 Well, class, welcome to episode 12. This episode is on the fundamental theorem of algebra, and it's sort of a continuation of what we saw in episode 11. I'm Dennis Allison, and I teach mathematics in the math department at UVSC. I'd like to begin today with our objectives. We ought to actually should go to the objectives. We want to look at the fundamental theorem of algebra, and then we'll look at complex roots of a polynomial, complex number of roots. Then we'll look at the so-called complete factorization of a polynomial. And finally, we'll look at a cubic formula. So I'd like to begin today with an example to sort of tie this episode in with the last episode. Let's take a look at this polynomial that I have on the green screen. I'd like to factor the polynomial, negative x cubed plus 7x squared minus 15x plus 9. And this is very similar to some problems we worked in the last episode. If you remember, we use a rule and we use a theorem. The rule is Descartes' rule of signs. And how does that go? Who can remind us? Descartes' rule of signs. Steven? The number of sign changes in the polynomial equals the number of positive roots. Right. So we have a sign change here. We have a sign change here. And we have a sign change there. We go from negative to positive, positive to negative, and then negative to positive. So three sign changes, what does that tell me about the roots of this polynomial? There are either three or one positive roots. Right. So over here on the side, I'll say for positive roots, there are either three or one. And of course, we're referring to real number roots. When we say positive or negative, those are real numbers we're talking about. And how do I find out the possible number of negative roots? Substitute negative x in for x. Right. Substitute in negative x for x. So if I substitute in a negative x, we'll count the sign changes after this. So this will be minus negative x quantity cubed plus 7 times negative x squared minus 15 times negative x plus 9. And when we reduce this, this will be a positive x cubed and a positive 7x squared and a positive 15x and a positive 9. How many sign changes do we have there? Zero. We have no sign changes. And therefore, there are no negative roots. Now, that's particularly helpful because now I know I don't need to check any of the negative rational roots that we're going to come up with in just a moment. So up here, for negative roots, I'll say there are none because there are no sign changes. Are there any questions about that? Does everyone agree? Yeah. On your positive roots, how did you come up with it? See, they're going to be three or one. Three or one? Well, what we do is we count the sign changes. There were three sign changes up here. And then we take either that number or we reduce it by two. So either three or one. And we can't reduce it by two more would go into a negative number of positive roots. You have to have at least zero or larger positive roots. Why did you choose to reduce it by two? Because the positive roots come in pairs. So if you remove one positive root, you have to take them out two at a time. OK, now that we know, now that we have an idea anyway of what the roots look like, we go to the rational root theorem, which says that we look at the divisors of 9 and we call that little p. And we look at the divisors of the coefficient negative 1 on the highest term. We call those divisors q. And then I'll look at the ratio p over q. And I'll put the p over q over here on the side out of the way. Well, p has to be a divisor of 9. That would be plus or minus 1, plus or minus 3, or plus or minus 9. And what would be the choices for q? What are the divisors of negative 1? Negative 1? Well, plus or minus 1, actually. So because either 1 or negative 1, either 1 divides those. We're talking about integer divisors, not fractional divisors. So if I take my choices p over q, I could have plus or minus 1. That's where I take plus or minus 1 over a plus or minus 1. And then p over q, 3 over 1. That's plus or minus 3. And then plus or minus 9. But from what we just learned up here about the roots using Descartes's rule of signs, there are no negative roots. So I know that I can actually reduce this list to only the positive choices, 1, 3, and 9. So I really only have three numbers to check. And if there is a rational number, that is a fraction or an integer, that will make this polynomial 0, it has to be one of these three integers. Any other root would have to be an irrational number or maybe a complex number. Well, let's check these one at a time. I want to find out if 1 is a root of this polynomial. And that's equivalent to saying, let me write that question down, is 1 a root or a 0 of p of x? And there's an equivalent question I could ask. And that is, is x minus 1 a factor of p of x? And the reason I know these two statements would be equivalent to each other is by the factor theorem from a couple episodes back. So to find out if x minus 1 is a factor or not, I'm going to divide by x minus 1 into the polynomial. And if I divide by x minus 1, I'll use synthetic division and I put these coefficients inside. And I'm going to divide by x minus 1 outside. But you remember if I throw away the x and call it a negative 1, and then if I change the sign to a plus 1, I can actually add out here. So we're just sort of reviewing some of the concepts we've seen earlier. So it ends up that I'm actually going to be dividing by the very same number, plus 1. And I bring down the negative 1 and I multiply and I get negative 1 and I add and I get 6. And then I multiply. 