12 - The Factor Theorem, Part 1 (Factoring Polynomials in Algebra) - By Math and Science
Transcript
| 00:00 | Hello . Welcome back today , we're going to conquer | |
| 00:03 | one of the most important theorems in all of algebra | |
| 00:05 | . I think it's right up there with the quadratic | |
| 00:08 | formula or any of the other most famous equations you | |
| 00:10 | can think of in math and that is called the | |
| 00:12 | factor theorem . In fact we've been using the factor | |
| 00:15 | theorem from the very beginning of our study of polynomial | |
| 00:18 | . I just we didn't really called it the factor | |
| 00:20 | theorem , but now that we have enough um learning | |
| 00:23 | under our belt , especially in terms of how to | |
| 00:25 | divide polynomial is we can formalize what we have already | |
| 00:28 | been using and we call it the factor theorem . | |
| 00:30 | What we're basically going to be doing is talking about | |
| 00:33 | , you know how we have polynomial is like uh | |
| 00:36 | x squared plus two X plus five or something . | |
| 00:38 | And we know how to factor those polynomial is when | |
| 00:41 | they're fact trouble . And if you want to solve | |
| 00:43 | the equation equal to zero , you factor those polynomial | |
| 00:47 | is if you can and you set each little parentheses | |
| 00:50 | equal to zero . So we've been factoring and we've | |
| 00:52 | been setting equal to zero for the purpose of what | |
| 00:54 | we call it , Finding the roots . When you | |
| 00:56 | set the polynomial equal to zero , you set the | |
| 00:59 | little parentheses equal to zero . We call it finding | |
| 01:01 | the roots . So we've been dealing with the concept | |
| 01:04 | of the roots of a polynomial and also the factors | |
| 01:07 | of the polynomial , the parentheses terms . We've been | |
| 01:09 | dealing with those for dozens of of lessons by now | |
| 01:14 | . Since ever since we started talking about quadratic . | |
| 01:16 | Now we're going to formalize it and talk about the | |
| 01:19 | concept of the factor theorem . How do you know | |
| 01:22 | when something really is a factor of a polynomial ? | |
| 01:24 | Also if you have one factor of a polynomial , | |
| 01:28 | how do you find more factors ? Because obviously factoring | |
| 01:31 | things is a really easy way to solve things . | |
| 01:33 | So we want to know how to find as many | |
| 01:35 | factors as possible . So first let's write the factor | |
| 01:38 | theorem down . Then we're going to explain it , | |
| 01:40 | talk about it , do some examples and go from | |
| 01:42 | there . So we have this thing again , one | |
| 01:45 | of the most important uh theorems and all of all | |
| 01:48 | of algebra , certainly theorem , whoops , can't spell | |
| 01:51 | theorem , right ? And I have to write it | |
| 01:55 | down . Unfortunately , it's going to take a second | |
| 01:56 | . It's not super long , but I have to | |
| 01:58 | write it down because we have to talk about so | |
| 02:00 | we talk about the polynomial polly P of X . | |
| 02:05 | That means the polynomial P of X has the factor | |
| 02:09 | X minus C as a factor . Remember these factors | |
| 02:15 | are little parentheses terms all multiplied together to give you | |
| 02:17 | P fx right ? So it has x minus C | |
| 02:20 | sees a number , this means x minus two , | |
| 02:22 | X plus three . X minus four . Or anything | |
| 02:24 | like that . It has this whatever it is as | |
| 02:26 | a factor if and only if So it's a necessary | |
| 02:33 | and sufficient condition if and only if the number C | |
| 02:38 | is a root of the equation . Mhm . P | |
| 02:47 | of x . The polynomial equal to zero . In | |
| 02:49 | other words , what this means is that the polynomial | |
| 02:52 | evaluated at sea must be equal to zero . All | |
| 02:55 | right . This is a lot of words and I | |
| 02:56 | know a lot of students look at words like this | |
| 02:58 | and then their brain freezes up and they don't know | |
| 03:00 | . They don't even read the words because they look | |
| 03:02 | really scary . What this is saying is that if | |
| 03:04 | you know the roots of a polynomial , which we've | |
| 03:07 | been finding roots forever . We can get the roots | |
