WEBVTT mathematics/math-analysis/selhorst-jones 00:00:00.000 --> 00:00:02.200 Hi--welcome back to Educator.com. 00:00:02.200 --> 00:00:05.100 Today, we are going to talk about partial fractions. 00:00:05.100 --> 00:00:14.000 Consider if we had two quotients of polynomials that we wanted to add together, like, say, 7/(x - 5) and 3/(x + 2). 00:00:14.000 --> 00:00:18.900 We could put them over a common denominator, then combine them; we have x - 5 here and x + 2 here, 00:00:18.900 --> 00:00:26.400 so we can multiply the one on the left by x + 2, so we get (x - 5)(x + 2) on the bottom. 00:00:26.400 --> 00:00:29.700 Its top will get also multiplied by that (x + 2). 00:00:29.700 --> 00:00:35.000 For the other one, we will have (x - 5) multiplied on it, so it has (x - 5) multiplied on the top, 00:00:35.000 --> 00:00:39.100 so we have (x - 5)(x + 2)--a common denominator--on the bottom now. 00:00:39.100 --> 00:00:48.800 We multiply this out; we get 7x + 14 and 3x - 15; we combine those, and we get 10x - 1 on top. 00:00:48.800 --> 00:00:53.800 divided by (x - 5)(x + 2), expanded to x² - 3x - 10. 00:00:53.800 --> 00:01:03.900 So, it is not a bad idea; we can get from here to here by putting it over a common denominator, and then just adding things out and simplifying. 00:01:03.900 --> 00:01:10.000 But if we wanted to do the reverse process--what if we started with a fraction involving large polynomials, 00:01:10.000 --> 00:01:13.700 and we wanted to break it into smaller fractions made of the polynomials' factors? 00:01:13.700 --> 00:01:20.600 If we wanted to do this process in reverse, if we started at (10x - 1)/(x² - 3x - 10), 00:01:20.600 --> 00:01:28.500 and we somehow wanted to be able to get that into 7/(x - 5) and 3/(x + 2), what process could we go through? 00:01:28.500 --> 00:01:35.200 We call the smaller fractions on the right partial fractions, because they are parts of that larger fraction. 00:01:35.200 --> 00:01:41.000 So, each of these here (7/(x - 5) and 3/(x + 2))--they are each called a partial fraction. 00:01:41.000 --> 00:01:45.200 And the process to break it up is called partial fraction decomposition. 00:01:45.200 --> 00:01:50.900 So, whatever this method is that we haven't explored yet, to get from that larger polynomial quotient 00:01:50.900 --> 00:01:55.800 into these smaller partial fractions, is called partial fraction decomposition. 00:01:55.800 --> 00:02:04.300 So, how do we do it? To understand this lesson--to understand how we do it--you will need some familiarity with solving systems of linear equations. 00:02:04.300 --> 00:02:08.300 Previous experience from past algebra classes will probably be enough to get through this. 00:02:08.300 --> 00:02:12.000 But if you want a refresher, if you are a little confused by how this stuff is working later on, 00:02:12.000 --> 00:02:18.100 check out the lesson Systems of Linear Equations--that will help explain this stuff if it confuses you right now. 00:02:18.100 --> 00:02:21.500 You also need to know how to factor polynomials, and you will need to understand them in general, 00:02:21.500 --> 00:02:26.800 along with the ability to do polynomial division; you will need all of that for some of the stuff in here, as well. 00:02:26.800 --> 00:02:32.200 Now, sadly, we won't be able to see the application of partial fraction decomposition in this course. 00:02:32.200 --> 00:02:37.500 However, it is quite useful in calculus, where it will allow us to solve otherwise impossible problems. 00:02:37.500 --> 00:02:42.600 Partial fraction decomposition is this really useful thing that lets us break up these complicated things 00:02:42.600 --> 00:02:46.100 that we couldn't solve, and turn them into a form that is actually pretty easy to solve. 00:02:46.100 --> 00:02:47.600 It is really handy in calculus. 00:02:47.600 --> 00:02:51.200 We won't be able to see its use in this course; we won't see it any time soon. 00:02:51.200 --> 00:02:56.600 But it is helpful to practice it now, just like we practiced stuff about factoring complicated polynomials in algebra, 00:02:56.600 --> 00:02:59.600 before we really had a great understanding of what it was all about. 00:02:59.600 --> 00:03:02.300 We are practicing something so that we can use it later on. 00:03:02.300 --> 00:03:10.900 And also, it just is a great chance to flex our brain and get some mental "muscles" that are kind of difficult ideas, but really require some analytic thought. 00:03:10.900 --> 00:03:14.600 All right, let's get to actually figuring out how to do this. 00:03:14.600 --> 00:03:19.500 If we have a polynomial fraction in the form numerator polynomial divided by denominator polynomial, 00:03:19.500 --> 00:03:26.400 a normal rational function format, there are two possibilities in regards to the degrees of the top and bottom polynomials. 00:03:26.400 --> 00:03:33.700 We call them proper (proper is when the degree of the numerator polynomial is less than the degree of the denominator polynomial-- 00:03:33.700 --> 00:03:42.400 the degree of n is less than the degree of the denominator) and improper (when the degree of our numerator 00:03:42.400 --> 00:03:45.900 is greater than or equal to the degree of the denominator). 00:03:45.900 --> 00:03:51.600 To decompose the fraction, it must be proper; we have to be in this proper format. 00:03:51.600 --> 00:03:58.700 We have to make sure that the degree of our numerator is less than the degree of our denominator, if we want to do partial fraction decomposition. 00:03:58.700 --> 00:04:00.800 So, what do we do if the fraction is improper? 00:04:00.800 --> 00:04:05.100 We have to turn it into a proper thing; so if the fraction is improper, and we want to decompose it, 00:04:05.100 --> 00:04:08.800 we have to make it proper through polynomial division. 00:04:08.800 --> 00:04:16.100 We can make that numerator smaller by being able to break it off and divide out the part that isn't smaller and the part that is smaller. 