08 - Learn Synthetic Division of Polynomials - Part 1 - By Math and Science
Transcript
| 00:00 | Hello . Welcome back to algebra . The title of | |
| 00:02 | this lesson is called synthetic division . Part one . | |
| 00:05 | We're gonna have several lessons here increasing the problem complexity | |
| 00:08 | . But the most important point here in the beginning | |
| 00:10 | is to really understand what synthetic division is , why | |
| 00:13 | we use it and to understand where it comes from | |
| 00:16 | . I could just give you the kind of the | |
| 00:18 | procedure and you could do it , but I want | |
| 00:20 | you to know where it's actually coming from . So | |
| 00:22 | in the last few lessons we learned how to divide | |
| 00:24 | polynomial . And if you're anything like me or like | |
| 00:27 | most students , you pretty much find long division of | |
| 00:29 | polynomial is to be a pain . Uh it's just | |
| 00:31 | because there's a lot of things to write down , | |
| 00:33 | a lot of cumbersome subtractions and so on . So | |
| 00:36 | there is a method to do that long division in | |
| 00:39 | a simpler way for certain kinds of problems . I | |
| 00:42 | need to say that one more time . There's a | |
| 00:43 | way to do that long division in a much much | |
| 00:46 | simpler way , but only for certain kinds of problems | |
| 00:49 | . There's a certain type of division where you can | |
| 00:52 | use this uh concept called synthetic division to make it | |
| 00:55 | simpler . Specifically , if you're dividing by the very | |
| 00:58 | specific form , if you're dividing something by like X | |
| 01:02 | minus a number or X plus a number , then | |
| 01:05 | you can use this synthetic division . So for instance | |
| 01:07 | , if you're dividing some polynomial , if you're dividing | |
| 01:10 | it by x minus one , you can use synthetic | |
| 01:12 | division . If you're taking some polynomial divided by x | |
| 01:14 | minus 10 , you can use synthetic division . If | |
| 01:17 | you're taking some polynomial , dividing it by X plus | |
| 01:20 | three or X plus seven or X plus or anything | |
| 01:23 | like that , X plus a number or X minus | |
| 01:25 | the number . If you're dividing by that , you | |
| 01:27 | can use synthetic division and it's much much faster uh | |
| 01:30 | than the long division . So you cannot use synthetic | |
| 01:33 | division for problems where for instance you're dividing by X | |
| 01:37 | squared plus to the X squared , kills it . | |
| 01:39 | You can't use synthetic division . If you're dividing by | |
| 01:41 | X cubed minus nine , you can't use synthetic division | |
| 01:44 | . So when you're divided by X plus a number | |
| 01:47 | or X minus a number , we can make our | |
| 01:48 | lives much much much easier by doing this . So | |
| 01:51 | what we're gonna do now is I'm gonna do a | |
| 01:53 | long division problems to kind of keep it in our | |
| 01:55 | mind and we're gonna do that long division problem in | |
| 01:58 | parallel by using the synthetic division procedure . So you | |
| 02:00 | can see where it comes from . And then we'll | |
| 02:02 | work a few problems , additional problems in synthetic division | |
| 02:05 | so that you can get some practice . So before | |
| 02:08 | I can introduce the synthetic division , we need to | |
| 02:10 | do one quick problem in long division . And I | |
| 02:12 | know that we just did this uh in the last | |
| 02:15 | lesson . But it's really really really really helpful to | |
| 02:18 | have a problem in your mind when we do the | |
| 02:21 | synthetic division . So this one is not going to | |
| 02:23 | be hard . So we're gonna take the polynomial two | |
| 02:27 | X cubed minus seven X squared plus zero X plus | |
| 02:33 | five . Notice we have a zero X . So | |
| 02:35 | remember when we do division , we have to pad | |
| 02:37 | any missing powers . We have an X cubed X | |
| 02:40 | squared X . To the first . And then no | |
| 02:41 | X at all . We have to pad any missing | |
| 02:44 | X's variables there with uh with a zero there and | |
| 02:48 | we're gonna divide that by x minus three . Now | |
| 02:50 | notice right away that this problem is set up . | |