1 times 6 is 6. And I add and I get negative 9. And then I multiply 1 times negative 9 is 0. So what that tells me is x minus 1 is a factor because I got a remainder of 0. And in fact, p of x can now be factored as, let's see, it's going to be x minus 1 times what polynomial? Negative 1x squared. Negative x squared, OK. Plus 6x. Plus 6x. Minus 9. Minus 9, very good. Now let me just erase this synthetic division and I'll move this back up above there so we can keep the factorization together. And this is x minus 1 times negative x squared plus 6x minus minus 9. And in order to factor this, I think what we should do is maybe factor out the negative 1 to make that a positive x squared term. So negative the quantity x minus 1 times x squared minus 6x plus 9. And now this trinomial factors again. Perfect squares. It's a perfect square. Yes, how does that factor? X minus 3 squared. X minus 3 squared. And so all together I would say that the roots of this polynomial are 1, 3, and then 3 again. I'll list it twice because it came from two factors. And you see those were two of the numbers over here that I was going to check out of the three. And this is the factorization of the polynomial. So p of x factors into negative 1 times x minus 1 times x minus 3 squared. And I can use this information to now draw the graph. So let's just put the graph over here on the side. And if I want to graph this polynomial, I'll go over to 3 and I'll put a dot there. And I'll put a 1 right there. And who can tell me what the graph will look like? Should I come in from above toward 3 or should I come in from below? From below. From below, exactly. Stephen, why should I come in from underneath? Because the x cubed term has a negative on it. Yeah, you see this was a negative x cube. And so that tells me that this graph has been inverted. And rather than going up on the right, it goes down on the right. So my graph is going to come down like this. And when I come to 3, what's going to happen? Will I cross at 3 or will I turn and come back? Turn and come back. Turn and come back because this has multiplicity 2, which means this looks something like a parabola right here at 3. It's going to turn. It's rounded right there. It comes back down. We don't really know exactly how far down it would to go. We're just making a rough sketch. But eventually, I'll have to turn and come back toward 1. And at 1, what's going to happen? Will it pass through or will it turn around? It'll pass through. It'll pass through because this has multiplicity 1. So it passes through and this graph now goes up on the left-hand side. Could anyone tell me what the y-intercept is right here where this function crosses the y-axis? 9. It's 9. Yeah, Susan, how did you know that? Plug 0 in for x from the original function. Exactly. The way you find y-intercepts is you let x be 0. And if I plug in 0 for x, Susan just did that in her head. She got 0 plus 0 minus 0 plus 9 is 9. So this is actually 9 on the y-axis. And you can see that my scale obviously must be different on the y-axis than it is on the x-axis. And I wouldn't guarantee that there's any accuracy in how low this point is right here. It could actually go lower or maybe not quite so low. But this is the general shape of the polynomial function p. So I'll just put a big p beside it to say that's the graph of p. So you see what we're doing is we're taking polynomials that are of larger degree than we're used to factoring. And we're using Descartes rule of signs and the rational root theorem to find factorizations to get the roots. And the roots are the x-intercepts. I plot the x-intercepts and I use multiplicity so that I can draw their graphs. Now, you know, in the process of doing problems like this, I think it's easy to lose track of some real shortcuts that we may overlook in the middle of all these theorems and rules. So let me show you a couple problems that have shortcuts that we might otherwise overlook if we were only thinking about the rational root theorem and the rule of signs. Take, for example, this polynomial. I think I'll call this one f of x. And let's say this is x to the fourth minus 5x squared plus 4. Now, I'm thinking rather than counting the number of possible positive and negative roots this may have, I think we can actually factor this by sight because this is in a quadratic form. And you see, if I think of x squared as my basic term, this is the square of x squared. So I could put an x squared here and an x squared here. And I don't think we need the rational root theorem to factor this. Does anyone see how to complete that factorization? Minus 4 and minus 1. Minus 4 and minus 1. Yeah, so rather than looking for sign changes and the choices for little p and little q, let's just factor this thing if we can do it. And x squared minus 4 factors as the difference of two squares in the x plus 2 and x minus 2. And x squared minus 1 factors in the x plus 1 and x minus 1. And therefore, what are the roots of this polynomial? Plus and minus 2 and plus and minus 1. Right, plus or minus 2, plus or minus 1. Those are exactly the values we would have come up with if we had used the rational root theorem. But this is a lot quicker. This is a lot quicker, absolutely. So we don't want to forget about the easy ways to factor polynomials when we're thinking about these other methods. Let's go ahead and graph this while we're here, by the way. I'm going to put an x intercept at 2 and at 1 and at negative 1 and at negative 2. And because this is a positive x to the fourth, I know my graph comes in from up