00:04:16.100 --> 00:04:21.300 Remember, when you divide your polynomial division, the remainder from the division 00:04:21.300 --> 00:04:25.200 goes back onto our original denominator, which we will then decompose. 00:04:25.200 --> 00:04:27.200 We will be able to decompose the part that is the remainder. 00:04:27.200 --> 00:04:30.800 The other part that comes out cleanly--well, that is just going to be there. 00:04:30.800 --> 00:04:36.300 That is just going to be a polynomial that is then going to be added to whatever ends up coming out of our decomposition. 00:04:36.300 --> 00:04:40.600 We will see an example of this in Example 3. 00:04:40.600 --> 00:04:44.100 Once we have a proper polynomial fraction to decompose into partial fractions 00:04:44.100 --> 00:04:49.200 (remember, proper: our numerator has to be smaller than our denominator before we can enact this process), 00:04:49.200 --> 00:04:53.000 once we have done that, we can then factor the denominator. 00:04:53.000 --> 00:04:55.600 Our next step is to factor the denominator. 00:04:55.600 --> 00:04:59.800 After the denominator is broken into its smallest possible factors, we are ready to decompose. 00:04:59.800 --> 00:05:02.500 Now, there are two types that the smallest possible factors can come in. 00:05:02.500 --> 00:05:09.400 They can come in linear factors, which is forms ax + b, and they will be raised to some power, because we might have a multiplicity of these factors. 00:05:09.400 --> 00:05:15.900 There might be multiple of a given factor, so if we have multiple ax + b's, we will have m of them, (ax + b)^m. 00:05:15.900 --> 00:05:19.900 So, we will have linear factors in this; or we could have irreducible quadratic factors, 00:05:19.900 --> 00:05:25.400 that is, things ax² + bx + c that can't be broken up further into linear factors. 00:05:25.400 --> 00:05:27.900 There is no way to break them up further in the reals. 00:05:27.900 --> 00:05:31.600 So, ax² + bx + c...and once again, it will be raised to some power; 00:05:31.600 --> 00:05:36.200 there will be m of them, so it is raised to the m, because there are m of them multiplying together. 00:05:36.200 --> 00:05:39.500 Remember: "irreducible" means it can't be broken up further in the reals. 00:05:39.500 --> 00:05:46.100 That means quadratics like x² + 1 or 5x² - 3x + 20, because they have no roots. 00:05:46.100 --> 00:05:55.300 x² + 1 has no roots; x² + 1 = 0...there are no solutions, in the reals, at least. 00:05:55.300 --> 00:05:58.600 If we allow for the complex, there are; but we are not working with the complex. 00:05:58.600 --> 00:06:03.700 There are no solutions in the reals; so, since there are no solutions in the reals, it can't be factored any further. 00:06:03.700 --> 00:06:06.500 x² + 1 is irreducible, therefore. 00:06:06.500 --> 00:06:11.200 The next step of decomposition will behave differently, depending on which flavor of factor we have-- 00:06:11.200 --> 00:06:17.400 whether we are at a linear factor or we are at an irreducible factor; and we will look at the two, one after another. 00:06:17.400 --> 00:06:25.300 Linear factors: the partial fraction decomposition must include the following for each linear factor that is in this form (ax + b)^m. 00:06:25.300 --> 00:06:33.900 So, it is going to have this in its decomposition: A₁/(ax + b) + A₂/(ax + b)² +... 00:06:33.900 --> 00:06:38.100 notice that we are doing this where it is going to keep stepping up with this exponent on the bottom, 00:06:38.100 --> 00:06:42.000 as we keep putting these things in, until eventually we get up to our mth step, 00:06:42.000 --> 00:06:45.800 because we had m of them to begin with, so we have to have m of them in the end. 00:06:45.800 --> 00:06:51.300 And each one of these A's on top (A₁ up until A<font size="-6">m</font>) are all just constant real numbers. 00:06:51.300 --> 00:06:55.400 We are using A with a subscript, this A₁, A₂, A₃ business, 00:06:55.400 --> 00:07:00.300 because we just need a way of being able to say m of them, and we are not quite sure that we are going to go only out to m. 00:07:00.300 --> 00:07:03.400 Anyway, the point is that they are all going to be constant real numbers. 00:07:03.400 --> 00:07:07.800 For example, if we had (x + 7)³, then we have some stuff up top. 00:07:07.800 --> 00:07:11.400 We don't really care what the stuff is right now, because we just want to see how it will break out. 00:07:11.400 --> 00:07:13.100 We will deal with the stuff later. 00:07:13.100 --> 00:07:18.900 If we have stuff over (x + 7)³, then it is going to break into 3 of these, because of this "cubed." 00:07:18.900 --> 00:07:24.500 We will have A/(x + 7) + B (we can switch into just using letters in general, because we know 00:07:24.500 --> 00:07:29.200 that each one of these capital letters just means some constant real number; we will figure them out later-- 00:07:29.200 --> 00:07:37.000 that will come up later--don't worry)...A/(x + 7) + B/(x + 7)² + C/(x + 7)³. 00:07:37.000 --> 00:07:42.000 Notice: we started cubed, and so we have three of these; we step up each time with the exponent, 00:07:42.000 --> 00:07:45.500 until we have gotten to the number of our multiplicity that we originally started with-- 00:07:45.500 --> 00:07:48.800 whatever the exponent was originally on the factor. 00:07:48.800 --> 00:07:54.800 If we have multiple linear factors, we do it for each one of these; so if we have (2x + 3)², x¹ 00:07:54.800 --> 00:08:03.500 (since there is nothing there), (x - 5)¹ (since there is nothing there)...we have A/(2x + 3) + B/(2x + 3)² 00:08:03.500 --> 00:08:10.100 (because remember: it was squared to begin with, so it has to have 2 of itself) + C/x. 00:08:10.100 --> 00:08:14.800 And there is only to the 1, so there is only going to be one of it; plus D/(x - 5). 00:08:14.800 --> 00:08:17.800 And it is only 1, because there is only one of them to begin with. 00:08:17.800 --> 00:08:23.200 And each one of these A, B, C, D...they are each just constant numbers. 00:08:23.200 --> 00:08:26.800 We are going to figure them out later on; don't worry. 00:08:26.800 --> 00:08:31.100 Irreducible quadratic factors: the method is very similar for irreducible quadratic factors. 