| 02:52 | I'm going to do it in the long division format | |
| 02:54 | . But because we're dividing by X minus and number | |
| 02:57 | or alternatively X plus a number . I'll show you | |
| 02:59 | in a few minutes . Then we know we can | |
| 03:01 | use synthetic division . So we will do that in | |
| 03:03 | just a minute . Um So in order to crank | |
| 03:07 | through the long division procedure , what do we do | |
| 03:09 | ? We take a look at the first term here | |
| 03:11 | in the first term here , X times something has | |
| 03:13 | to give me this . So there's a one out | |
| 03:15 | here . So one times two is two and X | |
| 03:18 | times something is X cubed . This has to be | |
| 03:20 | an X squared . All right now we go backwards | |
| 03:23 | and we multiply the two times the one gives me | |
| 03:26 | a two and the X squared times the X . | |
| 03:29 | Gives me X to the third power . And then | |
| 03:33 | the two times the negative three is gonna give me | |
| 03:35 | negative six . And then the X squared times one | |
| 03:37 | is X squared . Now I need to subtract both | |
| 03:40 | of these from their partners above . So I'm gonna | |
| 03:41 | put a minus sign . I'm gonna put a circle | |
| 03:43 | around it to remind me that I'm subtracting directly from | |
| 03:46 | above . And by the way if you have none | |
| 03:49 | of this looks familiar to you then it's just because | |
| 03:51 | you haven't watched my lessons on long division . So | |
| 03:54 | if this looks crazy already then go back and watch | |
| 03:56 | those lessons . I have entire lessons explaining every little | |
| 03:58 | part of this process . So we have the two | |
| 04:01 | X . Cubed minus two to execute . That's gonna | |
| 04:03 | give me a zero . So I don't need to | |
| 04:05 | really write it there . But this negative seven X | |
| 04:07 | squared subtracting a negative six X squared really becomes negative | |
| 04:12 | seven minus minus six . So negative seven minus a | |
| 04:16 | negative six is basically negative seven plus six . So | |
| 04:19 | it's gonna be negative one . I'll put it like | |
| 04:22 | this -1 X . Squared . So you have to | |
| 04:25 | be careful when you do this attraction negative seven minus | |
| 04:27 | a minus becomes negative seven plus six . So once | |
| 04:30 | I've done the subtraction I take my next digit and | |
| 04:33 | I bring it down which is zero X . And | |
| 04:36 | then I start again X . Times something will give | |
| 04:38 | me the negative one X . Square . It has | |
| 04:40 | to be a negative one X . So I take | |
| 04:43 | and multiply this direction I'm gonna get negative one X | |
| 04:46 | . Squared . And then the negative one X . | |
| 04:48 | Times the negative three is positive three X . Again | |
| 04:52 | I need to subtract I'll put a negative sign there | |
| 04:54 | with a circle to remind me that I'm doing both | |
| 04:56 | terms . This minus this gives me 00 minus three | |
| 05:00 | gives me negative three X . Gives me negative three | |
| 05:04 | X . And then after I've done this I dragged | |
| 05:07 | my next term down which is a five . And | |
| 05:10 | I look at the first terms X times something gives | |
| 05:12 | me negative three X . I have to have a | |
| 05:14 | negative three here and then I multiply . This is | |
| 05:16 | gonna give me negative three X . And then the | |
| 05:18 | negative three times the negative gives me positive nine . | |
| 05:21 | Almost done here , we subtract both terms . These | |
| 05:24 | give me 05 minus nine , gives you negative four | |
| 05:28 | . Okay , Gives You -4 . So we have | |
| 05:32 | done this numerous times before and now that we have | |
| 05:35 | the whole number of times this can be divided in | |
| 05:37 | and the remainder part , the answer to this problem | |
| 05:41 | is what we have written above here two x squared | |
| 05:44 | minus x minus three . And then we have a | |
| 05:47 | fractional remainder part which is negative four over what we | |
| 05:51 | have out here X minus three . So this is | |
| 05:53 | kind of the remainder is a fraction of what you're | |
| 05:55 | dividing by . So this is kind of the fractional | |
