 Welcome back. Today, we're going to talk about solving polynomial equations. Solving polynomials using synthetic substitution and also identifying what's called multiplicity of a certain solution. We'll talk about that here in a minute. Now, some of your polynomial equations are too difficult to solve by using grouping. I do have a previous video on how to solve polynomial equations by using factoring. If you want to check that out, go ahead and do that. But some of those equations are actually too difficult to use by factoring. In this example, I'll show you that you actually cannot use factoring to solve this. And so instead of factoring, we're going to use something else to help us with that. We're actually going to use synthetic substitution. So here's the situation. You're trying to solve this polynomial. You're trying to solve it, set equal to 0. Now, your first instinct, since you're solving polynomial equations, your first instinct is always try to factor. So now that your first instinct is to try to factor by grouping. So in this case, I have my group here on the left and my group here on the right. This group here on the left, I notice they have an x squared as a common factor. So when I factor that, I get x squared and then x plus 6 is what I get after I factor. This here, I notice that they have a 4 in common. So I'm going to take a 4 out. That means I'm left with 3x plus 2. Hmm. OK, now I factored correctly. I know I did. But the idea behind factoring by grouping is that this group and this group should be the exact same. They should be the same, because then I factor out that group. I would be able to factor out this group and then I would have an x squared plus 4 left over. But this group here is not the same. So I can't factor that. There's no way I can factor that. OK, so now here's the situation. You try to factor by grouping, but it just doesn't work. So we've got to have another strategy. We've got to have another way of solving these polynomial equations. So that didn't work. I've got to try something else. So this is where synthetic substitution actually comes in. So you learn synthetic substitution way back when you talked about synthetic division, when you talked about long division, polynomial long division. So this is where synthetic substitution is useful. So now what I'm going to do is I'm actually going to look at this number 8 right here. That number 8, and actually this number 1 that's out here, this one that we usually don't write, but that's right out there, those numbers are actually going to help us to figure out what we can divide by. Now in this case, what we're going to do is we're going to use factors of 8. We're going to use factors of 8 to try to figure out what to use for synthetic substitution. So now I'm going to set up synthetic substitution so we know a little bit more of what I'm talking about. I have a number here, and then I write my coefficients right here. So my coefficients are 1, 6, 12, and 8. No gaps, 3, 2, 1, 0. No gaps. And then I set this up. So now that I have set up my synthetic substitution, I need to know what number I'm substituting in. This number right here that's going to the box that I multiply with, that number I need to figure out what that is. In this case, that number is going to be factors of 8, factors of 8. That's kind of an easy way to figure out what numbers we can use. Now factors of 8, there's actually a lot of factors of 8. To list them out really quickly, factors of 8 could be, well, 8, 4, 2, and 1. But then also positives and negatives. Whenever we think about these numbers, we always have to think about the positives and negatives. So actually, there's eight different factors of 8. And as you can well imagine, as the numbers get bigger, we get more and more factors. So I need to test out a couple of these factors to see what's going to work. Now, this is where, if you're really good with synthetic substitution, you can just try a few numbers to figure out what might work. In this case, I'm going to try 2. I'm going to try 2 to see what's going to work. So now, here we go. Now I'm going to see if 2 is going to work. So I'm going to bring down this 1. 1 times 2 is 2. So 6 plus 2 is 8. 2 times 8 is 16. Add that together to get 28. That times 2 is going to be 56. 