00:08:31.100 --> 00:08:36.500 The decomposition must include the following for each irreducible quadratic: ax² + bx + c to the m. 00:08:36.500 --> 00:08:39.800 Once again, this is the same thing of stepping out m times. 00:08:39.800 --> 00:08:48.400 A₁x + B₁ over ax² + bx + c: notice how, previously, it was A over some x plus a constant, 00:08:48.400 --> 00:08:52.200 because we had that the degree of the top was one below the degree on the bottom. 00:08:52.200 --> 00:08:57.300 So, the degree on the top here is a linear factor on top, because we have a quadratic factor on the bottom. 00:08:57.300 --> 00:09:01.100 Previously, we had a constant factor on top, because we had a linear factor on the bottom. 00:09:01.100 --> 00:09:07.800 So, A₁x + B₁ + A₂x + B₂, over...that should be a capital... 00:09:07.800 --> 00:09:11.900 ax² + bx + c, now squared, because we are doing our second step... 00:09:11.900 --> 00:09:18.400 we do this out m steps, until we are at our m constants, and we are at the mth exponent on the bottom. 00:09:18.400 --> 00:09:23.900 So, A₁...all of these capital letters...B₁ to B<font size="-6">m</font>...they are all just constants. 00:09:23.900 --> 00:09:25.500 And we will figure them out later. 00:09:25.500 --> 00:09:31.100 For example, if we had (x² - 4x + 5)² on our bottom, then we have some stuff-- 00:09:31.100 --> 00:09:35.900 it doesn't really matter right now--x² - 4x + 5, squared; so we are going to have this happen twice. 00:09:35.900 --> 00:09:39.900 The first one will be just to the one; the second time, it will be squared. 00:09:39.900 --> 00:09:45.000 And on the top, it is going to be Ax + B and Cx + D. 00:09:45.000 --> 00:09:49.900 So, A, B, C, D...these are all just constants; we will deal with them later. 00:09:49.900 --> 00:09:52.800 If we have multiple irreducible quadratics, we just do each of these. 00:09:52.800 --> 00:09:59.600 For example, if we have x² + 2x + 3, that is an irreducible; and x² + 4, squared... 00:09:59.600 --> 00:10:05.500 then we have Ax + B, x² + 2x + 3...there is only one of them, so it only shows up once, 00:10:05.500 --> 00:10:12.200 + Cx + D over x² + 4...it is going to show up twice because of that 2, so it shows up a second time here. 00:10:12.200 --> 00:10:20.500 And we have Ex + F; so we just keep doing this process with capital letter, x, + capital letter, plus capital letter, x, plus capital letter, 00:10:20.500 --> 00:10:26.800 until we have worked out all of the times that we had things showing up--a total of 3, because we had one there and two there. 00:10:26.800 --> 00:10:29.000 So, we have a total of 3. 00:10:29.000 --> 00:10:34.500 If we have both types, a linear and an irreducible quadratic, mixed together in the denominator, that is OK--perfectly fine. 00:10:34.500 --> 00:10:37.100 We just decompose based on both of the rules. 00:10:37.100 --> 00:10:45.000 So, for example, if we had (x + 9)², then the x + 9's are going to follow this A, this single-constant format, 00:10:45.000 --> 00:10:50.200 because linears just had one constant on the top; x + 9, and then (x + 9)². 00:10:50.200 --> 00:10:55.800 And then, we switch to the other rule, x² + 1; now we are dealing with an irreducible quadratic. 00:10:55.800 --> 00:11:01.200 So, x² + 1...and we have Cx + D on top, because we have to be able to have... 00:11:01.200 --> 00:11:04.900 since we have a quadratic on the bottom, we now have linear factors on the top. 00:11:04.900 --> 00:11:11.000 When we are dealing with linear factors on the bottom, even if it is multiple linear factors, it is just constants on the top. 00:11:11.000 --> 00:11:14.200 All right, now, finally: how do we figure out what those numerators are? 00:11:14.200 --> 00:11:16.700 We have mentioned that the numerators are built out of constants. 00:11:16.700 --> 00:11:25.000 Remember: A for linear factors (constants for linear factors); Bx + C...linear factors for irreducible quadratic factors. 00:11:25.000 --> 00:11:29.800 But we haven't talked about how we actually find out what these capital letters' values are. 00:11:29.800 --> 00:11:32.900 We solve for the constants by doing partial fraction decomposition. 00:11:32.900 --> 00:11:38.400 We set them up in this format, and then we multiply each side of the equation by the original denominator. 00:11:38.400 --> 00:11:41.400 It will make more sense as we work through examples; let's look at this example. 00:11:41.400 --> 00:11:49.300 If we have (3x² + 3x - 4)/(x + 3), one factor, times (x² - x + 2), an irreducible quadratic factor, 00:11:49.300 --> 00:12:01.300 then we would be able to break it into A/(x + 3) (we get this right here)...and x² - x + 2 would be Bx + C/(x² - x + 2). 00:12:01.300 --> 00:12:06.400 So, we know, from what we just talked about (this process that we just went through) that that is how it breaks up. 00:12:06.400 --> 00:12:08.400 But how do we figure out what A and Bx + C are? 00:12:08.400 --> 00:12:14.200 Well, notice: we can just work things out the way we would a normal algebra thing. 00:12:14.200 --> 00:12:17.300 We multiply both sides, because we want to get rid of this denominator. 00:12:17.300 --> 00:12:25.900 We will multiply by x + 3 on the left side, and x² - x + 2 on both sides. 00:12:25.900 --> 00:12:39.600 So, we do the same thing on this side, as well: (x + 3)...we will color-code this...and (x² - x + 2). 00:12:39.600 --> 00:12:44.000 So, the (x + 3) will distribute to the one on the left and the one on the right. 00:12:44.000 --> 00:12:48.100 It has to hit both of them, because it is distributed, because there is this plus sign right here. 00:12:48.100 --> 00:12:56.300 So, (x + 3) distributes to both of them; similarly, (x² - x + 2) will distribute to both of these, as well. 00:12:56.300 --> 00:13:03.400 Now, notice: on the left side, we have (x + 3) on the bottom, and we are multiplying by (x + 3); so the (x + 3)'s cancel out. 00:13:03.400 --> 00:13:08.100 (x² - x + 2) is right here and here; they cancel out. 00:13:08.100 --> 00:13:15.900 What about on the left side? Well, we have (x + 3) and (x + 3); so here, it is going to cancel this and this. 00:13:15.900 --> 00:13:19.100 But it will still have the (x² - x + 2) coming through. 