| 05:57 | part of the answer we talked about that before . | |
| 06:00 | This is kind of the whole number part of the | |
| 06:02 | answer as we talked about before . So this is | |
| 06:04 | the answer now , why am I doing this ? | |
| 06:05 | Because in order to explain what long division is you | |
| 06:08 | have to have in your mind a concrete example from | |
| 06:11 | , I'm sorry , in order to explain what synthetic | |
| 06:13 | division is you really need to have in your mind | |
| 06:15 | . What a problem feels like from long division . | |
| 06:18 | We can all agree that this is a real pain | |
| 06:20 | . What are the real problems with this ? The | |
| 06:23 | first problem is I have to write all these exes | |
| 06:25 | down everywhere . There's excess everywhere . It's just completely | |
| 06:27 | filled with X X squared x cubes . And so | |
| 06:30 | uh I know we're doing it to keep track of | |
| 06:33 | it , but that's obviously something that's not good . | |
| 06:35 | Another thing that's not good is that uh this subtraction | |
| 06:39 | process is really cumbersome because you have to be careful | |
| 06:42 | when you're doing negative six minus a minus six . | |
| 06:45 | It's really negative . I'm sorry negative seven minus of | |
| 06:47 | minus . It's really negative seven plus six . And | |
| 06:50 | so it's very very easy to make sign errors when | |
| 06:53 | you're subtracting negative numbers and they're all spread out like | |
| 06:56 | that . That's the number two things . So synthetic | |
| 06:58 | division is a faster way to do this process . | |
| 07:01 | It's actually lightning fast . You don't have to worry | |
| 07:03 | about the crazy subtraction . You don't have to do | |
| 07:06 | that . And also as a bonus you don't have | |
| 07:08 | to write all those X . Is everywhere . So | |
| 07:10 | the problems don't take up the whole page . So | |
| 07:13 | what we're gonna do now is we're gonna talk about | |
| 07:15 | the concept of synthetic division . Send fucking spell synthetic | |
| 07:24 | division . Right ? So that take division . Right | |
| 07:29 | ? Alright . So I've already told you a few | |
| 07:31 | times , but basically you can only use it . | |
| 07:33 | It's a faster process . But it lets us to | |
| 07:35 | divide when the divisor that means what you're dividing by | |
| 07:41 | must B . I'm gonna write it as X minus | |
| 07:46 | a number . But keep in mind this number can | |
| 07:48 | be positive or negative . So when I say x | |
| 07:50 | minus a number , it can really be like x | |
| 07:53 | minus two . I can divide by this . I | |
| 07:55 | can divide by X minus 19 . That's fine . | |
| 07:58 | I can also divide by X plus three . I | |
| 08:00 | can also divide by X plus , you know whatever | |
| 08:02 | I want 18 . Right ? So when you see | |
| 08:04 | x minus see it doesn't mean it has to be | |
| 08:06 | x minus something . Consider that C can be positive | |
| 08:09 | or negative . So really it's better to say that | |
| 08:12 | you can use synthetic division when you divide by X | |
| 08:14 | plus a number or x minus a number . All | |
| 08:17 | right . So what does it look like first one | |
| 08:18 | ? I'm not gonna show you how to do it | |
| 08:20 | yet . I'm gonna show you what it will look | |
| 08:21 | like and then I'm gonna show you how to do | |
| 08:23 | it . Here's what you do this three out here | |
| 08:27 | . See how it's X minus of three . In | |
| 08:29 | synthetic division , it's X minus three . You're dividing | |
| 08:31 | by but you don't write the minus three . You | |
| 08:33 | write it as a positive three and then everything on | |
| 08:36 | the inside that you're dividing by . You write all | |
| 08:38 | the numbers down , right ? But you don't write | |
| 08:40 | these these variables now . So I'm gonna show you | |
| 08:43 | how to do it and we're gonna talk about it | |
| 08:44 | . Uh Just a second . So the to the | |
| 08:46 | negative 70 and the five . And then instead of | |
| 08:50 | doing a traditional division bar you kind of go upside | |
| 08:53 | down and kind of right it upside down . You'll | |
| 08:55 | see why in just a minute . So you can | |
