 So, let's say the question was the following. They wanted us to factor this expression, right? Factor this polynomial. And it's 2x to the power of 5 plus x to the power of 4 minus 2x minus 1. So what we're going to do is lay this out in the synthetic division form. We'll probably do it down here because it's transferred numbers down. And the possible factors of this are going to be possible factors of 1 divided by possible factors of 2. So possible factors of 1 are just plus or minus 1 and possible factors of 2 are plus or minus 1 and plus or minus 2. So we're very limited as to the possible factors that we can try for this. And keep in mind these are just the factors that we can try manually. There may be other factors there but we just can't do them manually using synthetic division anyway. So possible factors of 1, plus or minus 1 divided by possible factors of 2 and we're going to lay down the whole thing here. What we got is possible factors of 1 which is plus or minus 1 divided by possible factors of 2 which is plus or minus 1 and plus or minus 2. Now divided by plus or minus 1 it's just whatever the possible factors of the last term are. We lay down our division statement here or our synthetic division coefficients, right? For the synthetic division. And keep in mind everything has to be in descending order, right? So that's x to the power of 5, x to the power of 4. We're missing x cubed and we're missing x squared so we're going to have to put place markers here. So we put a 0 for x cubed, 0 for x squared and we got negative 2 and negative 1. So we actually have 1, 2, 3, 4, 5, 6 terms. We always have one extra term than whatever the power is, right? So what we're going to do is just lay down our synthetic division statement the lines here and we're going to choose x is equal to 1 and go through the synthetic division form. What we got here is we're going to try x is equal to 1 and that means we're trying to see if x minus 1 is a possible factor of this guy, the polynomial, right? Which is what we're trying to do. We're trying to factor this expression here. So what we're going to do is take the 2, drop it down, multiply it by 1, goes up here add those 2 together, bring it down, multiply it by 1, comes up here and just do the zigzag and I'm just going to lay the whole thing out right now, right? And every time we're going across this way, we're multiplying whatever was here by 1 and we're going to see what we end up with. So during this long division, we just go zigzag, zigzag, zigzag and keep in mind I mentioned this in when we're doing the polynomial long division with putting the place markers in where for polynomial long division we really didn't have to put place markers in. It's good to do it because it prevents you from making mistakes but for polynomial long division it's fairly simple. You just can't add terms that don't have the same power, right? You can only add like terms. So if you keep that in mind, especially from series 2, we talked about that if you keep in mind that you can only add like terms especially for polynomial long division then you won't really make too many mistakes of adding different powers together. But we still ended up putting place markers because we want to line everything up. That's what we're doing here too. We're putting the place markers just to line everything up just to make sure everything's in descending order. The problem arises with the place markers is if this power up here was extremely large, let's say it was power to, you know, x, it was 2x to the power of 27 and then we went to the power 4. Now you're not going to sit down and put place markers for, you know, x to the power of 27 and 26, 24, 25, 24 all the way down to, you know, x to the power of 4. So you're going to have, you know, your sheet's going to be gigantic and have zeros all over the place. You're not going to do that. What you can do is realize that whatever you're multiplying this guy with when you're going across, it just multiplies. When you're adding them, it's just the same term coming down, right? So what you could do is keep in mind where, let's say this guy, this is 3, this is 3 going up to a 0, right? If you add those two guys, that's 3 again, and that multiplies whatever is here. So if that was a 2, the 2 multiplies here. If that was a 0 again, you add it. And if you add another 0 here, you're multiplying it by 2 again. So what you could do is multiply, figure out how many terms you're missing and take whatever this is and take it to that power and then multiply it by these guys. Now, I'm not going to get into that, because I've never seen problems like that coming up. They're not going to give you, you know, in general, you're not going to get an expression where, you know, x to the power of 27 missing all the middle terms and, you know, goes down to x to the power of 4. So if this comes up, if people have problems with it, if this question comes up, I will make a video of how you go ahead and do that and figure out what you have to multiply, you know, the last term that you had here before you started having to place markers until you