a place marker there, right? So if this was negative three X squared plus five, that would have been if you were missing the middle term, the X term, you would have had to put a zero instead of the five here. Because you need your polynomials, your numerators to be in descending order, first of all, and you need them to be in order. So if you have, you know, X to the power of five, the next one has to be X to the power of four, X to the power of three, X to the power of two, X to the power of one, and your constant term. If you're missing any of the powers, any of the terms, you're going to have to put a zero marker there, okay? And we'll do a, you know, we'll do a couple of examples that way where you see what the place markers are. Right now, we're not missing any terms, it's all in descending order, and we don't have to put any place markers. So you take all the numbers, you lay them down there. This is the way synthetic division works. You're going to take the first number, the one in the coefficient in front, the highest power, and you're going to put it down here. This number here, so it's negative three is going to come down here. Negative three is going to multiply this guy. So this guy here multiplies all the terms, and the multiplication result goes up here, and you add up the two, these terms here, and that comes here, and you continue to do the same thing. So basically the motion that you're going through is this guy coming down going like this. So it's sort of a zig-zag, ding, ding, ding, ding, ding. So what we do is the negative three comes down, negative three times negative one over three goes up there. Now negative one over three times negative three is going to be positive one. Five plus one is going to be six. Negative one over three times six, that's going to go up there. Now negative one over three times six is going to be negative two, and we put that there, and two minus two is just going to be zero. So what that means is that guy is a factor of that guy because our remainder is zero. Our remainder is zero. That means if we sub and x is equal to negative one over three for all the x's in the numerator, the answer is going to equal zero, which means this guy is an x-intercept. That's where it crosses the x-axis. That's where y is equal to zero. Now we've talked a lot about doing the synthetic law, doing the polynomial long division, and doing the factoring stuff, right? So this should be fairly familiar to you, just a little quick recap. So what ends up happening if we think about the polynomial long division? Our terminology, this guy is our d of x, right? It's our divisor, right? This guy up here is our dividend, is our product, is p of x. That guy up there, right, is our polynomial. This guy is our remainder, whatever ends up being the last term in the synthetic division, right? That's going to be our remainder. Right now our remainder is equal to zero. And these guys, negative three and six, that's our quotient. And the way it works is right now we took an x to the power of two and divided an x into it, right? That means we've reduced this power by one x. This guy becomes negative three x plus six. If this guy was x to the power of five in the top and we're dividing an x into it, this term would have been negative three x to the power of four. So every time you do a synthetic division, every time you divide an x into the top, what it does, it kicks down the power by one order. And all you do is these numbers here, no matter how many you get, all they are, you start off with the next order, descending order, right? So if that was an x to the power of five, the first one would be x to the power of four, and then x to the power of three, x to the power of two, x to the power of one, x to the power of zero, which is your constant, right? Right now this one started off with x to the power of two, divided by an x. So that just means there's an x here, right? So it's negative three x plus six. This becomes our quotient. What this means is if you took, if you brought the negative three, negative one over three over, this would be x plus one over three. So if you multiplied x plus one over three by this guy, you would get your original polynomial. If you multiplied negative three x plus six by x plus one over three, you would get back our original polynomial, right? Here's the problem when you have, when you're doing synthetic division, when the number in front of the x is not one, but some other number. Now we got this quotient out, we got this result out with the remainder of zero. This is the problem. If we took this guy and multiplied it by negative three x plus six, we wouldn't get that guy up there, because negative three x times negative three x is going to be nine x squared, which obviously is not what's in the top. So this is what you really have to watch out for. Whenever you're doing synthetic division, when the number in front of the x is not one. When you do this division, you might end up with a quotient that if you multiply that quotient with your original expression here, you're not going to get the top guy. Because we ended up manipulating this guy to get if x is equal to negative one over three, right? We moved everything over, so our A term, when we set up the, when we talked about what we're going to use synthetic division for, our A term is now fraction. If you took this guy and multiplied by this, yes, you're going to get the numerator. If you took this guy and multiplied by this, you're not going to get the numerator. When you get something like this, you have to visually check to see if you multiply this by this, if you're going to get that guy. And if you realize that this guy times this guy doesn't give you the