above. And because each one of these roots has multiplicity 1, every one of those factors is different. I know that every time this function passes through, it's intercept. And it looks like this. This is the graph of F. I would guess that the graph has a higher maximum between negative 1 and 1 than it does have a minimum between 1 and 2 or negative 2 and negative 1. Why would you guess that point would be higher than these are low? Because the space in between it is bigger than that. Yeah, just sort of from a layman's point of view, because the space is wider, we have more time to wander away before we come back. So I would guess that it will go higher there than it goes down here. Can anyone tell me what that y-intercept is right there? Four. It's four. Yeah, it's the constant term on the end. Because if you let x be 0, you get four. Looking back at this function, I noticed this is an even power, x to the fourth. This is an even power, x squared. And this is an even power, because we could think of it as being x to the 0 power. That's an even power. And when you add or subtract multiples of even powers of x, what kind of a function do you get over here? An even. You get an even function. Yeah, and it's been a while since we talked about even functions. But an even function is symmetric about the y-axis. And you notice, if you were to fold this graph across the y-axis, it would be perfectly symmetrical. And so as soon as I see the graph on this side, I know the graph is just the mirror image on the other side. So using properties of symmetry, it helps me to see what this graph would look like. OK, one more example. This is another one that I think we could actually factor without going to the rational root theorem. And I'll call this g of x. And this is the function 2x cubed minus 3x squared minus 2x plus 3. Now, if we were not in our toes, we'd probably say, let's count the sign changes using Descartes' rule of signs and see what would be the possible number of positive and negative real roots, then look for the rational roots, but using the rational root theorem. But I think we can actually factor this. Does anyone see a way to factor this? Grouping. Factor you by grouping, exactly. So if we group these two terms together and group these two terms together, that's including the negative, so I'll make that a little bit wider. In the first pair of terms, what can I factor out? x. More than x. 2x. 2x. Not 2x. x squared. Well, an x and another x would be an x squared. Yeah, so it's x squared times 2x minus 3. And what can I factor out of the second pair? Negative 1. A negative 1. And then that gives me a 2x minus 3. You see, my goal is to try to produce the same factor in both terms there. And now I do have a common factor, so factoring by grouping allows me to express this as 2x minus 3 times what? x squared minus 1. x squared minus 1, right. And then we can factor that further. So we have 2x minus 3 times x plus 1 times x minus 1. And we got our three roots, but we didn't have to appeal to the rational root theorem, because this problem was a rather rare one where we could factor by grouping. In this case, the roots are, let's see, what would be the root for this one? 3 halfs. Yeah, you see, the way you find the root, technically, is you set it equal to 0 and you solve for x. And x is equal to 3 halves. So the number that makes this 0 is 3 halves, which means this graph has an x-intercept at 3 halves. And then the roots for these two factors are negative 1 and plus 1, so I'll just say plus or minus 1. And then if I draw this graph, I think we can squeeze it in over here, I'm going to locate plus 1, minus 1, and plus 3 halves. Now that's got to kind of squeeze that in between the 1 and the 2. And this was a positive x cubed. Therefore, my graph is going to be going up on the right-hand side. And each root has multiplicity 1. So it's going to pass through each x-intercept. And this is the general shape of the graph. The highest point is, well, I should say the y-intercept is at 3, but I don't really know that that's the highest point. In fact, I wouldn't be surprised if this graph doesn't actually look more like this. So that the highest point is actually not quite on the y-axis, but maybe skewed to one side or the other. You could check this on a graphing calculator to find out. But we wouldn't expect that there would be perfect symmetry between the negative 1 and the plus 1. So it may actually have its maximum value just off the y-axis. Now, on an exam, when you drew this graph, if you put its maximum right at 3, I wouldn't count off for that. Because without further investigation, you're just trying to sketch a rough sketch of the graph. And that would be sufficient for me. Now, in all the examples that I've been considering, the coefficients that I've been putting on my variables have been integers, like a 2, a negative 3, a negative 2, and a 3. What if I had put fractions as a coefficient? Suppose we had something like this. f of x equals 1 half x cubed plus 1 third x squared plus 5 sixth x plus 1. Now, of course, this makes the problem look much more complicated, but I can get it down to integers if I factor out the fraction with the least common denominator, namely 1 sixth. If I factor out a 1 sixth, this would be 3 x cubed, because 3 sixth is a half. What would the next term be? 2 x squared. 