 Welcome back to our lecture series, Math 4230. I've checked out for two for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. So in our conversation for lecture 20, we're naturally gonna continue what we were talking about in lecture 19, which just as a reminder at the very end of lecture 19, the very last video, we proved that a polynomial ring whose coefficients come from a field always forms a Euclidean domain. Therefore, there is a Euclidean norm, and we claimed that this is the division algorithm, but it's somewhat suspicious to call the Euclidean norm, the division algorithm, if we don't have an algorithm, right? There should be a process, there should be steps that we should take. But in the proof, in the proof where we prove that a polynomial ring over a field is in fact the Euclidean domain, we actually were able to construct the quotient and remainder recursively. And this is in fact the division algorithm that we know from previous college classes, like college algebra is a setting where I would teach students about polynomial division math 1050. It also comes up in maybe intermediate algebra, math 1010, or high school algebra, whatever. But the algorithm that we use in these classes, such as math 1050, is exactly the algorithm that was embedded inside of the proof that a polynomial ring over a field is in fact a Euclidean domain. So let's talk about, let's be explicit about the algorithm now instead of more of the proofy approach to it, right? Let's take two polynomials, f and g, which belong to our polynomial ring, f adjoined x there, where again, f is a field. And let's suppose that g is not the zero polynomial, otherwise we really couldn't divide by it. And so then we proved that given f and g, there exist unique polynomials, q and r, that satisfy this equation f of x equals q of x times g of x plus r of x. And r is either the zero polynomial or it has a polynomial degree, strictly less than g. And so I should perhaps amend what I said earlier. When we prove that f of x here is in fact, a Euclidean domain, we didn't construct the Euclidean norm, we argued that the degree function was the Euclidean norm in that setting. And the process of doing so, we recursively were able to construct these polynomials, q and r. So let's be explicit in this example here. So what I want us to do is we're just gonna use the setting, we're gonna use the setting of, well, basically the rational number. So q would join x, like so. This would be a Euclidean domain by what we've seen before. We have the Euclidean algorithm, but the division algorithms course what we're focused on right here. And just to make life easier, I'm actually just gonna assume that my coefficients are actually just integers in this situation. Cause we will see by the end of this video, the factorizations of polynomials over the integers are actually the same thing as factorization of polynomials over the rational numbers. But for the moment be in this video, we'll assume these are rational polynomials cause they are. And so take the polynomial 6x squared minus 26x plus 12 and we're gonna divide it by the polynomial x minus four. And so the process that we developed in the proof was the following. I'm just gonna look at the leading terms for a moment, right? And I'm gonna ask myself, how many times does the leading term of the divisor divide into the leading term of the dividend in this situation? And in this situation, well, I mean, x is gonna go into x squared x many times and then the coefficient here is once that makes it a lot easier. So six divided by one is gonna be six. And so we can do that calculation, six squared, oh excuse me, six x squared over x is equal to six x like so. We then record that number on the top of our bar right here. That was the quantity we computed generally in our theorem there. And so what we did is we then took this number and we times the divisor by this partial quotient we have right here. Six x times x minus four is going to give you six x squared minus the 24x. So we can write it like this. So we have the six x squared minus 24x. That's just six x times x minus four. But then we took that quantity and we subtracted it from the polynomial. So we took, so again, this right here is just the six x times x minus four. We subtracted it from the polynomial above. So if I distribute that negative sign we end up with six, negative six x squared plus 24x. In this situation, because we chose this number perfectly, these leading terms are gonna cancel out. Six x squared minus six x squared gives you zero x squared, it's gone. Then we also have to consider negative 26x plus 24x that's gonna give us a negative two x and then we'll just bring down any other terms that are left. And so now the dividend has been reduced to a degree of, to a polynomial of strictly smaller degree. And this is then where the algorithm recurses itself. We then consider dividing this dividend by x minus four. The process repeats itself. How many times does x divide into negative two x for which we get negative two x over x? That's just gonna be negative two. We record that number up here. We then are gonna take that number negative two and times it by x minus four. That's gonna give us negative two x plus four but we're subtracting this, excuse me, plus eight, two times four. But we're just gonna subtract this from above. So if you subtract this, it becomes a positive two x minus eight like so. And then combining like terms, the leading terms cancel out again. You have a 12 minus eight, which leaves you a four. And this is now gonna give you the remainder here because notice four as a constant polynomial has a degree strictly smaller than x, which in this case it's a leading polynomial, its degree is one. So that means the division algorithm doesn't fact terminate in this situation. And so therefore we end up with the following. We have actually the formula we were looking for, f of x, which was six x squared minus 26x plus 12, it factors as the quotient q of x, which is six x minus two, times the original divisor x minus four plus the remainder four. And so just like we had promised, the fact that the polynomial ring f adjoined x is a Euclidean domain does in fact give us this division algorithm. And like I said before, I was doing this over the ring q