| 03:09 | from the quadratic formula . We can get the roots | |
| 03:11 | from completing the square . We can get the roots | |
| 03:13 | for factoring right . But however you get them , | |
| 03:16 | if you know what the roots are , then you | |
| 03:18 | automatically know all of the factors of the equation . | |
| 03:22 | In other words , all those little parentheses terms , | |
| 03:24 | Those are called the factors . They're all multiplied together | |
| 03:26 | to equal a polynomial . It's obviously very important for | |
| 03:29 | us to be able to find the factors of polynomial | |
| 03:31 | in order to be able to solve them . If | |
| 03:33 | you know the routes . In other words , if | |
| 03:35 | you know the the values that make the polynomial go | |
| 03:39 | to zero , IFC happens to be a root meaning | |
| 03:42 | if it's a root it means you stick it in | |
| 03:44 | the polynomial and it makes it equal zero , then | |
| 03:47 | x minus whatever values you have as those roots are | |
| 03:50 | all factors right ? So that's really , really , | |
| 03:53 | really important because if you remember we've been factoring these | |
| 03:55 | polynomial forever using all these different methods and we even | |
| 03:59 | developed the completing the square in the quadratic formula to | |
| 04:01 | find all the routes to be able to solve these | |
| 04:04 | things . And along the way we've been having a | |
| 04:06 | factor of these things as well . So the factor | |
| 04:09 | theorem ties together what we've been doing before , but | |
| 04:11 | it formalizes it . So what I want to do | |
| 04:13 | is continue talking about this whole thing , forget about | |
| 04:16 | these words . For now we're gonna we're gonna bring | |
| 04:18 | it down with some concrete example . So here's an | |
| 04:20 | example of what this factor theorem really says . Let's | |
| 04:24 | say you have a polynomial and it's very concrete , | |
| 04:27 | it's X squared plus four X plus three . So | |
| 04:31 | we know we can try to factor this and set | |
| 04:34 | it equal to zero and find the roots . Right | |
| 04:36 | ? And so if we're going to find the roots | |
| 04:39 | , that's exactly what we're gonna do . Let's find | |
| 04:41 | these routes the roots . How do we typically do | |
| 04:45 | it ? Well we always you can use the quadratic | |
| 04:47 | formula if you want , but we always try to | |
| 04:49 | factor this thing if we can . So we're gonna | |
| 04:52 | say X squared plus four X plus three equals zero | |
| 04:56 | . That's what it means . To find the roots | |
| 04:57 | . You set the polynomial equal to zero . Then | |
| 05:00 | we do our little binomial thing that we always do | |
| 05:03 | . We have an X squared . So it's X | |
| 05:04 | times X . Three . Can only be multiplied by | |
| 05:08 | one times three to give you three . And then | |
| 05:09 | we have to have plus signs here . Inside gives | |
| 05:12 | you X . Outside gives you three X . They | |
| 05:13 | add to four X . And so you know , | |
| 05:15 | we've been doing this enough that this is the factored | |
| 05:17 | form of this . So in this example I'm going | |
| 05:20 | in a different direction to show you um in this | |
| 05:24 | case we already know how to factor them . So | |
| 05:26 | when we say , what are the factors of this | |
| 05:28 | polynomial ? What we're asking is this is a factor | |
| 05:32 | and this is a factor a factor . Remember just | |
| 05:35 | recalling from the past . A factor is just a | |
| 05:38 | fancy word . That means , tell me the things | |
| 05:40 | that multiply together to give you that thing that you're | |
| 05:43 | looking at . So the factors of 12 , there's | |
| 05:45 | lots of factors of 12 , but three is a | |
| 05:47 | factor of 12 , and four is a factor of | |
| 05:49 | 12 . Why ? Because three times four is 12 | |
| 05:52 | , One is a factor of 12 , and 12 | |
| 05:54 | is a factor of 12 . Why ? Because one | |
| 05:57 | times 12 is 12 to is a factor of 12 | |
| 05:59 | , and six is a factor of 12 . Why | |
| 06:01 | ? Because two times six is 12 . You see | |
| 06:03 | factors are just the things that you can multiply together | |
| 06:05 | to give you the thing you're talking about . So | |