top guy, take a look at this guy and take out a GCF. And you're going to have to make a decision of taking out something that's going to convert this to an expression where if you multiply it by what you had in the denominator, it's going to give you the original expression. So there's two different ways, and actually there's multiple different ways you can think about this and do the corrections necessary to present your answer, the correct answer if you're asked to do polynomial long division like this, or when you want to factor large polynomials, polynomials of higher degree than degree of two, and polynomials that are bigger than just simple trinomials, anything that has more terms than three terms. Basically the two different methods that you can think about in presenting the right answer, getting the right answer after you do your synthetic division are the following. The first one is all you got to do is look at the number in front of the X when it's not a one, take this number, and divide your quotient by this number. So the first way you can do this is the following. You can take negative three X plus six and divide it by negative three. And what happens is if you take out a GCF from the top, it's going to be, well, the GCF that you need to get rid of your negative three. You could take out a three from this, but what you would end up is you would have negative X plus two and there's going to be a three up front and that's not going to get rid of your negative three in the bottom. So what you're going to do is sort of a decision that you're going to have to make. There's a lot of decisions that you can make when you're doing mathematics, when you're doing simple simplification and running through algorithms that make your life easier. And the more you do, the easier it gets. So initially, if you take out the wrong thing, don't worry. What's going to happen is you're going to have to take out another factor to get rid of whatever's going to be left in the denominator. But the more of these you do, the better at it you're going to be. So right now, we have a negative three in the denominator because that was our B term, whatever our coefficient in front of the X. So what we're going to do is take out a negative three from here. So this turns out to be negative three comes out and if you take out a negative three from negative three X, it becomes X and if you take out a negative three from positive six, it becomes negative two. So negative three up top can kill negative three in the bottom. So this guy kills this guy. Your final answer for your quotient, when you take this polynomial and divide it by this polynomial, is going to be X minus two. Right? And that's the first way you can think about it. The second thing you can do is take a look at what we have here. It's X plus one over three. Then what you can do is take out a GCF from this and just take that GCF and multiply it with this guy, which is really the same thing that we did here. But it's just another way of thinking about it. So what you can do is the following. You can take negative three X plus three plus six and take a look at this guy and you know there's a three in the denominator here, right? So you want to get rid of that three and what you're going to do is GCF out the largest thing possible and in general for the quotients, you want the X term or four terms like this. If you can get this guy the X to be positive then that's what you want to do, right? So you're going to take out a negative three X and again that comes out to a sort of a decision that you're going to have to end up making because if you only take out a three and multiply it by this guy the three is going to come out here and multiply this guy and that's going to be three X plus one but that's not what we had in the denominator here. So you're going to have to go back and factor out a negative one and redo the thing. So again the more these things you do the better you're going to become at it so it's a decision that you're going to have to make, right? So you're going to have to look at this guy and say this is what you want to get out from this guy because this is really what you're dividing that guy with, right? So if that's negative three X minus one what you want this guy to be is negative three here so what you're going to do is take out a negative three from this guy. So what ends up happening is you have your quotient you've got negative three X plus six and this guy times this guy this guy times this guy gives you original polynomial but this guy doesn't appear here anywhere, right? So what you're going to do is try to convert this into this and the way you're going to do it is you're going to factor out GCF out a negative three from this so you factor out a negative three from this guy that's going to be negative three coming out of this again it's going to be X negative three coming out of this again it's going to be X and negative three coming out of six is going to be negative two and then what you're going to do is going to take this guy take this guy and multiply it inside this expression and what's going to happen is you're going to get X minus two and negative three times X is going to be negative three negative three X and negative three times one over three is going to be negative one And now this expression is equivalent to this expression, and this guy times this guy would give you your numerator up top there. And to write your answer for this, this was straight out the vision, this divided by this, your answer would be x-2, x-2. It would not be this guy because this guy times that guy does not give you that, okay? Watch out for these types of questions. Be careful with these. You might get them.