20 - The Rational Root Theorem, Part 1 (Rational Roots of Polynomials) - By Math and Science
Transcript
| 00:00 | Hello . Welcome back to algebra . The title of | |
| 00:02 | this lesson is called finding rational roots of polynomial . | |
| 00:06 | Part one . I could also title this the rational | |
| 00:09 | root theorem Part one . It's a really important theorem | |
| 00:12 | of algebra Again we're talking mostly about solving polynomial . | |
| 00:15 | You all know that we can solve quadratic polynomial is | |
| 00:18 | just fine . Oftentimes we can solve cubic polynomial is | |
| 00:21 | if we can guess one of the roots and then | |
| 00:23 | we remember we did it when we learned the factor | |
| 00:25 | here and we can divide that route in to find | |
| 00:28 | what's left over and solve what's left over and find | |
| 00:30 | the other two . But when you start getting into | |
| 00:32 | higher order polynomial is like fifth and six order polynomial | |
| 00:35 | . We don't have good tools to find all of | |
| 00:37 | those roots . Concretely we did learn in the last | |
| 00:40 | lesson . Deckert rule of signs which tells us the | |
| 00:43 | different possibilities we have for the positive and the negative | |
| 00:47 | and the complex roots . But it didn't help us | |
| 00:49 | actually find the roots . Okay , so here's kind | |
| 00:51 | of the other half of that . We're going to | |
| 00:53 | learn a technique to help us start to find some | |
| 00:57 | of the roots of these higher order polynomial and it | |
| 01:00 | still doesn't get us all the way there , but | |
| 01:01 | at least it gets us pretty close . All right | |
| 01:04 | . So this is called the rational root theorem . | |
| 01:06 | And the reason that it's called that is because it | |
| 01:08 | helps us find the rational roots of polynomial . It | |
| 01:11 | doesn't help us find the complex roots . It doesn't | |
| 01:14 | help us find the irrational roots . It just helps | |
| 01:16 | us find the rational roots . So let's review for | |
| 01:19 | a quick second uh that you are with me and | |
| 01:23 | you understand what a rational number is basically a rational | |
| 01:29 | number as you've learned before , is just a number | |
| 01:31 | . Any number that can be written as a fraction | |
| 01:33 | . If it can be written as a fraction , | |
| 01:35 | it can be written it is called a rational number | |
| 01:38 | . Right ? So the number two is actually a | |
| 01:40 | rational number . Why ? Well because you can write | |
| 01:43 | to as to over one , you can write that | |
| 01:45 | as a fraction . So it's a rational number . | |
| 01:47 | The number seven is a rational number . The number | |
| 01:50 | negative three is a rational number . Why ? Because | |
| 01:52 | I can write it as negative 3/1 . Right ? | |
| 01:54 | And then of course you have the actual fractions that's | |
| 01:56 | obviously written as a fraction . So it's a rational | |
| 01:58 | number one half negative one half whatever positive negative doesn't | |
| 02:02 | matter as long as it can be written as a | |
| 02:04 | fraction . So any whole number that you can think | |
| 02:07 | of can be written as it is a rational number | |
| 02:09 | . Any fraction of course is a rational number . | |
| 02:12 | So most numbers are rational to some degree or other | |
| 02:15 | . But we have of course other numbers that we | |
| 02:17 | have learned about called irrational numbers . And those are | |
| 02:23 | special numbers like pi this number goes on and on | |
| 02:26 | and on . The decimal never repeats . And so | |
| 02:28 | you cannot write pi as a fraction I should say | |
| 02:31 | . You can't write pie is a fraction of whole | |
| 02:33 | numbers . When I say you can write these as | |
| 02:35 | a fraction I'm saying can we write them as a | |
| 02:37 | fraction of whole numbers . Pie of course is the | |
| 02:39 | ratio of a circle's , You know , circumference and | |