| 06:07 | this polynomial , we've already known how to factor it | |
| 06:10 | forever and we now know that this is a factor | |
| 06:12 | and this is a factor . But what the factor | |
| 06:14 | theorem is telling you is going the other direction . | |
| 06:18 | The factors that you get are only factors if you | |
| 06:22 | find the roots of the polynomial and then the factors | |
| 06:25 | are always going to be able to form X minus | |
| 06:27 | whatever routes you have . So let's continue going down | |
| 06:30 | the road here . We know how to factor this | |
| 06:32 | and we can then say that X plus one is | |
| 06:34 | equal to zero , so that one of the roots | |
| 06:36 | is equal to negative one . And then we know | |
| 06:38 | that X plus three is equal to zero . So | |
| 06:40 | the other route Is equal to -3 . Okay , | |
| 06:44 | so what we know from doing these problems over and | |
| 06:47 | over again , we all we always told you factor | |
| 06:49 | at first and then set this equal to zero and | |
| 06:52 | then set this equal to zero . And those are | |
| 06:54 | going to give you the two routes , right ? | |
| 06:56 | But more importantly than that , the factor theorem basically | |
| 07:00 | says that X is equal to , I shouldn't say | |
| 07:03 | the factors here . Um let me just go down | |
| 07:05 | here . We've found the roots , we know that | |
| 07:07 | ex uh is X equals negative one is a route | |
| 07:13 | . Why ? It's a route ? Because if we | |
| 07:15 | stick it into the polynomial , you should get zero | |
| 07:18 | . That's the whole point . We solved them and | |
| 07:19 | we found them by factory . Okay . But if | |
| 07:21 | I just gave you if I didn't do these middle | |
| 07:24 | steps , if I said this is a route and | |
| 07:25 | this is how would you figure it out ? You | |
| 07:27 | would stick it in here and see if it equals | |
| 07:29 | zero . That's what you would do . So you | |
| 07:31 | put the -1 in here and you get -1 quantity | |
| 07:34 | squared Plus four times -1 . And then plus the | |
| 07:38 | three . Right ? And so what you get right | |
| 07:42 | here is a one minus four plus three . You | |
| 07:45 | can see the one plus three is a four and | |
| 07:47 | then minus four is you get a zero . So | |
| 07:49 | we know that X is equal to one is a | |
| 07:51 | root . Yes , we did it by the way | |
| 07:52 | that we've learned in the past , but we also | |
| 07:54 | know that it's a route because when we put the | |
| 07:56 | number into the polynomial , it makes it equal zero | |
| 07:59 | . That's what the definition of a route is . | |
| 08:01 | All right . We know that X is equal to | |
| 08:04 | negative three is also a route . Because when we | |
| 08:08 | stick that number into the equation , we also get | |
| 08:10 | zero . So negative three quantity squared plus four times | |
| 08:14 | negative three plus three . What do we get ? | |
| 08:16 | This is negative three square , so that's nine and | |
| 08:19 | then minus 12 plus three . And you can see | |
| 08:22 | that the nine plus the three is 12 minus 2 | |
| 08:24 | . 12 . You get zero . So these are | |
| 08:26 | both routes . How do you know their roots ? | |
| 08:28 | Of course we solved it the way we've always done | |
| 08:30 | . But the real way that you know you have | |
| 08:33 | these routes is because it's the only two numbers that | |
| 08:36 | I can put into this equation and get zero as | |
| 08:38 | an answer there . The crossing points , if you | |
| 08:40 | were to graph this quadratic where they cut into the | |
| 08:43 | X axis where Y is equal to zero . All | |
| 08:46 | right , So the factor theorem is going a step | |
| 08:49 | beyond that . The factor theorem is telling you if | |
| 08:52 | you know the roots of the equation , you know | |
| 08:55 | the values of C . That make this thing when | |
| 08:58 | you put them in their equal to zero . In | |
| 09:00 | this case the value of C is negative one and | |
| 09:02 | the other value of C is negative three . You | |
| 09:04 | have those two values of C . Then you automatically | |
| 09:07 | know the factors of this polynomial X minus C . | |
| 09:10 | And then the other factors also X minus C . | |