56 plus 8 is 64. No, that doesn't work. How do I know that doesn't work? Remember, this box right here, that tells me that's a remainder. I have a remainder of 64. Now technically, what I'm doing is I'm dividing by x minus 2. I'm trying to figure out if that's a factor, and it's the positive 2 that I would plug in. So that means that x minus 2 has a remainder of 64, which tells me that that does not divide evenly. So I got to try again. So that one didn't work for me. So this 2 right here did not work for me. So I got to do this over again. So now, this is where we become really good at synthetic substitution. We do this faster. So instead of a positive 2, let's try subbing in a negative 2. So in this case, I'm testing to see if x plus 2 is a factor for this polynomial. So let's try this again. So bring down the 1, multiply here to get negative 2. Multiply that to get what is that, 4? No, yep, 4. Negative 2 times 4 is negative 8. Add that to get 4. Negative 2 times 4 is negative 8. Hold on, remainder of 0. Yes. What that tells me right there with a remainder of 0, I just found that x plus 2 is a factor of this polynomial, which means it divides evenly, which means that I can factor this. Now, I'm going to use this right here. This synthetic substitution is going to help me factor. So the first factor is x plus 2. This factor right here, using negative 2 for a synthetic substitution, told me that it divided evenly. Now, if I were to divide it, if I take this polynomial divided by x plus 2, this right here coincidentally is my answer. So not only am I checking to see if I don't have a remainder to see if it divides evenly, this also tells me what the answer is. Well, part of the answer. It tells me my next step in factoring. So this actually tells me that this is my squared. This is going to be x squared, then x, and then constant. So it's going to be x squared plus 4x plus 4. So this actually tells me this piece right here, I can factor out an x plus 2, and then this would be the remainder. Nice. All right, so that's why we can use synthetic substitution, because, well, heck, factoring by grouping just wasn't working, so I had to find something different. So I was looking at all these different factors. I tried to plug in a positive 2, didn't work, so I tried to plug in a negative 2, and it did work, which tells me what one of my factors is. So that factor is sitting right there, and that's going to help me to factor the rest of this. Now, this is actually manageable now. After that, notice here, x squared plus 4x plus 4, I can still factor that. That I can do in my head. I don't need another fancy process. I don't need to do another synthetic substitution to try to figure out this. So actually, I can factor this one more time. Let me actually bring this down a little bit farther. x plus 2, and then factor this. This is going to be numbers that multiply to get, let me set this up a little bit better, x is here, numbers that multiply to get 4 add to get 4. Well, that's just plus 2 and the x plus 2. Wow. Now notice that I have x plus 2, x plus 2, x plus 2. All these are the exact same, kind of interesting coincidence. But now notice there's actually two things we're doing here. Solving polynomial is using synthetic substitution and identifying something called multiplicity. Now the thing is, we have one solution here. In this case, x is equal to negative 2. Negative 2 works, negative 2 works, negative 2 works. I'm only going to write it once. Negative 2 has a multiplicity of, and you might be able to guess it, how many times does this negative 2 show up? It shows up three times. Now the multiplicity part, that tells you a lot about the graphing. In this video, I'm not going to go over with the graphing because I don't have enough room for all the graphs or any of that kind of stuff. But the multiplicity actually tells you a lot about what the graph is going to look like. Hopefully, your teacher at the time, if you're going through this, will show you using graphs what multiplicity means for a graph. Alrighty, that is just one example of solving polynomials using synthetic substitution. I do have another example that I'm going to use. So this video is going to be a little bit long. So here's my second example. Solving polynomials using synthetic substitution and identifying multiplicity. So this is just another example. I'm going to go through this again. So in this case, now I know, I'm going to save a little bit of time, I know that this is not going to work by factoring by grouping. So if I'm trying to solve this equation, I know that factoring by grouping is not going to work. So I'm going to go straight in, straight in, to using synthetic substitution. So I'm going to set that up right away. 4, 3, 2, notice x to the first. I don't have that, so I have a gap there. So when I set up my coefficients, I'm going to have a gap of 0. I don't have an x to the first power term. So that means I have a 0 there. You've got to put that in there. Now here's the trick. I've got to figure out what number to multiply by here. Now the number I'm going to multiply with here is going to be a factor of 27. So that's going to be numbers like 27, 9, 3, or 1, something like that. Let's start with something small. Let's just start with 3. Let's just start with a positive 3. Let's see if this one works or not. Now 3 or negative 3, those are pretty good numbers to start with. Or even 1, one might be easier to work with. But I want to start with 3. It doesn't really matter what you start with. Just try to start with something simple. And so if you make a mistake, if it's not the right factor, you can very quickly go back and change it up. OK, so let's start the process. Bring this one down. 3 times 1 is 3. 