 All right, welcome back. We are going to talk about synthetic division. Okay, so in a previous video I talked about polynomial long division and this is going to be polynomial synthetic division. This is a process that it's basically a quick way of dividing polynomials. It doesn't really look like the polynomial long division that I did in a previous video. It looks much, much different, but it is a much faster way of doing division. You'll see that here in just a moment. Now the process itself does, it looks very confusing. It really, really is, but the one thing you got to do is just stick with me to the end of the episode. Stick with me to the end of the video and we'll get through this and you'll see actually how much easier synthetic division is than your normal polynomial long division. Okay, so the first thing that we want to do, identify kind of what we're dividing here. In this case, we're taking 3x squared minus x to the third plus 5x minus one, this polynomial, this quartic four term polynomial divided by this linear binomial x plus two. So we want to see, we kind of see what we get when we divide. Now if you saw from the polynomial long division video, it can get a little bit hairy about what all you're multiplying and that kind of stuff. This synthetic division is actually a simpler process. There's a little bit of setup to it though. So here's the setup. Now on this side, what we have, this is your setup right here. You have a number over here in this kind of little half box. Then you have these numbers that enter row right here. You have a gap right here. We're going to put some more numbers right down here. You have this line and then you have this box right here. We're going to have numbers down here also. There's going to be numbers all over the place. But this is kind of the setup. You have your box here, row of numbers, you have a gap right here. We have our line and then we have this little kind of three quarters box. I guess you could call it down here. Now where do all these numbers come from? Now we're dividing by x plus two and so the number that's going to go in this box is always going to be the opposite. Opposite of this number. So notice it's x plus two. So the number that we actually put inside this box is going to be a negative two. That's going to happen every single time. You just use the opposite of that number. Then these other numbers here, what are these numbers come from? Well notice I have this up here. These numbers are our coefficients, coefficients. Now these are the coefficients of our dividend. The number is going to be on top right here. But notice that there's actually something extra. We have this zero x squared. Notice that there's a gap right here. There's an x squared variable that is missing there. We're not necessarily missing. We just don't have one there. But for the synthetic division process, we have to have all of the coefficients, even the ones that are not there. Sounds kind of weird. We have to have all the coefficients, even the ones that are not there. So if a number is not there, we assume it to be zero. So this is actually zero x squared. So then my coefficients, I got from up here and from this little gap. So these numbers are just the coefficients of three, negative one, zero, positive five and that negative one right there. So that's where those numbers come from. Now we actually get, that's the setup. Now we get into the actual synthetic division process. So now what we're going to do, your first step is actually we're going to start here on the left side. We're going to start here working on the left. This three is what we're going to start with. Your first coefficient. I'm going to take this number and I'm going to just bring it down. I'm just going to bring this number down. Then this right here, this row of numbers that I'm going to get are going to be my answers. These are going to be my answers. Well that answers, plural, my answer, singular. Anyway, so the numbers that I get here are going to be part of my answer. Now actually writing the answer is going to be a little bit different, but we'll get to that here in a minute. So take this three and move it on down. Then what we're going to do is we're going to take this number that's inside the box and we are going to multiply it times any number that we bring down here. We're going to multiply it times any number that we bring down here. So in this case, for this number, I'm going to take negative two times three to get a negative six and that negative six, we are going to put up here in this gap. I mentioned this earlier. We have a gap right here. So again, negative, we take the three, brought it down. We're going to use negative two. We're going to multiply that times three to get a negative six. So that's what I have up here. All right. And then we just continue and then we just kind of repeat this process. I'm going to bring these numbers down. This is a negative one and a negative six words are going to combine to get a negative seven, you can also think add down, you can also think just add these numbers down to get negative seven. And then just continue with the process. I'm going to take negative two. And I'm going to multiply that times negative seven to get positive 14. Forgot the number. There we go five positive 14. And then you just continue the process, continue the process. So I'm going to add down to get 14, multiply up to get negative 28. Add down to get negative 23, and then multiply this is okay, negative two times 20, negative 23 is a positive 46. Okay. And then I'm going to take negative two. I'm going to take negative two. Oh, shoot, nope, I'm done. I'm done. I just have to add this down. So negative one plus 46 is a 45. There we go. That's better. Alright, so that's the synthetic division process. I didn't do my arrows for each one of these, but you can't we kind of get the idea. So let's recap. We're not going to write the answer quite yet. So let's recap a little bit. So I take, I start right here, I start with this three, I bring it down. And then from there, I take negative two times this three to get negative six, add down to get negative seven, then I take negative two times negative seven to get a positive 14. Then I add these numbers down. Okay, take negative two times 14 to get negative 28. Okay, then add these numbers down to get negative 23. Then I take negative two times negative 23 for a positive 46. Then I add those numbers down to get 45. Okay, now, I now I've done I've done the entire process. And that's a relatively straightforward process, add down, multiply up. Once you have everything written out, it's add down, multiply up, which actually is kind of simple. Okay, but now from here, we need to rewrite our answer. Okay, now this number in the box over here, this one is unique. This is the remainder remainder. Okay, so if you remember from our long division example, our long division or excuse me, our polynomial long division example, our remainder gets written like this, I'm going to write this down here. Okay, we write as 45 over over what we're dividing by. Now, in this case, we are dividing by x plus two, okay, we're dividing by x plus two. Okay, so that's what our remainder is. Okay, now what I'm going to do is I'm going to actually write these numbers in reverse. Okay, now you'll see why I'm in the reverse here in just a moment. I'm going to take this 23. This is my constant number. So I don't have a variable with this. This is going to be negative or minus 23. Okay, then I'm going to move on. Now in my answer down here, the numbers are simply just going to increase the variables are going to increase as I go up to the left here. So this 14 is my linear terms, that's going to be plus 14x. Next is negative seven, this is going to be my quadratic term, which is negative seven squared. Next is three, this is going to be my cubic term of three x to the third. Okay, now notice the progression x to the third, x squared, x, no x's remainder. Okay, notice the progression down the line. And that right there, there's our answer. So you kind of realize why I did this backwards, I started with the remainder, and then worked my way up from having no x's to one x to two x's to three x's. Okay, alrighty. That is synthetic division, that's the whole process right there. Yeah, and next video that I do will be synthetic substitution, which actually uses this synthetic division process to do some substitution stuff, which is actually kind of neat. Anyway, that's it for this video, I hope you enjoyed it, and thank you for watching.