| 09:13 | Okay , so the factor theorem says the factor theorem | |
| 09:19 | says That simply because we know that X is equal | |
| 09:24 | to negative one is a route , then we know | |
| 09:27 | just because of that fact without having factored it , | |
| 09:30 | without having knowing anything about it , that x minus | |
| 09:35 | a negative one . Or we say X plus one | |
| 09:40 | is a factor . Mhm . Right . Because the | |
| 09:44 | form of it is if you know what value of | |
| 09:47 | C makes the thing go to zero , then it's | |
| 09:48 | x minus that number . This number happened to be | |
| 09:51 | negative , so it's x minus that number , which | |
| 09:52 | means it's X plus that number . All right . | |
| 09:54 | And the factor theorem also says in this case because | |
| 09:57 | we know that a route is X is equal to | |
| 09:59 | negative three is a route , then X minus negative | |
| 10:04 | three , which means X plus three is a factor | |
| 10:10 | . All right . So the bottom line is if | |
| 10:13 | I were to give you if I never gave you | |
| 10:17 | the polynomial at all . If I never gave you | |
| 10:20 | the polynomial , if you didn't know that this polynomial | |
| 10:22 | have those roots , and I just said There is | |
| 10:26 | a polynomial out there . It has a root of | |
| 10:28 | -1 . It has a root of -3 . Right | |
| 10:32 | down the polynomial . What would you do ? The | |
| 10:34 | factor theorem tells you that since , you know this | |
| 10:37 | is a route that X minus that , which is | |
| 10:39 | this is a factor . And then since you know | |
| 10:42 | this is a root X minus that , which is | |
| 10:43 | this is also a factor . And that because you | |
| 10:46 | know , factors are things that just multiply together . | |
| 10:49 | Then you know that the equation X plus one , | |
| 10:52 | X plus three . This is the polynomial that you | |
| 10:56 | seek . And when you multiply all this out , | |
| 10:58 | X times X is x squared inside terms X . | |
| 11:01 | Outside terms is three X . And last terms is | |
| 11:05 | three . And what you get is x squared plus | |
| 11:08 | four X plus three . Which is exactly what we | |
| 11:11 | started with . You see when we start learning algebra | |
| 11:14 | , we don't learn about the factor theorem . We | |
| 11:16 | teach you how to factor these , try no meals | |
| 11:18 | and set them equal to zero and find the roots | |
| 11:20 | . The factor theorem is telling you , I don't | |
| 11:22 | even care if you know how to factor anything . | |
| 11:24 | I don't care if . I've never taught you anything | |
| 11:26 | about factoring if you know the roots of a polynomial | |
| 11:29 | , you already know the factors . That's what it's | |
| 11:32 | saying . If you know the factor is negative one | |
| 11:34 | , negative three , I know that . These are | |
| 11:36 | the factors just by the factor theorem , factors are | |
| 11:38 | things that multiply together to give me whatever it is | |
| 11:41 | I'm studying . So if I multiply those factors together | |
| 11:44 | , this must be the polynomial in question . Or | |
| 11:47 | at least one of the polynomial in question . We've | |
| 11:49 | talked in the past about how there could be a | |
| 11:51 | family of polynomial with these same roots , but this | |
| 11:54 | polynomial certainly does have those routes . All right now | |
| 11:58 | , I told you that there's two main things I | |
| 12:00 | wanted to really emphasize about the factor theorem . The | |
| 12:03 | first one was that if you know the routes is | |
| 12:06 | exactly what we've been talking about . If you know | |
| 12:07 | the routes , then you know what the factors of | |
| 12:09 | that thing really are . Right ? That's the first | |
| 12:11 | thing , is probably the most important thing . But | |
| 12:14 | another part of the factor theorem that we're going to | |
| 12:16 | do when we get to the problems a little more | |
| 12:18 | , is that the other part of it is if | |
| 12:20 | you know only one of the roots , I'm sorry | |
| 12:23 | , only one of the factors , then you can | |
| 12:26 | actually find more factors by doing division . Let me | |
| 12:30 | say that again . If I only give you one | |
| 12:33 | of the root , so , you know one of | |