| 08:57 | see when we write the problem statement like this . | |
| 08:59 | All of the important information is contained because all polynomial | |
| 09:03 | are going to have these coefficients even if there is | |
| 09:05 | a zero . We have to include it when we're | |
| 09:07 | doing a long division . So really we're just not | |
| 09:09 | writing all these exes everywhere . But we know that | |
| 09:11 | this has to be a two X cubed because this | |
| 09:14 | has to be constant . So this has to be | |
| 09:16 | zero X . So this has to be negative seven | |
| 09:19 | X squared . So this has to be two times | |
| 09:20 | X cubed . So you kind of go backwards and | |
| 09:22 | you can reconstruct what the polynomial is just by the | |
| 09:25 | numbers . All right now , I'm not going to | |
| 09:28 | show you how to do it yet , but we | |
| 09:29 | go through this process that I'm gonna show you in | |
| 09:31 | just a minute . And what you're gonna end up | |
| 09:32 | with is with some numbers down here . Uh Right | |
| 09:36 | . And then you're gonna have a line here and | |
| 09:39 | then you're gonna have some numbers down below . Yes | |
| 09:41 | , I'm not I'm not expecting you to understand this | |
| 09:44 | yet . I'm just showing you how it is set | |
| 09:46 | up . So what happens is when you do the | |
| 09:48 | process , which is very simple . What you end | |
| 09:51 | up with is the first three numbers that you get | |
| 09:53 | , or I should say , let me let me | |
| 09:54 | back up and say this way , the last number | |
| 09:56 | that you get this is the remainder notice I got | |
| 10:02 | a negative for their the remainder in our long division | |
| 10:04 | was actually a negative for all of the numbers to | |
| 10:07 | the left of this are basically the whole number part | |
| 10:10 | . So this is the remainder , This is the | |
| 10:12 | rest of the guy here and this means two x | |
| 10:16 | squared minus x minus three . So notice this polynomial | |
| 10:20 | is what we got the two X squared minus one | |
| 10:22 | minus three . That's what we got up here . | |
| 10:24 | The remainder in long division is way down at the | |
| 10:26 | bottom and then you have to reconstruct the answer as | |
| 10:29 | we've done many times in synthetic division . When you | |
| 10:32 | do the process , the very last number you get | |
| 10:34 | on the right is the remainder and then the numbers | |
| 10:36 | to the left represent the whole the whole number part | |
| 10:39 | of the polynomial . The rest of the answer , | |
| 10:40 | basically which we can reconstruct here . All right . | |
| 10:44 | Um And then of course I've already said it to | |
| 10:48 | you before but I'll just kind of like pointed out | |
| 10:51 | 100% clear this represents two X cubed minus seven X | |
| 10:57 | squared plus zero X plus five . That's what that | |
| 11:00 | represents . And then this represents x minus three . | |
| 11:03 | So if you have X minus three , you put | |
| 11:05 | a positive three that you're dividing by . If you | |
| 11:07 | end up divided by X plus five , then you | |
| 11:09 | have to put a negative five . So whatever the | |
| 11:11 | sign is that you're dividing by . You put the | |
| 11:12 | opposite sign in your synthetic division . Alright . I | |
| 11:16 | don't expect you to know anything about how synthetic division | |
| 11:18 | works yet . I haven't taught you that all I'm | |
| 11:20 | showing you is that when you write all the numbers | |
| 11:22 | down , this is what they mean . The number | |
| 11:24 | on the outside of the little houses what you're dividing | |
| 11:26 | by . But you have to use the opposite sign | |
| 11:29 | . The numbers on the inside of the house is | |
| 11:31 | what you're dividing into . And then the numbers on | |
| 11:33 | the bottom represent the remainder , along with the rest | |
| 11:36 | of the answer . As we have done in long | |
| 11:37 | division . All right . So now it's time to | |
| 11:42 | you see where I want to do this . Mm | |
| 11:46 | I think now it's time to actually do this problem | |
| 11:52 | , right ? And show you where it comes from | |
