 In the previous video, we talked about how to do long division with polynomials. In this video, we're going to talk about how to do synthetic division with polynomials instead. So synthetic division is nice because it is a little bit shorter version of long division, but it only works if you're dividing by x minus k, or x plus k. The key is that there can't be a coefficient in front of the x other than one, and it can't be x to any higher power at all. So this example here will work. We're dividing 3x to the fourth plus 2x cubed minus x minus one by x plus four, and this actually is the same problem that we did at the end of the video of the previous video. So when you're dividing using synthetic division, I always make a little box here. So there's a few different ways to do it, but this is just how I usually set it up. And in this box goes your k value. So k is whatever we're subtracting from x. In this case, x plus four, you could think of that as x minus a negative four. So in the box, we'll put a negative four. And then next to that box in a row, we're going to put all of the coefficients of these terms here. So starting with 3x to the fourth, we'll start just listening to three, and then we'll go down to the x cubes. We have 2x cubes, and then this is where you have to be careful. We don't have any x squared, so we have to put a zero there. You need a column for each degree. So this is like our x to the fourth column. This is our x cubed. These are our x squareds. Then we go to our x's. We have a negative 1x, and then the constant, which is again a negative one. So you list those coefficients in order, making sure not to skip any. Okay, then leave a little space, and what we'll do is to complete the synthetic division. First, always bring your leading term down. That will always stay the same as a three in this case. Once you have a term down here, you'll multiply it by the term in the box. So 3 multiplied by negative 4 is a negative 12, and we write that in the space that we've left. And this is where it's different from the long division. We are going to add every time. So 2 plus a negative 12 is a negative 10, and then we multiply again by the number in the box. 40. And we add. 0 plus 40 is 40. 40 multiplied by a negative 4 is a negative 160. And again, we add going down. It's negative 161. And then we multiply by the number in the box. Negative 161 multiplied by 4, or by a negative 4, is 644. And then we add a negative 1 to that, and we get 643. Now, if you are noticing this, it's pretty similar numbers, actually all similar numbers to what we had before. So we've essentially just done the long division, but just without all of the writing out of each term. So what happens here is this number at the end, this is always the remainder. And then if you work backwards from there, that is going to be your quotient right here. So each of these terms, they're just dropping one degree as you go. So this was an x. This is now going to be your constant. This will now be an x, then an x squared, and then an x cubed as we go back. So the constant is negative 161, then we add in 1x as we work backwards. So that 40 goes with an x, the negative 10 will go with an x squared, and then the 3 will go with an x cubed. So each time your answer should have a degree that is one less than the degree that you started with, which in this case was x to the fourth. So this is the quotient, that's the remainder, 643, if we wanted to write it all together, we could write our final answer as this. Okay, let's look at one more example of synthetic division because it will help us save some time as we're doing this division when we're factoring. Okay, in this case we're going to make our box to start. Notice it works because we are only dividing by x minus 3, and then in the box we will put a 3. Then following the box we'll put just the coefficients, and here if you look it doesn't skip a degree at all. So we can just go 1, 3, negative 2, 6, negative 1. Leave some space, your leading coefficient will always stay the same. Then multiply by the number in the box, 1 multiplied by 3 is 3, and each time we're adding here. 3 plus 3 is 6, 6 multiplied by 3 is 18, we add, get 16, 3 multiplied by 16 is 48, we add and multiply. 54 multiplied by 3 is 162, and when we add here we get 161. So this last number that we get is again our remainder, the 161. And then as we go backwards we're going to add in an x each time. So 54 will be the constant, the 16 will go with an x, 6 will go with an x squared, and then 1 will go with an x cubed. This here is the quotient, the 161 is the remainder, and we can write our final answer like that. Now one special property, it's called the remainder theorem that works well in order to check kind of the remainders and check your work in these cases. It says that if we are dividing by x minus k, we can plug that k value into this equation and that can give us the remainder. So in this case, let's call this function f of x. If we were to plug a 3, since we're dividing by x minus 3 into that equation, we should get 161. So what I'm going to do is I'll plug into my calculator 3 to the fourth plus 3 times 3 cubed minus 2 times 3 squared plus 6 times 3 minus 1. And when I do that I get 161. And that notice is what we had for the remainder. So anytime you divide by x minus a number, or x plus a number, you can plug that number in to your equation that you're dividing in order to find the remainder. And I find that that's a great property to use just to check your work as you're working through some of these difficult problems. But there is just a quick explanation of how to use synthetic division with polynomials.