2 x squared. And the next term would be? 5 x. 5 x. And the last term would be? 6. So if I wanted to graph this polynomial with the fractional coefficients, what I would do would be to graph this polynomial, just as I have been graphing the others. And then what does the 1 sixth do to the graph? Does it squash it? It's going to sort of squash it. Yeah, so when I draw it, I won't make it quite so elevated and wouldn't make the valleys quite so low. It would be a little bit calmer near the x-axis. So if you put fractions in here, it's not really a problem in going about graphing it. But now let's consider yet another possibility. What if I have a polynomial function, but this time I use a complex number as a coefficient? Suppose we have x squared plus ix plus 2. Now here's what's different about this problem is I have a complex number, in this case an imaginary number, as a coefficient for the polynomial. Now I wouldn't be able to graph this on the x-y plane because I can only graph real numbers on the x-y plane. And I have complex numbers introduced here. But if I wanted to find the roots of this polynomial, I could use the quadratic formula. Using the quadratic formula, let's see, x equals negative b plus or minus the square root of b squared minus 4ac all over 2a. What is the value of a in this problem? 1. What's the value of b? i. i, and the value of c is 2. And if I substitute those in, I get negative i plus or minus the square root of i squared minus 4 times 1 times 2. All over 2a, I left off an a there. All over 2a, which is 2 times 1. If I reduce this radical, what is i squared? Negative 1. And here I have negative 8. So we have negative 1 minus 8 all over 2. And if I simplify that, I have negative i plus or minus the square root of negative 9 over 2. What are we going to do with the square root of negative 9? Pull out an i. We can pull out i. Yeah, you see, we could write this as minus i plus or minus the square root of 9 times the square root of negative 1 over 2. I'm just separating the negative 1 from the 9. And how much is that when I reduce it? 3i. 3i. So we have negative i plus or minus 3i over 2. Does everyone agree with that? Now I actually have two answers there. One answer, one root, I'll call it root number 1 or x number 1 is negative i plus 3i over 2, which reduces to how much? i. i. And another root, I'll call it x number 2, is negative i minus 3i over 2, which is how much? Negative 2i. Negative 2i. So you see, I do have two roots for this quadratic function. But both of the roots in this case turn out to be imaginary numbers. And if I were to substitute either one of those numbers in to my original equation back here, either one of those should satisfy it. Why don't we just try checking the i just to see that it does? And the purpose of this example is to show you that the polynomial functions that we could deal with are now even larger, it's an even larger class of functions, because we can now use complex and imaginary coefficients. Is there an imaginary graph that you could? There isn't on the x, y coordinate plane because, well, in this case, we have two roots, i and negative 2i. And there's no place to locate them on the real number line on the x-axis. You could locate 1 and negative 2, but not 1i and negative 2i. So there's no coordinate plane that maybe wouldn't be x, y that you could graph? Well, as a matter of fact, in a course called complex variables, there is a way that you can do something like this, but we won't get into that in this course. So right now, all we can do is just find the roots of the polynomial without actually drawing a graph of it. If I substitute in i into this function, I'm going to get i squared plus i times i plus 2. Now, does that make it 0? Let's see. Well, what is i squared? Negative 1. Negative 1. And here's another i squared. Negative 1 plus 2 is 0. Yeah, so that's exactly what we were expecting to happen. And if you substitute in a negative 2i, you'll get 0 also. So those are the two roots. So if I were to put complex coefficients, complex or imaginary coefficients on the variables, I may get complex or imaginary roots for the problem. Well, this leads us to a theorem that is on the next graphic. Let's bring up on the screen. And this is called the Fundamental Theorem of Algebra. This says that every polynomial of degree n, 1, or larger, with complex coefficients, sort of like what we just saw a moment ago, has at least one root. And that root may be a complex or an imaginary number, just as it could be a real number. Now, there's a generalization of this, which is in the next graphic that's called the Complete Factorization Theorem. And it says every polynomial p of x of degree n, 1, or larger with complex coefficients has n roots, which we might call c1, c2, c3, up to c sub n. So that p of x is equal to a times x minus c1 times x minus c2 times x minus c3 out to the factor x minus c sub n for some constant a. Now, let's just discuss those two theorems for just a moment. Suppose I have a polynomial with complex coefficients. Let's say my function is 1 plus i times x squared minus 3i times x minus 1 minus i. This would be a quadratic function with complex coefficients. This is certainly complex. This is complex. Negative 3i is considered a complex number because it's actually 0 minus 3i. So it has a real and an imaginary component. But it's also just an imaginary number. And according to the theorem, the fundamental theorem of algebra, this has at least one complex root. Well, let's see what that complex root would be. If I solve for x, I'll get negative b, which is the negative of negative 3i, plus or minus the square root of b squared, that would be negative 3i squared, minus 4 times a times c. Now, c has a negative in there, so that's going to be negative the quantity 1 minus i. All over 2a, so that's all over 2 times 1 plus i. If I reduce that, my first expression is just a 3i, plus or minus the square root. What would be the square of negative 3i? Negative 9. OK, I'm going to write that as 9i squared, and then I'll make it negative 9 on the next step. 9i squared, there's a negative in the parentheses here. I'm going to bring out and make that a plus 4 times 1 plus i and 1 minus i. And this is all over 2 times 1 plus i. Now, if I reduce what's under the radical, I think it's going to be quite a bit more compact. 