00:13:19.100 --> 00:13:21.600 What about the Bx + C over (x² - x + 2)? 00:13:21.600 --> 00:13:24.700 Well, the (x + 3) is still going to come through--it doesn't cancel out any factors here. 00:13:24.700 --> 00:13:29.400 But (x² - x + 2) is the same thing, so it will cancel out that part of the factor coming in. 00:13:29.400 --> 00:13:39.000 So, we will see that we have A times (x² - x + 2) and Bx + C times (x + 3) coming through. 00:13:39.000 --> 00:13:48.400 This part right here manages to come through on the Bx + C, and this part right here manages to come through on the A. 00:13:48.400 --> 00:13:54.400 And on the left, since we canceled out everything on the bottom, we just have what we had on our numerator originally. 00:13:54.400 --> 00:14:05.000 That just makes its way down; so we have 3x² + 3x - 4 = A times x² - x + 2, plus Bx + C times (x + 3). 00:14:05.000 --> 00:14:10.800 Which...we can then expand everything on the right, and we can work things out by creating a system of linear equations. 00:14:10.800 --> 00:14:15.000 The examples are going to help greatly in clarifying how this works, so check them out, 00:14:15.000 --> 00:14:19.500 because we will actually see how this process of setting up our system of linear equations works, 00:14:19.500 --> 00:14:23.100 and solving for what these values are, in each of these examples. 00:14:23.100 --> 00:14:25.800 Not Example 2, but we will see it in a lot of the examples. 00:14:25.800 --> 00:14:28.300 And it will make a whole lot more sense as we see it in practice. 00:14:28.300 --> 00:14:33.400 So, let's go check it out: the first example is a nice, simple example to get things started with. 00:14:33.400 --> 00:14:38.300 Find the partial fraction decomposition for (4x + 3)/x(x + 3). 00:14:38.300 --> 00:15:00.000 So, we have 4x + 3, over x(x + 3); that is going to be equal to...we have two linear factors on the bottom, so A/x, plus our other linear factor, B/(x + 3). 00:15:00.000 --> 00:15:12.300 Now, at this point, we can multiply everything by x and (x + 3). 00:15:12.300 --> 00:15:17.400 Great; so, on the left side, that is just going to cancel this stuff out; and on the right side, it will end up distributing. 00:15:17.400 --> 00:15:26.100 So, on the left side, we have 4x + 3 = A...now, notice: we have the x multiplying here; that is going to cancel out. 00:15:26.100 --> 00:15:30.900 So, the x's parts get canceled out; but we are left with x + 3 multiplying on it. 00:15:30.900 --> 00:15:39.600 Plus B times (x + 3)...so the x + 3 part cancels out; I will switch colors here. 00:15:39.600 --> 00:15:46.400 x gets through; but the (x + 3) cancels out the denominator; great--we have managed to get rid of all of our denominators. 00:15:46.400 --> 00:15:48.900 That is great; it will make things easier to see what is going on. 00:15:48.900 --> 00:16:04.700 And we have some stuff multiplying through; so at this point, 4x + 3 on the left equals Ax + 3A + Bx. 00:16:04.700 --> 00:16:10.300 Now, notice: we see here that we have a 3 constant. 00:16:10.300 --> 00:16:12.400 Now, what constants do we have on the right side? 00:16:12.400 --> 00:16:15.900 We have to have all of our x's match up and all of our constants match up. 00:16:15.900 --> 00:16:22.100 Well, the only thing that is actually a constant on the right side is this 3A, because we have Ax and Bx. 00:16:22.100 --> 00:16:25.900 But we don't have any B just as B; we don't have any constants of B. 00:16:25.900 --> 00:16:32.600 So, it must be that 3 and 3A are equal, because they are all of the constants that show up on either side of our equation. 00:16:32.600 --> 00:16:36.300 The same thing has to be on the left and the right; otherwise, it is not an equation. 00:16:36.300 --> 00:16:41.400 That tells us that 3 = 3A; so what is our A? 00:16:41.400 --> 00:16:46.700 We divide both sides by 3; we get 1 = A; so now we have figured out what A is. 00:16:46.700 --> 00:16:59.500 What about the other part? Well, we have that 4x; 4x is equal to...let's give it a special thing; we will give it a curly around... 00:16:59.500 --> 00:17:03.100 4x is equal to all of our x's on the right side, put together. 00:17:03.100 --> 00:17:14.500 So, that is our Ax, combined with our Bx; so it must be the case that 4x is equal to Ax + Bx, 00:17:14.500 --> 00:17:18.700 because the same number of x's must be on the right side as on the left side. 00:17:18.700 --> 00:17:25.000 So, 4x must equal Ax + Bx, because these are our sources of x on the right side. 00:17:25.000 --> 00:17:35.300 4x = Ax + Bx; we just figured out what A is, so we can plug that in over here; so we have 4x = Ax + Bx... 00:17:35.300 --> 00:17:43.500 Now, notice...actually, before we do that, even, we have 4x = Ax + Bx; so if that is the case, let's just get rid of these x's. 00:17:43.500 --> 00:17:51.900 All right, we know that it must be the case that, since 4x = Ax + Bx...well, that has to be true for any x that we plug in, at all. 00:17:51.900 --> 00:18:02.200 So, it must be the case that 4 = A + B; we can divide the x out on both sides, and we cancel that to 4 = A + B. 00:18:02.200 --> 00:18:08.400 Now, we can substitute in that 1; that will make it even clearer what is going on: 4 = 1 + B. 00:18:08.400 --> 00:18:14.500 Subtract the 1; we get 3 = B; so at this point, we have figured out what A is; we have figured out what B is. 00:18:14.500 --> 00:18:17.900 So, we see that we can now go back to our partial fraction decomposition. 00:18:17.900 --> 00:18:31.400 We can plug in actual numbers, and we can have what that is equivalent to; and we have 1 over (x + 3) over (x + 3). 00:18:31.400 --> 00:18:33.800 And there is our answer; great. 00:18:33.800 --> 00:18:40.300 All right, the next one: Write out the form of the partial fraction decomposition, but do not solve for the constant. 