| 12:34 | the factors , but you don't know the other one | |
| 12:36 | , let's say I only give you that the route | |
| 12:38 | is this one , I never told you that 3 | |
| 12:40 | -3 was the route , right . It's possible to | |
| 12:43 | take this information . And by using long division , | |
| 12:46 | figure out what the other factor is , right ? | |
| 12:49 | And that's not so obvious at first and it's not | |
| 12:51 | even really written down in here , but it's really | |
| 12:54 | , really important . So , before showing you how | |
| 12:56 | to do it with the algebra , I want to | |
| 12:58 | talk about numbers for a second because it's gonna make | |
| 13:01 | it way easier to understand if you remember back , | |
| 13:05 | You should remember that three times four is equal to | |
| 13:08 | 12 . Right ? That's obvious . Right ? What | |
| 13:11 | three and 4 are these are called factors , factors | |
| 13:15 | or numbers that multiply together to give me what we | |
| 13:17 | care about . So , factors of 12 . Of | |
| 13:20 | course , there are more factors of 12 . Like | |
| 13:22 | I said , two times 61 times 12 , 3 | |
| 13:24 | times four , you know that kind of thing ? | |
| 13:26 | There's lots of factors , But it's perfectly fine to | |
| 13:29 | say that three and four of both factors , because | |
| 13:31 | you can essentially what it means to be a factor | |
| 13:34 | is you can divide it into 12 and get a | |
| 13:37 | whole number back . That's that's one way of looking | |
| 13:39 | at it . Another way is thinking that the factors | |
| 13:40 | can be multiplied together to give you 12 . That | |
| 13:43 | the reverse way of thinking about it is factors are | |
| 13:45 | things that can be divided into 12 evenly . And | |
| 13:48 | what I mean by that is think about what this | |
| 13:51 | means , that three times four is equal to 12 | |
| 13:53 | . What it means is that if I take 12 | |
| 13:58 | and I divided by three , you already know that | |
| 14:01 | 12 divided by three is four . Look at what | |
| 14:03 | happened here , If I only knew that three was | |
| 14:05 | a factor of 12 and I do this division , | |
| 14:08 | then I immediately figure out that four is also a | |
| 14:11 | factor of 12 because again , three times four is | |
| 14:14 | 12 . Right ? Also , If I only know | |
| 14:18 | for instance that four is a factor of 12 and | |
| 14:22 | I know that this divides and makes it three , | |
| 14:24 | then I found the other factors . So this allows | |
| 14:28 | us to I can find more factors . So Part | |
| 14:37 | one of the factor theorem is telling you about if | |
| 14:40 | you know what the route is , you know what | |
| 14:41 | the factors are . The other part is . Even | |
| 14:44 | if you don't know all the factors , you can | |
| 14:46 | still use division to find the remaining factors for the | |
| 14:50 | case of numbers . If I just told you literally | |
| 14:53 | , if I just said hey Um three is a | |
| 14:56 | factor of 12 , find another factor , then I | |
| 15:00 | would say , okay , well what a factor is | |
| 15:01 | is I mean you can think of it as multiplication | |
| 15:03 | but in reverse is also true . It factors of | |
| 15:05 | the numbers that when I divide in they give me | |
| 15:07 | a whole whole number of divisions with no decimals . | |
| 15:11 | So I divided and I get a four . So | |
| 15:12 | this must be a factor and I try it and | |
| 15:14 | I divided and I see that I get the three | |
| 15:16 | back , that's how multiplication works . Two times three | |
| 15:18 | is 63 times two is six . Six divided by | |
| 15:21 | two is 36 divided by three is to you see | |
| 15:23 | how it all works out . So if I know | |
| 15:25 | only one of the factors , I can find the | |
| 15:26 | other one just by division . Now it's easy to | |
| 15:29 | understand what numbers and I'm gonna use that to make | |
| 15:32 | it easy to understand in algebra . Okay , so | |
| 15:36 | for instance if I know if I know the following | |
| 15:42 | that this polynomial p of x is x squared plus | |
| 15:46 | four , X plus three . And I know that | |
| 15:50 | it factors as X plus one And X-us three . | |
| 15:54 | I know this why because we just did it right | |