adjoined x, but as we went through this entire process, we never actually used any fractions. It really was done inside of z. So is z adjoined x? Is it a Euclidean domain? And the answer of course is no, it's not. Because as we also know about Euclidean domains, Euclidean domains are principal ideal domains and z adjoined x is not a principal ideal domain. One example to see of course is if you take two and x, this is an ideal. So the ideal generate two and x, this is an ideal in the ring z adjoined x, but there is no single element that generates that entire thing. So it's not a principal ideal domain, so it can't be a Euclidean domain. But nonetheless, we can do division in this setting. Why is that? Well, because honestly when you do polynomial division, you might have to exit perhaps to the rational numbers, maybe you have to divide. But maybe you can fix that, maybe not. It turns out that division of course is related to factorization. And factorization over the integers is very similar to factorization over the rationales. Of course I'm talking about the polynomial ring and not the rings z and q themselves. And this is what we're gonna explore later on in this lecture. But what I wanna do first before I switch the slide here is I also wanna remind the viewer about this notion of synthetic division. Synthetic division is a trick that is commonly used in a setting like math 1050, college algebra, to help us factor larger degree polynomials. And synthetic division works in the following manner. What you do is you take your dividend and you write down all of its coefficients in descending order. So we'd say six minus 26 plus 12. And we don't need to put the plus, we'll just put 12. Like so. And if there was any terms that were missing, you put in a zero for placeholder. So it is important that you have this x squared spot, this x spot and this constant spot over here. All of those are necessary. You're gonna separate that with the bar. You're gonna draw another line here and then you're gonna record the number four up here. So you're dividing by x minus four and so you just record the four right here. The synthetic division only is applicable when your divisor has the form x minus something. You record that number here. Now the algorithm that we take in the place here is you're going to, whenever you see two numbers in a column, you add them together, which in this case is there's nothing down here. You're actually just adding a zero. That's how you initialize this. Six plus zero is just six. Then you're gonna take numbers on the bottom row and times it by this number right here. So four times six is 24, negative 20s. So then you add these together. Negative 26 plus 24 is negative two. You're gonna take negative two times four to give us a negative eight. And then you're gonna take 12 minus eight and that's gonna give you four. And this last number I'm gonna delineate it because it's the remainder. And then this right here gives us the quotient that we put back in the variables x. So we get six x minus two. And then we get that this is the quotient and then we have a remainder of four. So synthetic division can actually go through this process very, very quickly, a lot faster when your denominator is x minus a number C. Why does it even work though? I want you to, before I erase too much, right? I want you to see these numbers look strikingly similar, right? Notice what we had here. We ended up with, of course, the six right here, the 24, the negative two. Really I should write it more like this. We had this, whoops. Sorry about that, my four got deleted. We have a six right here, 24, a negative two, a negative eight and then a four, right? We're able to predict all these numbers. I forgot to mention the 12 as well. We could predict all these numbers and why does that work? So I'm gonna empty my tableau this time and think of the following. Okay, since the leading term here is just x, remember, this has to have the form x minus a constant. Since your first term is an x, how many times is it gonna go into the leading term? Well, it's gonna go on that, you're gonna reduce the power of x by one and then you're gonna have the exact same coefficient. So we just drop down the coefficient we started with. Then you're gonna take this six x right here, six times some power of x. You're gonna times this by that. You're gonna get six x times, excuse me, six times some power of x times x. That'll cancel out with the leading terms here. But then you're going to end up with this number, which in our case is four. You're gonna times it by the coefficient here, six. This is where the 24 comes from. Notice you have a 24, but shouldn't it be negative, right? Negative four times positive six x here. Oh, remember, you're subtracting it, right? So yes, six x times x minus four does in fact give you six x squared minus 24 x. But since you're subtracting it from above, when you distribute that negative sign, you end up with these terms canceling out, but then this becomes a positive 24, which is what we saw right here. You're then gonna add these numbers together because you're combining like terms. Negative 26x plus 24x ends up with the negative 2x, which is when we add these together, you end up with the negative two. We just aren't writing the powers of x because it's okay, but we get that negative two. So then we repeat this process. How many times does x divide into negative 2x? And this x could be any power of x potentially. Well, because again, this is a monic divisor, the leading coefficients one, all that's gonna happen is that the power of x will reduce by one, but the coefficient is gonna carry up here. Again, you might have some power of x, but it'll just be smaller. The coefficient's gonna be the same. And so when you take negative two times x minus four, this will give you the positive two because we're subtracting it, it cancels out the leading terms. But what happens to the other term? You're gonna take a negative four times a negative two, which gives you a positive eight, but you subtract it. So we end up with this negative eight right here. Notice a negative eight is just four times negative two. You end up with negative eight like so. And then we have to add these together. You end up with a four. And this process will continue