00:18:40.300 --> 00:18:43.600 That is good, because this would be really hard, if we actually had to solve it. 00:18:43.600 --> 00:18:51.200 So, first we have x, and then (x - 7)², and then 2x² + 1, an irreducible quadratic, cubed. 00:18:51.200 --> 00:19:02.500 All right, we have A/x, plus...the next one is a linear, so it is also just one constant up top, (x - 7); 00:19:02.500 --> 00:19:09.100 but it is squared in our original thing, so it gets squared; so another constant...C/(x - 7)...this time, 00:19:09.100 --> 00:19:18.600 it is two of them; it is squared; plus Dx + E, because now we are dealing with an irreducible quadratic. 00:19:18.600 --> 00:19:26.500 The first time it showed up, so 2x² + 1, plus Fx + G...just keep going with letters; 00:19:26.500 --> 00:19:32.900 I am just counting off the alphabet at this point, Fx + G...A, B, C, D, E, F, G...I don't want to make a mistake with my alphabet!... 00:19:32.900 --> 00:19:45.700 2x² + 1...now we are at the second time, so squared...Hx + I, over (2x² + 1)³. 00:19:45.700 --> 00:19:47.600 And that is the partial fraction decomposition. 00:19:47.600 --> 00:19:53.200 We haven't figured out what A, B, C, D, E, F, G, H, I are. 00:19:53.200 --> 00:20:02.400 But we could, if we multiplied, because we know that this thing here is equal to this thing here, once we get those correct constants in. 00:20:02.400 --> 00:20:05.700 So, we could multiply the left side and the right side by that denominator. 00:20:05.700 --> 00:20:09.500 And everything would cancel out, and we would be left with this awful, awful massive thing. 00:20:09.500 --> 00:20:15.300 But we could solve it out, slowly but surely; luckily, all it asked for was to just set things up-- 00:20:15.300 --> 00:20:19.100 set up the partial fraction decomposition, but not solve for these constants. 00:20:19.100 --> 00:20:21.900 So, happily, we can just see that this is how this breaks down. 00:20:21.900 --> 00:20:27.600 We break it down this way; we use each of the rules, based on whether it is linear or it is an irreducible quadratic. 00:20:27.600 --> 00:20:32.400 And they show up the number of times of each of their exponents. 00:20:32.400 --> 00:20:40.700 Great; the third example: Find the partial fraction decomposition for (x⁵ + 3x³ + 7x)/(x⁴ + 4x² + 4). 00:20:40.700 --> 00:20:48.500 The first thing to notice is that it has a degree 5 on top and a degree 4 on the bottom, so we start as improper. 00:20:48.500 --> 00:20:54.200 So, since it is improper, we have to use polynomial division to break it into a format that is proper, first. 00:20:54.200 --> 00:20:58.200 How many times does x⁴ + 4x² + 4 go into this thing here? 00:20:58.200 --> 00:21:10.600 So, let's set up our polynomial division: 4x² + 4 goes into x⁵ +...we have no x⁴'s... 00:21:10.600 --> 00:21:18.700 plus 3x³...we have no x²'s...+ 7x; and we have no constants. 00:21:18.700 --> 00:21:21.200 OK, how many times does x⁴ go into x⁵? 00:21:21.200 --> 00:21:33.700 It goes in x, because x times x⁴ gets us x⁵; and we do it on the other things...+ 4x³ + 4x. 00:21:33.700 --> 00:21:39.600 So, we subtract all of this; let's distribute that subtraction... 00:21:39.600 --> 00:21:42.800 x⁵ - x⁵ becomes 0--no surprise there. 00:21:42.800 --> 00:21:51.500 3x³ - 4x³ becomes -x³; 7x - 4x becomes positive 3x. 00:21:51.500 --> 00:22:03.800 We can now bring down everything else, but as soon as we do that, we realize that there is nothing else to be done here, 00:22:03.800 --> 00:22:09.100 because how many times does x⁴ go into 0x⁴? It goes in 0 times. 00:22:09.100 --> 00:22:11.200 It can't fit in at all, because it is just a 0 to begin with. 00:22:11.200 --> 00:22:17.700 So, we found our remainder; our remainder is what remains: our -x³ here, our 3x, and that is it. 00:22:17.700 --> 00:22:22.200 -x³ + 3x is what remains. 00:22:22.200 --> 00:22:30.400 That means that what we originally started with, x⁵ + 3x³ + 7x, 00:22:30.400 --> 00:22:45.700 all over x⁴ + 4x² + 4, is equal to...x came out of it, so x plus what our remainder was, 00:22:45.700 --> 00:22:57.300 -x³ + 3x, over that original denominator, x⁴ + 4x² + 4; great. 00:22:57.300 --> 00:23:00.000 We need some more room; let's flip to the next page. 00:23:00.000 --> 00:23:06.400 That is what we figured out: we figured out that what we started with breaks into this thing right here. 00:23:06.400 --> 00:23:10.800 So, now let's figure out how we can decompose this part on the right. 00:23:10.800 --> 00:23:19.700 We will work on decomposing this first, but we can't forget (circle it in red, so we don't forget) that x, because that is still part of that function. 00:23:19.700 --> 00:23:23.600 We can't actually decompose it without putting that back in at the very end. 00:23:23.600 --> 00:23:27.900 Otherwise, we will have decomposed a different function; we will have decomposed this function right here, 00:23:27.900 --> 00:23:33.100 but forgotten about what we originally started with, because what we originally started with includes this x. 00:23:33.100 --> 00:23:38.600 All right, -x³ + 3x...oh, that is one thing I forgot to do last time. 00:23:38.600 --> 00:23:53.800 We had x⁴ + 4x² + 4; I quietly factored that, without ever mentioning that it factors into (x² + 2)². 00:23:53.800 --> 00:23:57.300 I'm sorry about that; hopefully that isn't too confusing. 00:23:57.300 --> 00:24:06.200 -x³ + 3x...over (x² + 2)²...that is going to be equal to Ax + B, 00:24:06.200 --> 00:24:17.900 because it is an irreducible quadratic, divided by x² + 2, plus Cx + D, over (x² + 2)². 00:24:17.900 --> 00:24:29.900 At this point, we multiply both sides by (x² + 2)², (x² + 2)², (x² + 2)². 00:24:29.900 --> 00:24:37.900 So, that cancels out here, and we are left with -x³ + 3x =...what happens to the Ax + B? 00:24:37.900 --> 00:24:45.300 Well, its denominator gets canceled, but then it is still left getting hit with one more x² + 2. 00:24:45.300 --> 00:24:51.900 What about Cx + D? Well, its denominator just gets canceled; there is nothing left after (x² + 2)² hits, 00:24:51.900 --> 00:24:55.400 because its denominator was the same thing, so we are just left with Cx + D. 00:24:55.400 --> 00:25:03.000 So, let's expand that, and let's also put this in a different color, just so we don't get confused by the separation. 