3i plus or minus the square root of, this is negative 9, right here, negative 9, plus 4 times. Now, how do you multiply 1 plus i and 1 minus i? I'm thinking that's the difference of two squares. So when you multiply, what do you get? You'll get negative 2, won't you? You get 1 minus i squared. You get 1 squared minus i squared. a plus b times a minus b is a squared minus b squared. And I'll come back and reduce that in a moment. Over 2 times 1 plus i. And that will be 3i plus or minus the square root of negative 9 plus 4 times 2. I think that'll be a 2 right there. 1 minus a negative 1 makes it a 2. Over 2 times 1 plus i. And this is 3i plus or minus the square root of negative 1 over 2 times 1 plus i. Well, of course, the square root of negative 1 is i. So this is 3i plus or minus i over 2 times 1 plus i. So this is sort of an exercise in patience in working out all this algebra. So I actually have two roots. So let's separate the two roots. One of the roots I'll call x1 is 3i plus i over 2 times the quantity 1 plus i. That's going to be 4i over 2 times 1 plus i. So let's just make that 2i over 1 plus i. What I've done is to cancel off a 2 on top and bottom. The other root is x2 equals 3i minus i over 2 times 1 plus i. And if I cancel off a 2 on top and bottom, what will this be? Just 1 over 1 plus i. Just i over 1 plus i. Now, a person might argue, well, Dennis, this is not a complex number. This is a quotient of two complex numbers. So what I'd like to do is now convert these both into complex numbers as such. You remember the fundamental theorem said if you have a polynomial with complex coefficients, then you will have at least one complex root. I think these are both complex roots. So let's see what the complex roots are. I'm going to erase this middle portion and leave what's down below. And let's find out what root, what complex number this is. x1 is 2i over 1 plus i. Now, if I want to make this into a standard form complex number, what I do is multiply by the conjugate on top and bottom. And the conjugate is 1 minus i over 1 minus i. So I'm putting the conjugate on top and bottom. And what I do is just change the middle sign there in front of the i term. So when I multiply on top, I have 2i minus 2i squared. And on the bottom, I have the difference of two squares, 1 minus i squared. And if I reduce the denominator, what will that be? That'll be a 2. And if I reduce the numerator, how much is negative 2i squared? 2? No, 4. Actually, it's a plus 2 because it's negative 2 times negative 1 is plus 2. I'm going to put that in front. And then add the 2i on the back. So I'm just moving this guy over to the back side. That way, I have my real and my imaginary component in what we call standard ordering. So if I divide by 2, I get 1 plus i. So the number that I had down here that I called x1 as this ratio is actually just 1 plus i. And what is x2? Well, if I reduce i over 1 plus i, what I'll do is here is multiply by the conjugate on the top and bottom, just like I did up above. And the numerator is going to be i minus i squared over 2. 1 plus i times 1 minus i is 1 minus i squared or 2. And that's going to be, let's see, that'll be a plus 1 plus i over 2. And that's the complex number 1 half plus 1 half i. So the original function has two complex roots. It has 1 plus i and 1 half plus 1 half i. Now the fundamental theorem of algebra said there would be at least one complex root. But then in the second theorem called the complete factorization theorem, it says you will have as many complex roots as you have the degree on the polynomial. This is the second degree polynomial, so I get two roots. Sometimes I get the same root twice, and then I still count that as two roots. They just happen to be a multiplicity of multiplicity, too. So it is possible that you can have complex roots for a polynomial that has real or complex coefficients. And of course, I think a very reasonable question is, who would ever use this information? I mean, why are you telling us all this? Well, if you remember, in the first days of this course, we said that this is a surface course for many different disciplines. And those students who are going into engineering, physics, mathematics, and let's say computer science, computer engineering, these students will be using this information. If your major is history or art or psychology, this may not be of interest to you. But there'll be other things that are of interest for those disciplines. So we're just covering a number of bases for the many different disciplines that require their students to take this course. Now, let's see. Suppose I'm looking for a polynomial, p of x. And this polynomial is going to be of degree 3. In other words, it'll be a cubic polynomial. And suppose I told you that its roots, or its zeros, are 1, 1, and negative 4. Can anyone tell me a polynomial of degree 3 that has these roots? Steven? x minus 1 times x minus 1. OK, I'll just put a square on that. Times x plus 4. Times x plus 4, exactly. So if I go back and build