| 11:54 | . So I'm gonna rewrite everything in purple here . | |
| 11:56 | I'm not gonna write the answer down . I'm just | |
| 11:57 | gonna rewrite the problem statement here in purple . Uh | |
| 12:00 | as if we're dividing these polynomial together are into one | |
| 12:04 | another . So I have a three outside the house | |
| 12:06 | . And then I had a to a negative 70 | |
| 12:09 | and a five . And you draw like an upside | |
| 12:12 | down division , simple sort of right ? Here's how | |
| 12:14 | you do it . You write this problem statement down | |
| 12:17 | , and then you draw a line there and then | |
| 12:20 | you draw a little arrow . I do this anyway | |
| 12:21 | . And you need to drop down the first number | |
| 12:24 | . Okay , the first thing I want to show | |
| 12:26 | you actually I forgot to mention it to you is | |
| 12:29 | one very very important thing . I'm sorry I didn't | |
| 12:31 | mention this before when you're doing long division . Obviously | |
| 12:35 | everything written on here is important . But when you're | |
| 12:37 | doing this subtraction and you're working through this black part | |
| 12:40 | here , you see how the twos are subtracting . | |
| 12:42 | The negative ones are subtracting , the negative threes are | |
| 12:44 | subtracting and so on . The most important numbers . | |
| 12:48 | Um Are these numbers boxed in red ? Um You | |
| 12:52 | have uh you have a negative six . Let me | |
| 12:55 | just double check myself . You have a negative six | |
| 12:58 | . You have a three . Uh and you have | |
| 12:59 | a nine here . Why are these the most important | |
| 13:01 | numbers ? Because these are the numbers that you get | |
| 13:04 | basically uh when you when you do the multiplication down | |
| 13:08 | and then you need to subtract . Right ? So | |
| 13:10 | these numbers give you the next line Uh in the | |
| 13:14 | process . So notice that the -6 , the three | |
| 13:16 | and the nine are also present in the synthetic division | |
| 13:19 | here , but they're opposite signs , right ? So | |
| 13:22 | whereas when you were divided by X -3 , we | |
| 13:26 | had to subtract all of these polynomial . And that's | |
| 13:28 | why the subtraction was really cumbersome . When we instead | |
| 13:31 | of putting X -3 , when we write this as | |
| 13:33 | a positive three , then what happens is these numbers | |
| 13:36 | there become opposite signs of what they were here . | |
| 13:39 | So we don't have to do any subtraction anymore . | |
| 13:41 | We end up adding . In fact , you can | |
| 13:43 | kind of see I haven't really shown you the process | |
| 13:46 | yet , but you can kind of see right here | |
| 13:47 | that the negative 67 plus six gives me this this | |
| 13:50 | plus this gives me this and this plus this gives | |
| 13:52 | me this . So one of the main main advantages | |
| 13:55 | of synthetic division is that you no longer have to | |
| 13:57 | subtract . You can just add things together . Which | |
| 13:59 | is much much easier to do . Yeah . All | |
| 14:03 | right . So let's get to the process . We | |
| 14:05 | drag the first number down . That's our starting point | |
| 14:08 | , right ? And then what we wanna do , | |
| 14:11 | I'm not gonna do this for every problem . But | |
| 14:12 | just to explain it to you , I'm gonna draw | |
| 14:13 | a little arrow through here , right ? And in | |
| 14:17 | this era I'm gonna write three times two . Because | |
| 14:19 | what I'm doing is I'm taking three times two and | |
| 14:22 | I'm gonna write it in this position which is six | |
| 14:25 | . That's what this means three times to arrives in | |
| 14:27 | the sixth position . Then I just take negative seven | |
| 14:33 | . Plus six negative seven plus six is negative one | |
| 14:36 | . Right ? Once this number is in place , | |
| 14:39 | then I'm gonna again kind of carve through the one | |
| 14:42 | position here like this and say three times the negative | |
| 14:45 | +13 times the negative one which is negative three , | |
| 14:49 | which goes in the next position right here . And | |
| 14:53 | then I add these together . Zero plus a negative | |
| 14:56 | three is negative three . And then once I have | |