| 15:56 | . I've shown you that when you multiply all this | |
| 15:58 | out , that's what you get . So this is | |
| 16:00 | the factored form of this . So what this means | |
| 16:02 | is of course you can kind of mentally cover this | |
| 16:04 | up . What it's telling you is this polynomial here | |
| 16:07 | is exactly equivalent to this factored form . They are | |
| 16:11 | the same thing . It's like they look different obviously | |
| 16:15 | , but one half is exactly the same thing as | |
| 16:18 | 5/10 , which is exactly the same thing as 51 | |
| 16:22 | hundred's . You get the idea so they look different | |
| 16:25 | but they're the same thing . So this is exactly | |
| 16:27 | the same thing as this . So what this means | |
| 16:29 | because these things are exactly equivalent , this is an | |
| 16:32 | equation , right ? I could take this , I | |
| 16:33 | could divide by the x minus plus one if I | |
| 16:36 | want . So what it's saying is if I take | |
| 16:38 | the X squared plus the four X plus the three | |
| 16:42 | . And I do long division . I divide by | |
| 16:44 | this the X plus one . I mean remember this | |
| 16:47 | is an equation . What it's saying is if I | |
| 16:49 | divide here then I should get X plus three as | |
| 16:52 | an answer . Because I mean you can see it's | |
| 16:54 | an equation . If I divide it then this is | |
| 16:56 | what should be left . So let's divide it and | |
| 16:57 | see what happens . X times X is X squared | |
| 17:00 | . Do the multiplication . I'm going to get an | |
| 17:02 | X plus I'm sorry X squared here . Plus X | |
| 17:06 | . I do a subtraction like this . All right | |
| 17:09 | . Do a subtraction . 4 -1 . X . | |
| 17:11 | is going to give me three x . All right | |
| 17:15 | . And then I dropped my new number down which | |
| 17:17 | is three whoops . Making a plus three . Yes | |
| 17:20 | . And then X times what is three X ? | |
| 17:22 | I have to have a three . So I multiply | |
| 17:24 | get a three X plus three because I multiply this | |
| 17:27 | , I subtract these both subtract to give me zero | |
| 17:30 | , there is no remainder . So factors are things | |
| 17:34 | that divide in and there's no remainder . Right factors | |
| 17:37 | of 12 are numbers that divide into 12 with no | |
| 17:41 | remainder . Uh 12 divided by three is four , | |
| 17:43 | no remainder . 12 divided by uh sorry 12 divided | |
| 17:46 | by three is four , no remainder . 12 divided | |
| 17:48 | by four is three . No remainder 12 divided by | |
| 17:51 | six would be too with no remainder . 12 divided | |
| 17:54 | by one would be 12 with no remainder , same | |
| 17:57 | sort of thing . They're all factors because you can | |
| 17:58 | divide them in and you get no remainder . So | |
| 18:01 | because you already know the factored form in this example | |
| 18:04 | , I'm using it to show you and prove to | |
| 18:05 | you that you can take polynomial . If you know | |
| 18:07 | one of the factors , you can divide it and | |
| 18:10 | get the other factor . So what this is telling | |
| 18:13 | you is that if I only knew that this was | |
| 18:15 | a factor , I could divide it in and I | |
| 18:17 | would get the other factor X plus three . So | |
| 18:19 | these are both factors . This is a factor because | |
| 18:21 | I knew that to begin with and this is a | |
| 18:24 | factor , which I just found by division . Okay | |
| 18:29 | , let's go the other way and check it out | |
| 18:31 | and see what happens if I take this polynomial again | |
| 18:33 | . Uh Where do I want to do it ? | |
| 18:34 | Yeah , I'll do it right here . Now let's | |
| 18:36 | yeah let's do it right here . Um X squared | |
| 18:40 | plus four , X plus three here are divided by | |
| 18:43 | X plus one . Now we'll divide by X plus | |
| 18:46 | three . We're gonna divide this one and see what | |
| 18:49 | we get . We know that if we divide by | |
| 18:51 | X plus that we should get this thing back . | |
| 18:53 | So X times X gives me X squared , multiply | |
| 18:56 | get X squared plus three X . Subtract four minus | |
| 19:00 | three is one . X . Drop the three down | |
| 19:03 | X times one is X . And then plus three | |
| 19:08 | you can see again I get a zero remainder , | |