the factors, x minus 1, x minus 1 again, and then x minus negative 4, or x plus 4, this is a polynomial that has those roots. And then if I multiply this out, x squared minus 2x plus 1 times x plus 4, then I could write that polynomial as x cubed plus 4x squared minus 2x squared minus 8x plus x and then plus 4. So then if I combine terms, this is x cubed plus 2x squared minus 7x plus 4. Here is a polynomial of degree 3 that has these three roots. Can anyone think of another polynomial of degree 3 that has these same roots? There are actually many others. You could multiply the whole thing by 2, or by a coefficient? Yeah, multiply by 2, or by a coefficient. Multiply by negative 2, multiply by 10, whatever. So there are other choices. Let's say what if I call the next choice 2 a q. I'll just double everything and get 2x cubed plus 4x squared minus 14x plus 8. We've actually talked about these problems before. And the only difference in these two graphs is that q is just stretched more than p is, because we doubled it. So it's stretched vertically top to bottom, but it has the same intercepts. OK, well now suppose we change this problem to one that we haven't seen before. And I'm going to give you a complex number root. So this time I'm looking for a function p of x. And let's say this one is of degree 3. And suppose that its roots are, let's say, 1 and i. 1 and i. You notice I've only given you two roots there. So you could manufacture a third root on your own. So it looks like one of the factors would be x minus 1. Another factor would be x minus i. And another factor, for the sake of argument, since I didn't give you a third root, suppose I put in, let's say, negative 1, which wasn't given, but we'll just make that one up. So this gives me x plus 1 right here. Look what happens when I multiply this out. I am going to get a cubic polynomial, but the coefficients are going to be a little unusual. If I take x minus 1 and x plus 1, that product is x squared minus 1. And I multiply by x minus i. So this gives me x cubed minus, there's a minus i x squared minus x plus i. What do you notice about the coefficients on my polynomial answer? They're not real. They're not all real. Now there's a 1 here and there's a negative 1 there. Those are real, but this is an imaginary number, negative i. And here's a plus i. So what if I added the stipulation of this problem that I'm looking for a polynomial of degree 3, and it has these two roots, and it has all real coefficients? I want to have real coefficients. How can I do that? Well, the way I will adjust it is I'm going to put in, for my third factor, something other than the negative 1 that I just chose, I'm going to put in the conjugate of i. Let me write down that word, because you've seen this before when you talked about complex numbers. But when I say the conjugate of i, when I say i, that's 0 plus i. And the conjugate of i would be 0 minus i, like we were multiplying by a moment ago. That's going to be minus i. So from my third root, I'll put in a minus i right there. And now I'll have real coefficients. Let's see how this works. I'll put in a x plus i. So when I multiply this out, I'm going to begin by multiplying these two factors, because it's these two that make the difference of two squares. I have my x minus 1. And what is the product, x minus i and x plus i? x squared minus i squared? x squared minus i squared, exactly. And we're going to reduce that next. And what can we replace this with? x squared plus 1. x squared plus 1. Now you see we've gotten rid of all of our complex numbers. So I'm multiplying x minus 1 times x squared plus 1. And the product is x cubed minus x squared plus x minus 1. Now I have a polynomial of degree 3 with real coefficients. Now you see the reason that I put in the conjugate is because when you have real coefficients, the conjugate roots come in what they call conjugate pairs, like a plus bi and a minus bi. So if you put in the conjugate pair, you'll go back to the original real coefficients. Let's take another problem with different roots and see how we could construct that. Let's see. Suppose I'm looking for a polynomial this time of degree 4. And suppose the roots of the polynomial, or the zeros, are 2 and 0 and 1 plus i. Then I'm going to need to include an extra root with that because I said degree 4. I need an extra root. And if I also add the attachment that I want real coefficients, then what extra root should I put there? 1 minus i. 1 minus i. That's the conjugate of 1 plus i, exactly. So my factors would be x minus 2, x minus 0, x minus the quantity 1 plus i, and making a little more room here, x minus the quantity 1 minus i. Now, to reduce this, first of all, x minus 0 is just an x. So let's just put that out in front. And then here's my x minus 2 factor. I'll bring that down next. And for the other two factors, I'm going to distribute the negative and write this as x minus 1 minus i and x minus 1 plus i. Now, these first two are very easily multiplied together. That will give me x squared minus 2x. But to multiply the other two, I'm going to regroup the terms. And I'm going to write these with the x minus 1s grouped together. x minus 1 minus i and x minus 1 plus i. So you see what happened. Originally, I had the 1 and the i's grouped together. I distributed the negative. And then I regrouped the 1 with the x. x minus 1 minus i, x minus 1 plus i. The reason I do this is because this looks like the difference of two squares. This is a minus b and a plus b. So when I multiply those together, I'm going to get a squared. That'll