| 14:59 | this in position , then the same thing happens here | |
| 15:02 | . I go through here and it's gonna be three | |
| 15:05 | times a negative three , Which is -9 Like this | |
| 15:10 | . And then I add these guys together and it's | |
| 15:12 | -4 . So then I have arrived at the answer | |
| 15:15 | notice that's what I showed you here . I said | |
| 15:16 | I just didn't put all these arrows everywhere . I | |
| 15:18 | said you wrote the problem statement down there's some intermediate | |
| 15:21 | number six , negative three negative nine . That's where | |
| 15:23 | they come from . And then you have the answers | |
| 15:26 | on the bottom . The very last number here is | |
| 15:29 | the uh remainder . And then everything before that is | |
| 15:33 | the actual polynomial . Now , when you're writing the | |
| 15:36 | answer down from synthetic vision , you know , this | |
| 15:39 | is an X cubed polynomial because this is a constant | |
| 15:42 | term , This is an X term , This is | |
| 15:44 | an X squared term , This is an X cubed | |
| 15:46 | term . So because this is an X cubed term | |
| 15:49 | , you know that this has to be a two | |
| 15:50 | X squared term . You know , this has to | |
| 15:53 | be an X term . You know , this has | |
| 15:55 | to be a constant term and then this is the | |
| 15:57 | remainder which is negative four over whatever I divided by | |
| 16:01 | , which we know is x minus three because it's | |
| 16:03 | opposite signs what we put up there . How do | |
| 16:06 | you know that ? The answer you right , is | |
| 16:07 | actually squared here . It's because when you look back | |
| 16:10 | to the long division problem , that's why I wanted | |
| 16:13 | to do it first . We're dividing into a cubic | |
| 16:17 | and we're dividing by x minus something . So you | |
| 16:19 | always start have a answer that's gonna be one degree | |
| 16:22 | lower than what you're dividing into . When you're dividing | |
| 16:25 | by X minus something , right ? You can see | |
| 16:28 | the answer was X squared when we're dividing into an | |
| 16:31 | X cube . So when we're doing synthetic division , | |
| 16:33 | because we always have to divide by X plus number | |
| 16:36 | X minus and number the answer you get is always | |
| 16:39 | gonna be one degree lower than what you're dividing into | |
| 16:42 | . Okay , so here's the procedure , you write | |
| 16:45 | down the numbers of the polynomial , dividing into . | |
| 16:47 | You write down what you're dividing by but you have | |
| 16:49 | to put a different sign , the opposite sign of | |
| 16:52 | what you're dividing by . We were divided by X | |
| 16:53 | -3 . So we put a three there . We | |
| 16:56 | dropped the two down three times two is six . | |
| 16:58 | Okay great add these together we get a one . | |
| 17:00 | Okay , three times negative one is negative three . | |
| 17:03 | Great . We add these , we get negative 33 | |
| 17:05 | times a negative three is negative nine . We add | |
| 17:08 | these , we get negative four . All the numbers | |
| 17:10 | in the bottom represent your answer and notice we didn't | |
| 17:12 | have to do any subtractions . The reason we didn't | |
| 17:15 | have to do any subtractions is because we write the | |
| 17:17 | problem while we're dividing by we basically change the sign | |
| 17:21 | instead of X -3 . We divide by X-plus three | |
| 17:24 | . That ends up with the effect of making all | |
| 17:26 | of these numbers opposite in sign to the long division | |
| 17:30 | . So we don't have to subtract them anymore . | |
| 17:31 | We can just add them . That's the reason why | |
| 17:33 | it works . But procedurally you just multiply add . | |
| 17:37 | Multiply add . Multiply add . You have the answer | |
| 17:40 | here . This is the remainder divided by what we're | |
| 17:42 | dividing by . This is the fractional part of the | |
| 17:45 | remainder of the remainder and then this is the the | |
| 17:47 | whole part of the polynomial answer . All right . | |
| 17:51 | Now that you have the procedure in place , we | |
| 17:54 | can go crank through two more problems that are gonna | |