| 02:43 | diameter . But but those are not gonna ever be | |
| 02:45 | whole numbers to give you pie , you're not gonna | |
| 02:47 | be able to get pie by dividing two whole numbers | |
| 02:49 | . Uh some people say 22/7 is pretty close to | |
| 02:52 | pie . That's true , it's pretty close but it's | |
| 02:54 | not equal to pi right . Some other examples square | |
| 02:57 | root of two . You know we've gotten routes that | |
| 02:59 | are square root of two or minus square root of | |
| 03:01 | two and the polynomial these are irrational . Square root | |
| 03:04 | of three . I can go on and on . | |
| 03:05 | Lots of radicals end up becoming irrational . If I | |
| 03:08 | put these numbers on a calculator or their negative counterparts | |
| 03:11 | like negative square root of three , the decimals are | |
| 03:13 | gonna go on and on forever . They're never going | |
| 03:15 | to end . And when that happens you cannot write | |
| 03:17 | those numbers as fractions . So they're irrational . So | |
| 03:20 | they're different , slightly different class of numbers . The | |
| 03:22 | rational root theorem that we're going to talk about does | |
| 03:25 | not help us figure out routes that are radicals like | |
| 03:28 | this or pie or anything like that . Doesn't help | |
| 03:31 | us with that . It doesn't help us with complex | |
| 03:33 | numbers . It just helps us figure out what routes | |
| 03:35 | we have that can be written as a fraction , | |
| 03:38 | but that turns out to be a pretty common thing | |
| 03:40 | . So it is very useful . All right . | |
| 03:42 | So , we have to first talk about this theorem | |
| 03:44 | and then we're gonna do a problem to show you | |
| 03:46 | how to use it . So this thing is called | |
| 03:47 | a rational route theorem . Really important , especially as | |
| 03:57 | you get higher up in terms of of higher order | |
| 04:01 | polynomial , is because we don't have concrete ways of | |
| 04:03 | finding the roots of those higher order ones . So | |
| 04:06 | the theory goes like this for any polynomial polynomial with | |
| 04:13 | integral coefficients . What I mean by interval , I'm | |
| 04:20 | talking about the polynomial itself has to have , you | |
| 04:23 | know , one negative two negative five is coefficient . | |
| 04:26 | The polynomial can't have one half as a coefficient or | |
| 04:28 | 1/7 as a coefficient . But any polynomial that you | |
| 04:31 | have with fractional coefficients , you can just multiply through | |
| 04:34 | the whole thing and clear any of those fractions out | |
| 04:36 | . So you can always get it into the form | |
| 04:38 | of a polynomial with integral coefficients . So let's give | |
| 04:42 | an example of a polynomial like this . So for | |
| 04:45 | example a polynomial with integral coefficients , like this might | |
| 04:48 | be six x cubed plus seven X squared minus seven | |
| 04:54 | X minus three equals zero . So we might want | |
| 04:57 | to find the roots of this equation . We know | |
| 04:59 | we're gonna have three of them . Uh And I'm | |
| 05:01 | using a third order just to show you what we're | |
| 05:03 | doing . But this whole process works for any order | |
| 05:06 | of Polonia . You can have 1/10 order or 15th | |
| 05:08 | order or 39th order polynomial . I hope you never | |
| 05:12 | have to deal with anything like that . But of | |
| 05:13 | course it works for anything Integral coefficients . Why ? | |
| 05:17 | Because 6 , -7 -3 . They're all whole numbers | |
| 05:23 | . They're not fractions . Well there you can write | |
| 05:25 | them as 6/1 but I'm saying there's no obvious fractions | |
| 05:28 | are all whole numbers like this . Now what we | |
| 05:30 | have to do to figure out how to use this | |
| 05:32 | rational roots here . Um It all boils down to | |
| 05:34 | looking at the coefficient of the highest term in this | |
| 05:36 | case X cube and the coefficient of the lowest term | |