be the x minus 1 squared minus b squared. That's the i squared. In fact, I don't need that much room. Well, at this point, I'll get rid of the i's. I have x squared minus 2x times, what's the square of x minus 1? x squared minus 2x plus 1. Plus 1, right. And then minus i squared, I could write as what? Plus 1. Plus 1. So this is x squared minus 2x times x squared minus 2x plus 2. And you notice there are no imaginary or complex numbers left. I got rid of them when I squared the i squared in my last step. So now if I just finish this off by multiplying this out, I get x to the fourth. That's this product. How many x cubes will there be? Let's see. I'm going to get negative 2x cubed here and negative 2x cubed here. So that's a total of negative 4x cubed. How many x squares? Well, let's say I get 2x squared there. I get 4x squared there. Do you see any other x squares? I think that's it. So what would we say 2 plus 4 is 6x squared. And now let's count the x's. I have negative 4x there. And I think that's all of them, just negative 4x. So this is a fourth degree polynomial. It does have four roots, 2, 0, 1 plus i, and then the one that we created, 1 minus i. And it does have real coefficients. So we've been able to go back and reconstruct the polynomial with real coefficients, even when we had a complex root given. Let's go to the next example on the next graphic and look at another problem about polynomials with real coefficients. It says factor the following completely. And there are two polynomials given. Now when I say completely here, what I mean is to factor these down to linear factors. No quadratics, just linear factors. So let's take the very first problem. I'll write it here on my green screen before we take the graphic off. p of x is 3x to the fifth plus 24x cubed plus 48x. OK, let's look at this polynomial and see how we would factor it. It says factor completely. And what I mean by that is to factor it so that every factor is a linear factor, like x plus 2, x minus 1. Well, do you see any way we could reduce this, first of all? Take out an x. Take out an x. Is there anything else I could take out? A 3. A 3. Let's take out a 3x. And that gives me, well, who could tell me? What will I give you? I'll give you x to the fourth plus 6x squared. Plus 8x squared. It's 8, OK. Plus, jeez, I don't know, 48 to the ninth or 3? 12, 16, plus 16. Plus 16. OK, now, if we're lucky, we would be able to factor that without going to the rational root theorem. Can anyone think of a way to factor x to the fourth plus 8x squared plus 16? It's perfect square again. It is a perfect square. Very good. Jenny, what's it the square of? x squared plus 4. x squared plus 4, yeah, x squared plus 4. In other words, I have that factor twice. So far, I've been able to avoid using the rational root theorem and Descartes rule of signs because that's my alternative, but it would take longer. You know, if I write that factor separately, x squared plus 4 and x squared plus 4, then let's see how I could factor that to linear factors. Anyone have an idea how you could factor x squared plus 4 if you're allowed to use complex and imaginary numbers? I would set it equal to 0. OK, let's set it equal to 0 over here, and then do what? Subtract 4. OK, x squared is negative 4. And then take the square root of both sides. OK, so x is going to be, now we're looking for every possible root, so I'll put a plus or minus square to negative 4. That's exactly right. And what is that going to be? Plus or minus 2i. Plus or minus 2i. Plus or minus 2i. So I have two roots. One of the roots is 2i. The other root is negative 2i. So how does that tell me what the factors of this guy are going to be? x plus 2i and x minus 2i. OK, now you said x plus 2i. Which one of these roots were you using when you said x plus 2i, Steven? Negative 2i. Yeah, so if you take this root, it's x plus 2i. It's x minus that root. Why is that? Well, in the problems we worked before, if you remember, when we had a root of like 1, the factor was x minus 1. So this time, if the root is negative 2i, I take x minus negative 2i, which is x plus 2i. And then the other factor, Steven, would be what? x minus 2i. x minus 2i, yeah. I just want to make sure that everyone understands that when you look at 2i, that's not the one that goes with x plus 2i. It's the one that goes with this. You see, if I said what is the 0 produced by this factor, you'd set that equal to 0, and it would say x equals 2i. But now I have more roots produced by this one. But they'll be the same as this. So why don't I square this factor, square that factor? And so the factorization is 3x times x minus 2i squared times x minus 2i squared. I have five linear factors. When I say linear, I mean first power on x. Five linear factors for that polynomial. OK, we were able to get by that one using just fundamental factoring techniques. Now let's go to the other factoring problem that was on that screen, that graphic. This one I think we're going to have to go to the rational root theorem. It says q of t, q of t, I just went to the variable t just to change names a moment. And that was, let's see, can I see that once again? t to the fourth plus t squared minus 6t plus 4. OK, I don't see any obvious way to factor that. So instead, let's count positive roots and negative roots. And negative roots. Now when I say positive and negative roots, I'm talking about real number roots, not complex or imaginary. How many positive roots could we have? Two or zero. Right, sign change, sign change, two sign changes. So either two or none. Two or none. And for q of negative t, that'll be t to the fourth plus t squared, because when you square the negative, you get a positive t squared, plus 6t plus 4. Well, this time we don't have any sign changes. So there are no, there are no negative roots. So either two or none for the positive roots, no negative roots. That helps me to reduce the possibilities pretty quickly. What are my choices for p in this function? Well, p would be a divisor of 4. So that would be plus or minus 1, plus or minus 2, plus or minus 4. And we'll be restricting this to positives in just a moment. And q would be divisors of the coefficient of t to the fourth plus or minus 1. So for p over q, if I list only the positive roots, 1, 2, and 4. OK, let's take these one at a time. 1 divides into 1, 0, 1, negative 6, 4. And I'll be adding here. 1 added to 0 is 1. 