00:25:03.000 --> 00:25:06.900 I am not sure how much that will help us. 00:25:06.900 --> 00:25:15.300 3x =...Ax times x² becomes Ax³; Ax times 2 becomes 2Ax. 00:25:15.300 --> 00:25:24.500 Bx² + 2B...and we will also bring along + Cx + D. 00:25:24.500 --> 00:25:33.900 At this point, we can set things up so we can see it a little more clearly by putting everything together that comes in a single form. 00:25:33.900 --> 00:25:41.100 We have Ax³, and we will leave a blank for the x²'s...+ 2Ax. 00:25:41.100 --> 00:25:50.000 And then, we can write over here + Bx² + 2B...oops, sorry; that is going to have to leave a blank there, as well... 00:25:50.000 --> 00:25:58.900 + 2B...and then + Cx, and then + D; notice how this works. 00:25:58.900 --> 00:26:05.700 We have our cubes here (degree 3 things here); degree 2 things here... 00:26:05.700 --> 00:26:08.000 well, not quite "degree," because they are not the whole polynomial, 00:26:08.000 --> 00:26:13.300 but things with exponent 3 here, exponent 2 here, exponent 1 here, and exponent 0 here. 00:26:13.300 --> 00:26:18.100 So, now we can compare that over here, where we have exponent 3 here and exponent 1 here. 00:26:18.100 --> 00:26:27.300 So, all of the things that are going to connect to the -x³, which we can also see as -1 times x³, are just Ax³ here. 00:26:27.300 --> 00:26:33.400 So, it must be that -1 is equal to A; otherwise, we wouldn't have -x³ in the end. 00:26:33.400 --> 00:26:38.000 We now know that -1 = A; great. 00:26:38.000 --> 00:26:47.900 What about 2Bx²? Well, we have 0x²; we know that 0x² = Bx². 00:26:47.900 --> 00:26:55.200 Things of degree 2...well, we could have just written that as 0 = B, because our things of degree 2 have to line up with that number B. 00:26:55.200 --> 00:27:02.600 So, 0 = B; what about our D here? We can figure out our D. 00:27:02.600 --> 00:27:12.400 Well, we have a constant...+ 0 here; so it must be that 0 is equal to 2B + D. 00:27:12.400 --> 00:27:20.700 Well, we already know that B equals 0 here; so that just knocks out; so we are now told that 0 equals D, as well. 00:27:20.700 --> 00:27:30.000 Finally, we are left figuring out what C is; so we know that 3, here, equals 2A, because there are that many x, 00:27:30.000 --> 00:27:41.900 because we have 3x and 2Ax and Cx, so it must be that 3 = 2A + C, when we combine them all together. 00:27:41.900 --> 00:28:04.500 We plug in numbers: 3 = 2 times...what is our A? It is -1, so -1, plus C; so we get 3 = -2 + C, or we add 2 to both sides: 5 = C. 00:28:04.500 --> 00:28:11.700 All right, so 5 = C; so at this point, we have managed to figure out everything that goes into our decomposition. 00:28:11.700 --> 00:28:20.700 So, we can write that decomposition out: Ax + B...we have -1 as our A, so -x...what is our B? 00:28:20.700 --> 00:28:32.300 Our B is 0, so it just isn't anything; x² + 2, plus Cx + D...our D was 0; our C was 5. 00:28:32.300 --> 00:28:39.600 So, it is going to be 5x/(x² + 2)². 00:28:39.600 --> 00:28:48.500 And we can't forget that x that was originally there; so x plus that...that is our entire partial fraction decomposition. 00:28:48.500 --> 00:28:50.900 It breaks down into that one right there. 00:28:50.900 --> 00:29:01.600 All right, the final example: this one is a rough one, but this is the absolute hardest you would end up coming across in class and a test. 00:29:01.600 --> 00:29:06.400 So, don't worry: this is basically the top difficulty you will have to deal with, probably. 00:29:06.400 --> 00:29:10.000 You might see something a little bit harder, but this is really about as hard as it is going to get. 00:29:10.000 --> 00:29:16.200 All right, so happily, they already factored it for us, and we have (t - 2)²(t² + 2). 00:29:16.200 --> 00:29:22.400 So, this is degree 2; this is also degree 2; so that combines to degree 4 on the bottom, degree 3 on the top. 00:29:22.400 --> 00:29:29.400 So, we are proper; our bottom is already factored; we just need to actually work through the decomposition now. 00:29:29.400 --> 00:29:41.600 We have -t³ + 8t² - 6t, over (t - 2)²(t² + 2). 00:29:41.600 --> 00:29:52.200 How does that decompose? A over our linear factor, t - 2, plus B over our second version of that linear factor, 00:29:52.200 --> 00:30:01.400 (t - 2)², plus Cx + D over our irreducible quadratic of t² + 2; great. 00:30:01.400 --> 00:30:14.200 At this point, we multiply both sides by the denominator, and we will get (-t³ + 8t² - 6t) = 00:30:14.200 --> 00:30:21.900 A times...it was hit with (t - 2)² + (t² + 2), so it is going to cancel out one of those (t - 2)'s, 00:30:21.900 --> 00:30:28.000 and it will be left being multiplied by (t - 2) and (t² + 2). 00:30:28.000 --> 00:30:38.300 Plus...B: (t - 2)² was in its denominator, so it is going to cancel out all of the (t - 2)² that hit it, but not the t² + 2 at all. 00:30:38.300 --> 00:30:41.900 So, we have t² + 2. 00:30:41.900 --> 00:30:49.300 And then finally, plus Cx + D, and it is going to have to go in parentheses, because it was hit by another entire thing. 00:30:49.300 --> 00:30:56.600 And it is going to get (t - 2)², but it did cancel out the (t² + 2) that hit it, because it had that in its denominator. 00:30:56.600 --> 00:30:59.300 At this point, we want to start simplifying things out. 00:30:59.300 --> 00:31:03.100 We have A times...what is t - 2 times t² + 2? 00:31:03.100 --> 00:31:16.900 We get t³...t times 2 becomes 2t...minus 2t², minus 4, 00:31:16.900 --> 00:31:33.600 plus Bt² + 2B + (Cx + D) times (t - 2)², which becomes t² - 4t + 4. 00:31:33.600 --> 00:31:39.200 All right, move a little bit to the left, so we have more room; it is still the same thing on the left side. 00:31:39.200 --> 00:31:59.800 We have At³ - 2At² + 2At - 4A (oops, sorry--I just switched colors there) 00:31:59.800 --> 00:32:06.000 plus Bt² + 2B...break onto a new line now, so I can see what is going on... 00:32:06.000 --> 00:32:12.400 Cx times t² will become C...oops, I made a mistake right from the beginning. 