| 17:58 | be really , really fast . Now let's say that | |
| 18:00 | you're doing three x cubed minus five X squared plus | |
| 18:05 | x minus two . And we're dividing that by x | |
| 18:08 | minus two . Now , if you were doing long | |
| 18:11 | division , you would have to write this whole thing | |
| 18:13 | out and then do all the stuff with the subtraction | |
| 18:16 | . And it would be basically take the whole board | |
| 18:17 | , right ? But since we're doing long synthetic division | |
| 18:21 | , we don't do that . We take the numbers | |
| 18:23 | . We take away the numbers and strip away all | |
| 18:25 | the variables . And we write it under this house | |
| 18:27 | here . So three is the first number negative . | |
| 18:30 | Five is the next 11 is the third one and | |
| 18:33 | negative two is the last one . Now we have | |
| 18:36 | cube square first power , no power . So we're | |
| 18:38 | not missing any terms . So we don't need any | |
| 18:41 | zeros here . But if we were missing , for | |
| 18:43 | instance the X squared term , if it was just | |
| 18:45 | three X cubed plus X minus two and this one | |
| 18:47 | were gone , then we would have to put a | |
| 18:49 | zero here would be +3012 You have to pad the | |
| 18:52 | zero same as long division . Then you put an | |
| 18:55 | upside down house and you're dividing by X -2 . | |
| 18:58 | But you don't put a -2 here , you always | |
| 19:00 | flip the side . You can think of it like | |
| 19:02 | this in synthetic division were allowed to add instead of | |
| 19:05 | subtract . So when you're dividing by something you just | |
| 19:07 | flip the sign of whatever it is and notice it | |
| 19:09 | has to be X minus something or X plus something | |
| 19:12 | in order to even do this at all . Yeah | |
| 19:15 | . All right now we're ready to do a very | |
| 19:17 | simple process . We take the three . We draw | |
| 19:19 | a little era . We drop it down three times | |
| 19:22 | two gives me six . No need to draw those | |
| 19:24 | arrows . Now we know what we're doing three times | |
| 19:26 | two is six . Then we add these negative five | |
| 19:29 | plus six is just the number one , Then two | |
| 19:31 | times 1 is two . Then we add these to | |
| 19:35 | to give us three , then two times three is | |
| 19:37 | six . Then we add these to give me a | |
| 19:40 | positive four . Let me double check myself 3134 Now | |
| 19:44 | I think you can agree that this process is way | |
| 19:48 | more cumbersome and complicated than this process . So again | |
| 19:53 | , just one more time from the top . Drop | |
| 19:54 | the first number down three times two is six . | |
| 19:56 | Ad get a 12 times one is to add get | |
| 20:00 | a 33 times two is six . Ad get a | |
| 20:03 | four and you always know that this number at the | |
| 20:06 | end is the remainder , everything else before it is | |
| 20:10 | the rest of the answer basically . So the way | |
| 20:13 | you're going to write it is since you know that | |
| 20:15 | this is a cubic your this represents the cube term | |
| 20:18 | , the X squared term , the X term and | |
| 20:20 | the negative two term . Then the answer here has | |
| 20:22 | to be three X squared plus X plus three because | |
| 20:25 | it has to be one degree less than what you | |
| 20:28 | were dividing into . And then you have the four | |
| 20:31 | as a remainder divided by the x minus two . | |
| 20:34 | This is the whole answer . Three X squared plus | |
| 20:37 | x plus three Plus fraction for over X -2 . | |
| 20:42 | That's all done . Synthetic division is a really , | |
| 20:46 | really huge time saver . And there's also other other | |
| 20:49 | uses of synthetic division . We're gonna discuss a little | |
| 20:51 | bit later . Right now we're just doing it just | |
| 20:53 | to divide polynomial . But there's actually some other uses | |
| 20:56 | that we can use this technique for a little bit | |
| 20:58 | later . Now what we're gonna do is our final | |
| 21:01 | problem just to get one more bit of practice . | |
| 21:03 | Let's say that we have the polynomial we want to | |
| 21:06 | divide is as follows , X cubed Plus three , | |
| 21:10 | x squared minus two times x minus six . And | |