| 05:39 | which means the constant term negative three in this case | |
| 05:42 | . So we have special names for this . So | |
| 05:44 | this were basically for the purpose of this theorem . | |
| 05:47 | We're going to basically ignore everything in the middle there | |
| 05:50 | because we're dealing so much with the coefficient of the | |
| 05:53 | leading term and the trailing term . And we're not | |
| 05:55 | really dealing with anything in the middle of the polynomial | |
| 05:58 | . We have special names for these guys . So | |
| 06:01 | we're going to end up finding the factors of the | |
| 06:07 | leading coefficient . Right ? And we call these coefficients | |
| 06:15 | up here , we call them . Kay you'll see | |
| 06:18 | why because there's a whole list of them . Right | |
| 06:20 | ? So we're not going to care about anything in | |
| 06:22 | the middle of the polynomial . Only the first term | |
| 06:24 | in the last term . The last one here . | |
| 06:26 | The factors of these , the factors of the constant | |
| 06:32 | term . We call these H . Now your your | |
| 06:38 | book might not say or your teacher might not call | |
| 06:40 | them K and H . It might call them A | |
| 06:43 | and B . Or you know , whatever W and | |
| 06:45 | Z . It doesn't matter . It's just we have | |
| 06:47 | to have some label , we label all of the | |
| 06:49 | factors of the leading terms something and the factors of | |
| 06:52 | the trailing terms something . You'll see why in a | |
| 06:54 | second because it makes it easier to to uh to | |
| 06:58 | deal with them . So let's first of all figure | |
| 07:01 | out what the factors of this last term are , | |
| 07:05 | this is a negative three , remember what is a | |
| 07:07 | factor ? Anyway ? We've been factoring things in algebra | |
| 07:09 | forever , but what is a factor a factor means | |
| 07:12 | if you have something you're examining and you want to | |
| 07:14 | find the factors of that thing , you need to | |
| 07:16 | find all of the things that can multiply together to | |
| 07:19 | give you that thing , and we can typically find | |
| 07:22 | all the factors by doing a factor tree . But | |
| 07:24 | these are you've been doing it long enough , we | |
| 07:26 | don't need to find we don't need to to build | |
| 07:29 | a factor tree for this exercise , you're going to | |
| 07:31 | find out is pretty simple to do . But for | |
| 07:33 | instance , we're gonna take a number like three and | |
| 07:35 | we're gonna negative three in this case , and we're | |
| 07:36 | gonna figure out what are the factors of negative three | |
| 07:40 | ? Well , we know that one is a factor | |
| 07:42 | because one times three is three and we know that | |
| 07:44 | three is a factor three times one is three . | |
| 07:47 | But other than those two numbers , there really are | |
| 07:49 | no other numbers . But five for instance is not | |
| 07:51 | a factor of three because five times nothing really works | |
| 07:55 | to give you three and there are no other really | |
| 07:56 | numbers , so only one and three are factors there | |
| 08:00 | . But yeah , this these factors we're gonna call | |
| 08:04 | them , h I'm gonna open up a little curly | |
| 08:06 | brace , which means we have a set of numbers | |
| 08:08 | . Right ? So the factors of this last term | |
| 08:11 | is not just one in three , it's actually plus | |
| 08:13 | or minus one and plus or minus three . Why | |
| 08:16 | do we have factors of two of them ? Because | |
| 08:18 | the last term here is negative three . Right , | |
| 08:20 | So positive one times something can give me negative three | |
| 08:24 | positive one times negative three gives me that . But | |
| 08:27 | negative one is a factor as well because negative one | |
| 08:30 | times positive three gives me this same thing here , | |
| 08:33 | positive three times negative one gives me this negative three | |
| 08:36 | times positive one gives me this . So as long | |