00:32:12.400 --> 00:32:21.700 It should not be x; it should be (I just stuck with what I have been used to)...the variable we are dealing with isn't x; it is t here. 00:32:21.700 --> 00:32:31.200 So, that should have been a t the whole time; sorry about that, but that is the sort of mistake you want to catch on your end, too, 00:32:31.200 --> 00:32:34.700 because we are dealing with t as our variable. 00:32:34.700 --> 00:32:38.400 So, while the form was with x, as before, that is because the variable was x before. 00:32:38.400 --> 00:32:42.700 But now, we are dealing with t as a variable, so the form needs to switch to using t. 00:32:42.700 --> 00:32:52.800 So, let's work this out: Ct times t² becomes Ct³; Ct times -4t becomes -4Ct²; 00:32:52.800 --> 00:33:05.300 Ct times +4 is + 4Ct; + Dt² - 4Dt + 4D; OK, we have a lot of stuff here. 00:33:05.300 --> 00:33:13.400 So, we can say that this is (-t³ + 8t² - 6t) = 00:33:13.400 --> 00:33:19.800 and we can put this in that form, again, of thing with exponent 3, exponent 2, exponent 1, exponent constant; 00:33:19.800 --> 00:33:47.400 minus 2At² + 2At - 4A; the next one is + Bt² + 2B + Ct³ - 4Ct² + 4Ct, 00:33:47.400 --> 00:33:56.500 and then + Dt² - 4Dt + 4D. 00:33:56.500 --> 00:34:01.700 So, from this, we will be able to figure out that all of our t³'s, our line of t³'s here, 00:34:01.700 --> 00:34:03.700 has to be the same as our line of t³'s here. 00:34:03.700 --> 00:34:08.400 So, we will be able to get things like -1 = A + C. 00:34:08.400 --> 00:34:13.100 Now, this is a lot of stuff to have to work through; we have 4 different variables; 00:34:13.100 --> 00:34:17.600 we are going to end up having different simultaneous linear equations that we are going to have to solve through. 00:34:17.600 --> 00:34:21.900 If you have done much work with simultaneous linear equations, you know that that is going to be kind of a pain to work through. 00:34:21.900 --> 00:34:26.900 So, at this point, we might think, "Oh, I'm lazy; is there anything clever I can do?" 00:34:26.900 --> 00:34:31.000 Is there a clever way to work through this--some little trick I could use? 00:34:31.000 --> 00:34:36.000 Well, if we look back to what we originally had when we multiplied out that denominator, 00:34:36.000 --> 00:34:46.300 we might think, "Oh, look, there is a t - 2 here; there is a t - 2 here; but there is an absence of t - 2 on our B." 00:34:46.300 --> 00:34:53.000 So, if we realize that, we could say, "Well, if we plug in t = 2, that would cause..." 00:34:53.000 --> 00:34:58.800 now, none of this stuff up here is going to be true; but we are going to plug in t = 2; 00:34:58.800 --> 00:35:04.700 and we are going to realize that, if we plug in t = 2, that causes this to turn to 0, which knocks out our A terms. 00:35:04.700 --> 00:35:10.400 And it will cause this here to turn to 0, as well, and also knock out our C and our D terms. 00:35:10.400 --> 00:35:17.000 And we will be left with just B times t² + 2, where t is 2--only t is 2-- 00:35:17.000 --> 00:35:22.400 and -t³ + 8t² - 6t, when t is 2. 00:35:22.400 --> 00:35:26.900 Now, this is true for any t; so that means we can plug in any t we want, 00:35:26.900 --> 00:35:34.100 and if there is a convenient way to get certain things to disappear by plugging in a cleverly chosen t, it is fair. 00:35:34.100 --> 00:35:38.400 Anything we can do to help us work through this--anything that makes it easier--is fair. 00:35:38.400 --> 00:35:43.900 So, in this case, we notice that by plugging in this carefully-chosen t, we can get certain things to disappear. 00:35:43.900 --> 00:35:52.600 Now, it is true for any t; this equation, this whole thing here, is true in general for any t, before we cross stuff out. 00:35:52.600 --> 00:36:00.800 But in the specific case where we plug in t = 2, certain things will cancel out, and we will be able to figure things out in a very easy way mathematically. 00:36:00.800 --> 00:36:12.700 So, we plug this in; and we know that, on our left side, we are going to have -2³ + 8(2)² - 6(2). 00:36:12.700 --> 00:36:19.800 That equals (sorry, there is not quite a lot of room here; I am going to draw in a line, just so we don't get confused)... 00:36:19.800 --> 00:36:25.900 That equals B times 2² + 2; OK. 00:36:25.900 --> 00:36:48.000 So, we have -2² is 8, plus 8(2)² is 4, so 32, minus 6 times 2 is 12, so - 12, equals B times 4 + 2, or 6. 00:36:48.000 --> 00:37:01.300 Simplify the left side: -8 - 12 is -20; 32 - 20 gets us 12, equals 6B; we divide both sides by 6, and we get 2 = B. 00:37:01.300 --> 00:37:07.400 So, by use of some cleverness, we were able to skip having to do four simultaneous linear equations, 00:37:07.400 --> 00:37:10.000 to where eventually we are effectively going to bring it down to 3. 00:37:10.000 --> 00:37:16.600 And we will be able to not have to get confused by this most complicated column, where we have four different things all going together. 00:37:16.600 --> 00:37:22.100 We can just be done with B; B is already figured out, which will be really helpful and make things easier on us later on. 00:37:22.100 --> 00:37:27.000 We could have solved out those four simultaneous linear equations, because each one will produce things. 00:37:27.000 --> 00:37:33.200 Remember how the first column produced -1 = A + C; each one of these columns will produce an equation. 00:37:33.200 --> 00:37:39.100 And from four simultaneous linear equations, solving for four variables, we will be able to do it; it is just kind of a pain. 00:37:39.100 --> 00:37:43.600 So, we came up with this clever way, and we were able to figure out 2 = B by just happenstance. 00:37:43.600 --> 00:37:51.100 If we plug in t = 2, it made everything but the B terms disappear, and that just left us with a single equation that was pretty easy to solve. 