| 19:10 | I should always give a zero remainder back . So | |
| 19:13 | this thing was a factor because I knew it was | |
| 19:15 | a factor to begin with . And when I divided | |
| 19:18 | , I figured out that this was a factor extremely | |
| 19:22 | important for you to understand this concept because oftentimes a | |
| 19:26 | problem in algebra will go something like this , A | |
| 19:28 | polynomial , blah blah blah blah blah has a factor | |
| 19:32 | X plus one . Find the other factor . And | |
| 19:35 | you're gonna be like how do I do that ? | |
| 19:37 | I have no idea how to do that . All | |
| 19:38 | you have to do is take the polynomial and divide | |
| 19:41 | by the factor that you know the way that you | |
| 19:43 | can do that is You know that if it's a | |
| 19:44 | factor it must be that these things are all multiplied | |
| 19:48 | together . So find the other factor is a simple | |
| 19:50 | matter of division . Right ? Just like we do | |
| 19:52 | with numbers , we could do the same thing with | |
| 19:54 | polynomial . We're gonna do a lot of that in | |
| 19:56 | terms of the problem . So don't get too crazy | |
| 19:59 | worried about it uh yet because we're going to do | |
| 20:01 | it a lot . So , you know , I'm | |
| 20:03 | not done talking about it . But for our last | |
| 20:06 | problem of this lesson to kind of wrap it all | |
| 20:07 | up , I want to ask you a question is | |
| 20:10 | X plus one . A factor of the polynomial . | |
| 20:18 | X to the seventh power minus X . To the | |
| 20:21 | fifth power plus x to the third power by this | |
| 20:24 | X . So the question is , is X plus | |
| 20:26 | one a factor of this polynomial ? Right ? So | |
| 20:30 | half of the battle with this stuff is to figure | |
| 20:32 | out what it's saying . So what it's asking , | |
| 20:35 | it is asking the fallen uh if we can Right | |
| 20:45 | as the following , basically when it's a factor , | |
| 20:47 | what it's asking is , is the following true is | |
| 20:51 | P of X . This polynomial , Can it be | |
| 20:55 | written as X plus one times some other function of | |
| 21:00 | that ? Some other junk here . In other words | |
| 21:02 | , if it's a factor that it's going to be | |
| 21:05 | able to be multiplied by something else , so , | |
| 21:07 | what what the other factors are ? I don't even | |
| 21:09 | know . But it's gonna be able to multiply by | |
| 21:10 | other factors to give me this polynomial . That's what | |
| 21:13 | it's basically saying . When something is a factor cannot | |
| 21:16 | be written as a string of things multiplied together to | |
| 21:18 | give me what I'm asking for back . And the | |
| 21:20 | factor theorem tells us that the polynomial has this as | |
| 21:26 | a factor if and only if this value of C | |
| 21:29 | when it's x minus C evaluates that P evaluated at | |
| 21:33 | C is equal to zero . In other words it | |
| 21:34 | has to be a roof . What has to be | |
| 21:37 | a route ? Well there's a number one in there | |
| 21:38 | , but the theorem has written is x minus C | |
| 21:41 | . So what we have to figure out is P | |
| 21:43 | of negative one equals question mark zero . Why negative | |
| 21:47 | one ? Because the theorem is written in terms of | |
| 21:49 | x minus the number , but the number was positive | |
| 21:52 | . So it was it was like x minus a | |
| 21:54 | minus one , right ? Um And so we have | |
| 21:57 | to evaluate it . Let me put it another way | |
| 21:59 | . If the question said , is x minus one | |
| 22:02 | factor , then it's written exactly as the theorem says | |
| 22:04 | , and you would evaluate P of one because it's | |
| 22:06 | written as x minus one . But the the thing | |
| 22:09 | says X plus one , so you have to flip | |
| 22:10 | the sign . So however you want to think about | |
| 22:12 | it works for me . We have to put it | |
| 22:14 | in there and figure out if it equals zero . | |
| 22:15 | So we're gonna put the negative one , raise it | |
| 22:18 | to the seventh power negative one to the fifth power | |
| 22:22 | negative one to the third power negative one . Now | |
| 22:26 | when you have an odd power you're always on a | |
| 22:28 | negative guy , you're always gonna get a negative answer | |