to repeat itself until you've exhausted everything. And so synthetic division really is just capturing all of the information from long division. And this can be very, very useful, of course, when you're trying to divide by a linear factor of the form x minus four. We were doing this over the rational numbers, the rational field, but I want you to be aware that this makes sense for any, any polynomial ring with field coefficients. Because we can divide the scalars, we can divide the coefficients, we can divide these polynomials using this division algorithm. And we can always get our polynomial is equal to the divisor times a quotient plus a remainder. And then synthetic division also applies in the setting where you have x minus c of some kind. The algorithm is not gonna change. Now the arithmetic could change. Don't get me wrong, right? What does it mean to add two numbers together? What does it mean to multiply two numbers together? That's a very curious thing, of course. And so that's what one has to investigate in this situation. I also want to mention that synthetic division, of course, makes sense in any ring because, well, I should say any polynomial ring where the coefficients are somewhat irrelevant because when it comes to synthetic division, you add and you multiply, which every ring can do that. So whenever you're dividing by x minus four, excuse me, x minus c, you can use synthetic division with when it comes to, of course, to long division, you might need to actually have fractions but again, in a field that's a non-issue. So why am I talking so much about division and particularly synthetic division? What's the dealio? Well, I want to remind us about a very important result that a student often sees in a class like College Algebra Math 1050. And I want to convince us that this result is actually a general theorem of polynomial rings where again, f is a field. And so this is known as the remainder theorem. If f is a field and little f is a polynomial in the polynomial ring f of joint x, then if f of x is divided by x minus c, then the remainder f of c is actually the evaluation of the polynomial f of x at the number c itself. And so let's look at this to see how that happens. So this polynomial f of x with regard to the divisor x minus c, the division algorithm applies because we do have this Euclidean domain here. And so therefore there exists polynomials, unique polynomials q of x and r of x such that f of x equals x minus c times q of x plus r of x here. Now as the divisor is x minus c, the remainder here has to either be zero or it has to have degrees smaller than the divisor x minus c but the divisor is a linear polynomial. So the only thing smaller in degree would be degree zero. That is it has to be a constant polynomial. The zero polynomial of course is also a constant polynomial. So because r of x is a constant polynomial, we really can just call it a number. It will just call it r and not think of it as a polynomial anymore. And then I want us to consider what happens when we evaluate this function at x equals c. So at f of c here, we end up with c minus c times q of c plus r. Since r is constant, it doesn't depend on the c there. Now of course c minus c is zero and regardless of what q of c turns out to be, zero times q of c is equal to zero. And so zero plus r will give you r. And so this remainder theorem is a very critical important thing. f of c is just gonna equal r. And we did this over a polynomial ring with field coefficients but it does have to be field coefficients. Well, we just need to have a setting where this can happen, where there exists polynomials q of x and r of x such that this happens. Which that does happen over z of x as well. And this mostly has to do with the fact that it can be extended to its field of fractions like so. And so this remainder theorem has broad reaching scope. And so if you want to evaluate a polynomial, let's take the polynomial we saw a moment ago, right? Let's take f of x is equal to where did it go? It was 20, excuse me, 6x squared minus 26x plus 12. And so when we evaluate it at the number four, you're gonna end up with six times four squared minus 26 times four plus 12. Work through this thing here. We're gonna end up with four squared and of course it's 16. So we get six times 16. Over here we have 26 times four. If you don't know that up the top of your head, the nice thing about multiplication by four is you just double it twice, right? So if you double 26, you end up with 52 times two plus 12 right there. Here, what's gonna happen here is three times, three times 16 is 48. So you get two times 48. So then if we double that, we end up with 96. And then we're gonna subtract from that. We then have to take 52 times two, which is 104 plus 12, continuing on with this thing here. 96 take away 104. That is going to give us negative eight plus 12. That ends up with a positive four, like so, all right? Now that calculation wasn't too brutal, but I didn't even use a calculator, amazing, right? But you can see that with the exponents and such. This thing potentially get quite nasty if you have a much larger degree polynomial. But remember the synthetic division, okay? Six, negative 26, 12, you divide by four here. Well, by synthetic division, you bring down the six. Four times six is 24, minus 26 is negative two times eight, is negative eight plus 12 is four. And this is evidence of the remainder theorem, right? The evaluation is equal to the remainder of the polynomial divide by x minus four. For which when you look at synthetic division, I think most of us would agree that actually the synthetic division, you never have to deal with exponents whatsoever. It's just add, multiply, add, multiply, add, multiply, over and over and over and over again, right? And even the products never really got that big in comparison, right? You know, we had like a 96 over here and a 104 over here. But over here, the biggest number we ever had was negative 26, which was one of the coefficients. The remainder theorem does allow for us to much more efficiently compute evaluations of polynomials using division. And this is just the first of many applications we're gonna see of the remainder theorem. The factor theorem being a big one, which we'll talk about in the next video.