00:37:51.100 --> 00:37:55.300 And we figured out that 2 = B; all right, well, let's work through it now. 00:37:55.300 --> 00:38:01.800 So, we figured out B = 2, and now we have -t³ + 8t² - 6t = 00:38:01.800 --> 00:38:14.100 At³ + Ct³ - 2At² + Bt² - 4Ct² + Dt² + 2At + 4Ct - 4Dt - 4A + 2B + 4D; wow! 00:38:14.100 --> 00:38:20.700 OK, so at this point, let's put columns to columns; so -t³ goes to our At³ + Ct³. 00:38:20.700 --> 00:38:28.900 So, it must be that -1 is equal to A + C. 00:38:28.900 --> 00:38:36.400 We can do this column here; I accidentally cut through that negative sign--hopefully that wasn't too confusing. 00:38:36.400 --> 00:38:53.900 So, that one here lines up with the -6t; so we have that -6 (let's write it on a separate location) equals 2A + 4C - 4D. 00:38:53.900 --> 00:39:00.500 They all have t's showing up, so we just divide out the t's, and we are left with this simple linear equation that we can work with. 00:39:00.500 --> 00:39:07.200 And then, finally, the last one of constants: what is our constant? Our constant is 0. 00:39:07.200 --> 00:39:17.300 So, we have 0 = -4A + 2B + 4D; for extra credit, let's just see what this would have been. 00:39:17.300 --> 00:39:22.300 How many t²'s did we have? We had 8, so in our difficult-to-read yellow (I'll make it black, 00:39:22.300 --> 00:39:31.300 just because black is easier to read), 8 = -2A + B - 4C + D. 00:39:31.300 --> 00:39:34.800 We could have worked that one out, but I wanted you to see what it would have been. 00:39:34.800 --> 00:39:39.100 But we will actually end up having enough information, because we have three equations and three unknowns; 00:39:39.100 --> 00:39:41.700 so we have enough with the red, blue, and green things. 00:39:41.700 --> 00:39:48.900 So, at this point, let's figure out A first; so that means we need to solve for everything that isn't A, 00:39:48.900 --> 00:39:52.300 so that we can plug them in, get rid of everything else, and just have A. 00:39:52.300 --> 00:40:04.800 So, on the left side, we have -1 = A + C; so we solve for C, and we get -A - 1 = C. 00:40:04.800 --> 00:40:09.000 OK; on the green one, we plug in what we know for our B. 00:40:09.000 --> 00:40:30.200 We have 0 = -4A + 2(2 + 4D); 0 = -4A + 4 + 4D; 0 =...let's divide everything by 4, just to make things easier...-A + 1 + D. 00:40:30.200 --> 00:40:39.600 So, we move that stuff over, and we get: add A on both sides; subtract by 1 on both sides; A - 1 = D. 00:40:39.600 --> 00:40:59.700 So, we have A - 1 = D; we have -A - 1 = C; so we can plug this information into this equation, and we will be able to get something that is just using A. 00:40:59.700 --> 00:41:07.800 -6 = 2(A)...that actually stays around, because it was just A from the beginning; 4...what did we figure out C was? 00:41:07.800 --> 00:41:15.300 C is the same thing as -A - 1, minus 4 times...what did we figure out D is? 00:41:15.300 --> 00:41:29.300 It is the same thing as A - 1; simplify that out: -6 = 2A + 4(-A), so we will make that - 4A, minus 4... 00:41:29.300 --> 00:41:40.400 -4A - 4A...and -4 times -1 becomes + 4; so at this point, we see that we have -4 and +4, so those cancel each other out. 00:41:40.400 --> 00:41:53.600 We have -6 = 2A - 8A; -6 = -6A; and thus 1 (dividing both sides by -6) = A. 00:41:53.600 --> 00:41:58.000 With 1 = A in place, we can now easily go back and figure out everything else. 00:41:58.000 --> 00:42:13.000 We have -A - 1, so negative...plug it in, knowing that it is 1...minus 1, equals C; so we have -2 = C; great. 00:42:13.000 --> 00:42:24.800 And also, plug in over here; and we have that 1 - 1 = D; it turns out that D is just 0. 00:42:24.800 --> 00:42:29.400 So, at this point, we have figured out all of the things that we need. 00:42:29.400 --> 00:42:36.800 We have this set up as (t - 2) + (t - 2)² + (t² + 2). 00:42:36.800 --> 00:42:59.000 It was in the format -t³ + 8t² - 6t, all over (t - 2)²(t² + 2). 00:42:59.000 --> 00:43:05.700 OK, so that is the case; we have that as something over (t - 2). 00:43:05.700 --> 00:43:21.100 It was A/(t - 2) + B/(t - 2)² + (Cx + D)/(t² + 2). 00:43:21.100 --> 00:43:28.600 OK, at this point, we can actually plug in our numbers; we have that A is 1, so we have 1/(t - 2). 00:43:28.600 --> 00:43:44.000 B is 2, so we have 2/(t - 2)², plus...C was -2, so -2x, and 0 was D, so that part just disappears. 00:43:44.000 --> 00:43:53.100 So, we have -2x/(t² + 2); oops, once again, I did it again; I got to thinking in terms of x, as opposed to t. 00:43:53.100 --> 00:43:57.900 But we are using a different variable: -2t/(t² + 2). 00:43:57.900 --> 00:44:02.900 And there we are; we have decomposed it into its partial fractions. 00:44:02.900 --> 00:44:07.000 That is pretty long; it is pretty complicated; but that is pretty much as hard as it gets. 00:44:07.000 --> 00:44:10.700 As long as you break it down, and then you multiply everything out carefully, 00:44:10.700 --> 00:44:16.300 and you are careful with all of your algebra and your arithmetic, you can get it into this form right here, 00:44:16.300 --> 00:44:22.500 where you have this big block compared to these original things, and then you are able to figure out all of these linear equations. 00:44:22.500 --> 00:44:26.900 And you can possibly be clever and figure out some way to figure out B = 2. 00:44:26.900 --> 00:44:32.800 But you also can just work through a bunch of linear equations, solving for one thing at a time in terms of the other, 00:44:32.800 --> 00:44:35.200 plugging them all together, and then solving it out. 00:44:35.200 --> 00:44:37.700 Sometimes it takes a lot of work; sometimes it goes kind of slowly. 00:44:37.700 --> 00:44:40.600 But ultimately, if you just keep working at it, you can get it. 00:44:40.600 --> 00:44:43.000 And it is really, really useful in calculus. 00:44:43.000 --> 00:44:45.600 I know--I know that you aren't going to be using it immediately. 00:44:45.600 --> 00:44:51.500 But honestly, this thing makes a problem that would be, otherwise, totally impossible to do, just really easy. 00:44:51.500 --> 00:44:53.600 So, it is a really handy thing in calculus. 00:44:53.600 --> 00:44:56.000 All right, we will see you at Educator.com later--goodbye!