8 times 3 is, or see, 8 plus 3 is 11. That's going to be 33. Add that down to get 51. Multiply to get, what is that, 51 times 3 is 153. Add that down. It's not going to work. It's just not going to work. This could be, what is that, 459 or something like that. Yeah, it's not going to give me 0. It's not going to divide evenly. So right away, I can tell it's not going to work. So let's try something different. In this case, I'm going to try negative 3. What happens if I use negative 3? All right, so bring the one down. Multiply here to get negative 3. That's going to multiply to get 5. Multiply is to get negative 15. Adds to get 3. Multiply is to get negative 9. Adds to get negative 9. Multiply is to get 27. Oh, look you there. Just by changing the negative on it, I found something that actually divides evenly. So this actually tells me that x plus 3, remember when we got to change the sign there, x plus 3 is going to be a factor for, is going to be one of my factors. So x plus 3, and then this right here, tells me after I take this x plus 3 and factor it out, this tells me what the remainder is going to be, what's going to be in my second parentheses over here. So in this case, constant linear quadratic cubic. So this is going to be x to the third, 1x to the third, plus 5x squared plus 3x minus 9. And of course, equals 0. All righty. So that's one step of using the synthetic substitution. Now, here's the thing. As I look over here with my previous example, I looked at this parentheses, and I was able to look at it and just factor it. That's not the case now. I can't just look at this and factor it. Can I use factoring by grouping? I might be able to, but I don't want to mix up my different processes together. So you know what? I'm just going to do synthetic substitution again. I'm just going to use it again. So in this case, what I'm going to do is I'm going to try to figure out, take this piece of it. So I'm going to use a different color here. I'm going to take this piece of it right here, and I'm going to see what is going to divide evenly into that. So I'm going to do my work down here. So now, as we can see, the higher up you get in level of mathematics, the more organized you have to be. Because now I have 1, 5, 3, and negative 9. And actually, you can get those numbers from right there. But the thing is I want to figure out what factors to use. And so now what I'm looking at is I'm looking at negative 9. Factors of 9 are going to be 9 and 3 and that kind of thing. So you know what? Heck, let's just use this one again. Let's see if x plus 3 is going to be another factor. Let's just try it. I mean, it's a legitimate try. I can use a negative 3, a positive 3, negative 9, positive 9, negative 1, positive 1. I mean, any of them will work. So let's see if this is going to work. Bring the 1 down, multiply here to get negative 3. Add that to get 2. Multiply to get negative 6. Add that to get negative 3. Multiply to get positive 9. Oh, look at that. First try. Good job. First try. All right, so I know that that divides evenly. So here we go. x plus 3 is yet again another factor. So now I'm going to take this, bring it down. x plus 3 was the first factor. x plus 3 is yet again another factor. And then this part has been factored out. The x plus 3, here's the remainder right here of x squared, a 1x squared, plus 2x minus 3 equals 0. There we go. All righty. And you know what? I think I can factor this one more time. So I use synthetic substitution to find my first factor. This second part here, I use synthetic substitution to find it yet again another factor, which coincidentally happened to be the same as the first. Now over here, I'm going to factor this one more time. Now this factoring right here is actually going to be easier than all the rest. Now x plus 3 is a factor. x plus 3 is a factor. And then this right here, I could actually factor just your good old fashioned two parentheses factoring, numbers that multiply to negative 3 and add to a positive 2 are positive 3 and negative 1. So that factoring is actually easier. I didn't have to do my synthetic substitution again. So that saves me a little bit of time, a little bit of effort. Now everything is completely factored. And again, what am I doing here? I am solving. I'm trying to solve this equation up here. So now what I do is my two solutions are my solutions. Actually, oh, look at this. I have 3, 3, and 3. So I have x is equal to negative 3. And I have x is equal to 1. Those are my two solutions. But negative 3 has a multiplicity of how many times? 1, 2, 3, multiplicity of 3. So negative 3 actually has a multiplicity of 3, whereas x equals 1. It doesn't have a multiplicity. It just happens. It just occurs once. Awesome. So that's another problem of solving polynomials using synthetic substitution. I'm going to try to cut this short. Thank you for watching the video. And we'll see you next time.