 All right, welcome back. In this video, I'm going to talk about how to determine if a binomial, right up here, I'm going to determine if a binomial divides evenly into a polynomial. Now, what I'm going to be doing in this video is by using synthetic substitution, we are going to quickly determine if a binomial divides evenly into a polynomial. Now, why would we even need to know this? This is a process that we use for solving polynomials, for graphing. There's a couple of different uses for this. So anyway, the question that we have here is, is the binomial a factor of the polynomial? If it's a factor, that means it's going to divide evenly. So that's what I want to do. I want to see if x squared minus 3x plus 1, is it divisible by x plus 1? Is x plus 1 going to be a factor of this polynomial? And I can actually quickly figure that out using the synthetic substitution process. So here we go. What I'm going to do is inside the box, the number I'm going to multiply with is actually a negative 1. Remember, we use the opposite of this number here. And then the coefficients are the coefficients here, which are going to be 1, negative 3, and 1. And then I have a little gap here for some numbers I'm going to be adding with. And then right below my last number, I put this little box. That's where the remainder is going to go. Now, and that's the thing, is that if a polynomial divides evenly, there will not be a remainder. Just like dividing any other number. If I take 4 divided by 2, I get 2 with no remainder. But if I take 5 divided by 2, I get 2 and a remainder of 1 half. 2 goes into 5 twice with a remainder of 1. That doesn't divide evenly. This one here does divide evenly. So just for a very basic example. Anyway, now I want to go through this synthetic substitution process. So I'll bring this one down. Negative 1 times 1 is a negative 1. Add this down to get negative 4. Multiply here. Bring that up to get 5. Add that down to get 6. So what it looks like here is that I have a remainder of 6. Now, right there, that tells you that x plus 1 is not, I repeat, is not a factor. So that's a big n, oh, big no. All right, so that's basically what we're looking at here is that this x plus 1 does not divide evenly in this polynomial. So this is something later on when we use this for solving. This x plus 1 would not help me to solve. It will not help me to grab. It won't help me do any of that. So now let's do this for the next one down here. Next one down here, let me change colors here. So now what I want to do is I want to use synthetic substitution to see if this works. Now notice, x to the fourth, x to the third, x to the first. I have a gap here. I don't have an x squared there. So when I set up my synthetic substitution, I've got to remember that I'm going to have a gap there. So inside the box here, I'm going to have negative 2. That's the opposite of this number here. So I've got a negative 2 there. Then I have my coefficients of 3 and 6 and 0. Don't forget the gap, and negative 5 and negative 10. This one's a little bit longer than the last one. It's going to take a little bit more. But again, the process still remains the same. So what I'm going to do is add down, multiply up. Add down, or, yep, add down, multiply up, add down, and multiply up. Add down, a lot of zeros there. I'm going to double check this when I get the chance. And multiply up and add down. Holy cow, I've never seen so many zeros. OK, let's double check this. 3's and 6 and my gap of 0 and then negative 5 and then negative 10. So I've got those numbers right. I'm multiplying by a negative 2. Got that right. Bring the 3 down, negative 6. Add that to get 0. That sounds 0. Adds to get 0. Multiply this to 0. That's negative 5. That's going to be a positive 10. Yeah, that's right. OK, I just wanted to double check because this is actually the first time I've worked it, to be quite honest. And I wasn't sure if I was going to get that or not. But anyway, what does this mean? What are we trying to figure out? This tells me right here, I have a remainder of 0. A remainder of 0. So what that actually tells me is that x plus 2 is a factor. That means it is a factor of this polynomial here. x plus 2 is a factor. That means that I can divide by it. That means it's going to help me with the graphing. That means it divides evenly so that when I factor and do some solving, I can actually use x plus 2. So that's what you can use. That's using a synthetic substitution. There we go. This is using synthetic substitution to see if a binomial divides evenly into a polynomial. And there's just a couple of quick examples of that. All righty. Hope you enjoyed the video, and we'll see you next time.