get to the next term where you don't have a place marker. But right now, we're just going to leave it alone and just use place markers for it, okay? So what we ended up doing is, you know, doing our synthetic division, getting a remainder of 0, right? So that means x minus 1 is a factor of the top guy. And what we end up having down here is our quotient when we, you know, we're basically dividing this guy by x minus 1. And what we did was this was x to the power of 5. We took out an x from it. So what we've done is our quotient now starts with x to the power of 4. So all those numbers there, they're the constants, the coefficients in front of the variables. So I'm just going to write those down right where the numbers are because I don't want to continuously write down all the numbers again and write the x's there, right? So I'm going to write those down with pink chalk. So we're going to write it down with pink chalk and those become our variables for our coefficient and the last term is going to be our constant. So we've kicked down our original polynomial by 1 degree. And what we're going to have to end up doing is factoring this guy using synthetic division and the factors of this guy are possible factors of this guy divided by possible factors of that guy. Same deal as this guy. So what we have is it becomes 2x to the power of 4 plus 3x cubed plus 3x squared plus 3x plus 1. And we don't have any missing terms. So what we're going to do is just do straight out long synthetic division as well again. Now possible factors of this guy are possible factors of this guy divided by possible factors of this guy which is plus or minus 1 divided by plus or minus 1 or plus or minus 2. And what we're going to do is try positive 1 again. So x equals positive 1. And I'm just going to lay down the synthetic division statement, you know, the lines right here and just continue from there that way we're saving space. So what we're doing, we're trying to see if x minus 1 is a double factor of this guy. You know, a polynomial could have the same thing again. And all you would do if it was, if x minus 1 was a double factor of this guy, you would go x minus 1 all squared is a possible factor of this. So we're going to go ahead, I'm just going to go ahead and just do this whole thing and find out what the end result is. To see if x equals 1 is again a factor of this quotient here, this polynomial here. Which would mean it's also a factor of that guy. So what we end up having is x equals 1 is not a factor of this guy. So we already divided. We found out x minus 1 was a factor of this guy. Reduce that guy to a new polynomial, right? And we just found out x minus 1, x equals 1 is not a possible factor of that guy because the remainder is not 0. The remainder ends up being 12. So x minus 1 is not a factor of this guy. So we already checked, we already tried one of the possible factors of this polynomial. What we're going to do is try another one. Now, what I'm going to do is transfer all these numbers up here and, you know, try x equals negative 1, which means x plus 1 is, you know, we're trying out x plus 1. So we're going to do that one and see where that goes. So I'm going to go ahead and just do the synthetic division here and that's x is equal to negative 1. So the 2 comes down, multiplies negative 1, goes up here, add them together, multiply negative 1, continue, continue, continue to see what we end up. Doing the synthetic division, we ended up with a 0. So now we know x is equal to negative 1 is a possible factor of this guy, which means it's also a possible factor of this guy. What we have right now, so far we found 2 factors, which is x minus 1 and x plus 1 are factors of this guy. And what we've done is we've divided, you know, x to the power of 4. Is that x to the power of 4? It's x to the power of 4. We've divided x to the power of 4 by an x. So our next term here is x cubed. So that's going to be 2x cubed plus x squared plus 2x plus 1. And always remember, if any of these were negative, that means they're minus, they're not pluses, right? The sign in front of the number goes with the number. Now what we're going to do is continue our synthetic division and try to find possible factors of this. Possible factors of this are plus or minus 1 divided by plus or minus 1 and plus or minus 2, right? Same deal as here. I just picked the polynomial where it was going to be easy to figure out what the factors were, right? Now what I've done is I'm running out of blue chalk. So we're going to change colors. And we haven't used green on here. So green, you can picture green as being blue. So green is going to take over and we're going to do the long division for this or synthetic division for this using green, okay? And what we're going to do is try 1 again. Okay, let's start off with 1. Or you know what? 1 we know is not going to work because all of these are positive. If we multiply everything by 1, then everything's being added together. So there's no way we're going to get a number here where the remainder is going to be equal to 0. So what we're going to do is try x is equal to negative 1.