| 22:30 | . So this is negative one minus . This is | |
| 22:32 | an odd power . So this is a negative one | |
| 22:34 | , this is a plus sign , this is an | |
| 22:36 | odd power . So it's a negative one . And | |
| 22:38 | then this is a positive one because negative times negative | |
| 22:40 | positive . Now we have double negative here so we | |
| 22:43 | have negative one plus one . This becomes a negative | |
| 22:46 | one plus one . So this zero and this is | |
| 22:51 | zero . So when you get zero plus zero is | |
| 22:54 | zero . So what you did is you looked at | |
| 22:57 | the factor , you picked the number one out . | |
| 22:59 | If it was x minus one you would evaluated one | |
| 23:02 | . But since it was X plus one you have | |
| 23:04 | to evaluate negative one because the therapy is written in | |
| 23:06 | terms of x minus whatever . And when you put | |
| 23:08 | it in there you get basically you figure out that | |
| 23:11 | this is a route . So because it's -1 is | |
| 23:14 | a root of the equation , X plus one is | |
| 23:18 | a factor . If you were to have plugged in | |
| 23:22 | that value and gotten three or two or negative one | |
| 23:25 | or 17 or anything else then it would not be | |
| 23:27 | a factor . Alright , so we've done a lot | |
| 23:30 | in this lesson and there's actually going to be four | |
| 23:33 | including this 14 different lessons on this . We're not | |
| 23:36 | done with the factor theorem . As I said , | |
| 23:38 | it's one of the most important theorems in algebra because | |
| 23:40 | it talks about solving polynomial , which is one of | |
| 23:43 | the most common types of equations that we encounter in | |
| 23:45 | all of science and all of math and all the | |
| 23:48 | physics and chemistry and engineering . Right ? So what | |
| 23:50 | it's basically saying is if you know the routes the | |
| 23:53 | values that make this polynomial go to zero , then | |
| 23:55 | you automatically know the factors , Right ? If you | |
| 23:58 | know that one makes this thing go to zero , | |
| 24:01 | then X -1 is a factor . If you know | |
| 24:03 | that five makes this thing go to zero , then | |
| 24:05 | X -5 is a factor . If you know that | |
| 24:08 | negative one makes this thing go to zero , then | |
| 24:11 | x plus one is a factor , right ? Because | |
| 24:13 | x minus of minus one , it's got to be | |
| 24:15 | X plus one in that case . So if you | |
| 24:17 | know the routes , you know the factors and secondarily | |
| 24:20 | because what we talked about the concept of what a | |
| 24:22 | factor is and it's things multiplied together . If I | |
| 24:25 | only know one of the factors , I can divide | |
| 24:27 | and get the other one . So I showed with | |
| 24:30 | this here , for instance , we know we happen | |
| 24:31 | to know the factors here . But if I only | |
| 24:33 | gave you one of these uh roots , I'm sorry | |
| 24:36 | , one of these factors , I gave you the | |
| 24:38 | polynomial , I only gave you one factor and said | |
| 24:39 | find the other one , then you would say , | |
| 24:41 | okay , I'm gonna do division because that's what factors | |
| 24:44 | are when you divide , you're gonna you're gonna get | |
| 24:46 | uh You're going to get a remainder of zero , | |
| 24:49 | because it has divided even number of times , just | |
| 24:51 | like three , divides into 12 . And even number | |
| 24:53 | of times , you're always gonna get a zero remainder | |
| 24:55 | and whatever is up here is the other factor . | |
| 24:57 | We did it both ways to show how that works | |
| 24:59 | . So now that you have an idea of what | |
| 25:01 | the factor theorem is , follow me on to the | |
| 25:03 | next lesson where we're gonna crank through a bunch of | |
| 25:05 | problems . They're almost all gonna center around giving you | |
| 25:08 | the roots , and then you tell me the factors | |
| 25:10 | or maybe only give you one route and then you | |
| 25:12 | have to divide and you give me the other factors | |
| 25:14 | and so on . So let's go ahead and start | |
| 25:16 | tackling working the problems with the factors here . Um | |
| 25:19 | Right now |
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