1 and 1 is 2. 1 times 2 is 2. Adding, I get a negative 4. 1 times negative 4 is negative 4. So very quickly, we get a 0. So one of the roots is t equals 1. Now, shall I move on to 2 or should I test 1 one more time? Do it again. Let's do it again. Yeah, because it could be a root of multiplicity, too. So now I'm going to divide 1 into 1, 1, 2, negative 4. Because this is the polynomial that remains. 1, 1 times 1 is 1. 2, 1 times 2 is 2. 4, 1 times 4 is 4. Yep, sure enough, it's a root a second time. But what's even more important is not only have I found that this is a root of multiplicity 2, but I have my remaining factorization, t squared plus 2t plus 4. Now, I'll need a little room to write here. So let me just back up and take out this first portion. And here's what we have found. This factorization is t minus 1 squared times this polynomial, t squared plus 2t plus 4. Now, I don't believe this would factor by ordinary factoring techniques for real numbers. So instead, I'm going to go to the quadratic formula to factor that. t will be negative b plus or minus the square root of b squared minus 4 times 1 times 4 all over 2. So t will be negative 2 plus or minus the square root of, let's see, 4 minus 16 is negative 12 over 2. So t will be negative 2 plus or minus, how could I write this as an imaginary number? Negative square root of 12. I root 3? Let's see. Well, now you're going to have to take a square root. Yeah, there's a 2 times the square root of negative 3. So there's the square root of negative 3, and here's the square root of the negative 1. Now, sometimes people put the i over here on the right-hand side, but if you're not careful, that square root bar may look like it goes over the i. So what a lot of textbooks do is they put the i in front of the radical, so there's no doubt that it's outside the radical. And this becomes negative 1 plus or minus i square roots of 3. So when I go to write my complete factorization up here, I can now factor this last quadratic using these roots. I'll have t minus 1 squared times t minus the quantity negative 1 plus i square roots of 3 times, better erase this too, I guess, t minus the quantity negative 1 minus i square roots of 3. Well, that looks messy, so I think what we need to do is to distribute those negative signs. We have t minus 1 squared times t plus 1 minus i square roots of 3 and t plus 1 plus i square roots of 3. I have two linear factors here. I have a linear factor there, t plus or minus a complex number, and another linear factor there. I actually have four linear factors. That's called the complete factorization of that polynomial. Now, let's just go to the second graphic, the one on the cubic formula. Let me just mention this in passing. We only have a few minutes left here. But just as a side note, you won't have to know this for an exam. So if this is too quick of a discussion of this last item here, don't worry because it won't be on the test. But I just thought you should know that there is a formula for factoring cubic polynomials of the form p of x equals x cubed plus px plus q. You notice that polynomial has no quadratic term. It has a cubic term, a linear term, and a constant term. And to solve that, there is a formula, something like the quadratic formula. But this would be a cubic formula. And it says x is equal to the cube root of negative q over 2, q being the constant term, plus the square root of q squared over 4 plus p cubed over 27, plus an almost identical radical added onto it. What's the only change in the second radical? Second negative. There's a negative sign in front of the square root. And so there's an interesting history of this, but I don't think we have time to go into the history. OK, so this time I'd like to try solving this polynomial, which is in the form that was on that graphic. There is no quadratic term. p is negative 3 and q is 2. The formula says I should take the cube root of negative q over 2 plus the square root of q squared over 4 plus p cubed over 27. And then add on the cube root of negative q over q minus the square root of q squared over 4 plus p cubed over 27. Now when I reduce that, this is the cube root of negative 1 plus 0. And over here, this is the cube root of negative 1 minus 0, the square root of 0. So that's the cube root of negative 1 is negative 1. And the cube root of negative 1 is negative 1. We get negative 2. Now if you substitute in negative 2 for x in the original polynomial, it'll make it 0. I'm not going to ask you about this formula on a test, but if you have an opportunity where you'd like to use it on an exam, you certainly can. Well, let's see. We've talked about the fundamental theorem of algebra and complex roots today. We'll see you next time for episode 13.