| 21:14 | we're gonna divide all that by X plus three . | |
| 21:18 | So you have to ask yourself , am I divided | |
| 21:20 | by x minus a number or X plus a number | |
| 21:22 | ? I am divided by X plus the number . | |
| 21:24 | So , I'm allowed to use synthetic division . If | |
| 21:26 | I were allowed . If this problem were different . | |
| 21:29 | If it were all this junk divided by X squared | |
| 21:32 | plus three , then you can't do it . You | |
| 21:34 | can't use synthetic division . You have to go back | |
| 21:36 | and use the long division method . That works for | |
| 21:38 | everything . Okay ? If you're dividing by X cubed | |
| 21:42 | plus three three X minus two , you can't do | |
| 21:45 | it . It only works when it's X plus a | |
| 21:47 | number or X minus the number . All right . | |
| 21:50 | So then what we're gonna do is write down the | |
| 21:52 | relevant information . This is a cube , a square | |
| 21:55 | of first power . And in the no power we | |
| 21:56 | don't have any missing terms . So it's just gonna | |
| 21:59 | be one , three negative to negative six . Draw | |
| 22:03 | my upside down a little house and then we're divided | |
| 22:06 | by X . Plus three . You put the opposite | |
| 22:08 | sign of whatever you're dividing by their . That is | |
| 22:11 | really what allows us to do addition into this attraction | |
| 22:13 | . So that's the way to remember it . Whatever | |
| 22:15 | you divide by just change the sign of it . | |
| 22:17 | And then you draw your horizontal line and you drop | |
| 22:21 | the number one down here and then you say three | |
| 22:23 | times negative three times the one is negative three . | |
| 22:26 | I add these when I get a zero And then | |
| 22:29 | three times 0 is zero . Uh And I add | |
| 22:32 | these and I get a negative two and then the | |
| 22:34 | three times negative two is positive six . I add | |
| 22:37 | those and I get a zero . So I get | |
| 22:39 | 10 negative two . And zero . Now notice the | |
| 22:41 | interesting thing here is the remainder is actually equal to | |
| 22:45 | zero . So this is one of those examples where | |
| 22:47 | if I had done this long division way , I | |
| 22:50 | would get all the way down to the bottom . | |
| 22:51 | I would do my final subtraction . I would get | |
| 22:53 | a zero . So I would not have a fractional | |
| 22:55 | part because it would be the fractional part will be | |
| 22:58 | zero over over this , which is zero . So | |
| 23:02 | I can write the answer down even easier than before | |
| 23:04 | . I know that it's a cubic dividing by . | |
| 23:08 | So this has to be X squared one X squared | |
| 23:11 | plus three . X . I hope so . I'm | |
| 23:13 | writing the wrong thing down . I'm sorry about that | |
| 23:16 | . I'm sorry about that . What I have here | |
| 23:19 | is an answer is one X squared plus zero x | |
| 23:23 | minus two . This is the answer . Or you | |
| 23:25 | can just write it as X squared minus two , | |
| 23:28 | X squared minus two . So the answer is X | |
| 23:30 | squared minus two . All right . So that's the | |
| 23:33 | process of synthetic division . It makes our long division | |
| 23:35 | problems much much easier to solve . The only thing | |
| 23:38 | we have to do is remember that when we're dividing | |
| 23:40 | by whatever we're dividing by , it can only be | |
| 23:43 | X plus the number , or X minus the number | |
| 23:45 | , and we have to write the opposite sign in | |
| 23:47 | front of the synthetic division symbol . After that . | |
| 23:50 | This process is really , really fast , right ? | |
| 23:53 | And we have other uses for it . We have | |
| 23:54 | other theorems that use synthetic division . I'm gonna show | |
| 23:56 | you a little bit later , but for right now | |
| 23:59 | use it to solve your long division problems . Whenever | |
| 24:02 | you're dividing by X minus the number , or X | |
| 24:04 | plus a number , solve all of these problems . | |
| 24:06 | Follow me on to the next lesson . We're gonna | |
| 24:07 | get a little bit more practice with synthetic division . |
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