| 08:38 | as it can be multiplied by anything to give me | |
| 08:40 | the thing that I'm looking for , then it's a | |
| 08:42 | factor . So for this exercise , when you find | |
| 08:44 | the factors , it's always going to have plus or | |
| 08:46 | minuses in front , Right ? So you find the | |
| 08:48 | numbers that are factors and you slap positive minus in | |
| 08:51 | front of that . That's the set of factors of | |
| 08:54 | this term . We call them . H Now we're | |
| 08:55 | going to find this one . This is a longer | |
| 08:57 | set . We call this one K . So I'm | |
| 08:58 | gonna put a little curly brace , the set of | |
| 09:01 | numbers that are factors of this guy as plus or | |
| 09:04 | minus +11 time 6 to 6 . Right ? Plus | |
| 09:07 | or minus +22 times three is six . So that | |
| 09:10 | works plus or minus +33 times two is six . | |
| 09:13 | So that's a factor and plus or minus six . | |
| 09:15 | Because six times one of six noticed that something like | |
| 09:18 | four is not in this list because four times nothing | |
| 09:20 | works to give me six . Five isn't on this | |
| 09:23 | list because five times nothing works to give me six | |
| 09:26 | . In order to be in the list , it | |
| 09:27 | has to be a factor of the coefficient . Here | |
| 09:29 | , we call these the H factors and these the | |
| 09:32 | K factors . Now , why do we go through | |
| 09:33 | the trouble of naming them like this ? Uh And | |
| 09:36 | it's because of the following thing . So the only | |
| 09:42 | only okay possible rational roots of this polynomial are the | |
| 09:57 | following . Uh And basically I'm just gonna write them | |
| 10:00 | down as H divided by K . So what you | |
| 10:06 | have and I'm gonna write this down all possibilities , | |
| 10:12 | possibilities , possibilities . Sorry about that took me a | |
| 10:18 | minute to make sure I can spell possibilities . Alright | |
| 10:21 | . So basically what you're saying is you take a | |
| 10:23 | check over K . Which means all of the numbers | |
| 10:25 | in this set of list , divided by all of | |
| 10:28 | the numbers in this set of list . But you | |
| 10:29 | have a lot of combinations because don't forget I have | |
| 10:33 | you know , for instance 1/1 , that is one | |
| 10:36 | of the possibilities to be a route , right ? | |
| 10:39 | 1/2 . 1/3 . 1/6 . Those are all possibilities | |
| 10:44 | , however , I have negative one here , so | |
| 10:46 | it's negative 1/1 negative 1/2 . I have all the | |
| 10:49 | possibilities of all the numbers over here divided by all | |
| 10:52 | the numbers here . And I also have all the | |
| 10:54 | possibilities of all the signs positive over negative , negative | |
| 10:57 | or positive and so on . Now . That's not | |
| 10:59 | telling me that all of these possibilities are are actually | |
| 11:02 | roots . It's telling me all of the possible all | |
| 11:05 | the possible ones . So if you're you're going to | |
| 11:08 | find the rational roots of the polynomial , they have | |
| 11:10 | to be in this list of numbers that we're about | |
| 11:12 | to write down . So the easiest way to do | |
| 11:15 | it is to do something like this . We say | |
| 11:19 | the following thing . Let's see . We don't want | |
| 11:20 | to write this . I think I'm gonna go to | |
| 11:22 | the next board . So I don't crunch it up | |
| 11:24 | . So I have one in three and then 123 | |
| 11:27 | and six . All right , So the way we're | |
| 11:30 | gonna write it is this we're gonna say H over | |
| 11:32 | K . And that's going to be a and put | |
| 11:37 | a little coal in here . Right ? So we | |
| 11:38 | have one over 123 and six . Here's how we're | |
| 11:42 | gonna write it . It's gonna be plus or minus | |
| 11:45 | 1/1 plus or minus 1/1 then plus or minus 1/2 | |
| 11:50 | then plus or minus 1/3 then plus or minus 1/6 | |
| 11:53 | . The plus and minus catches all of the possibilities | |
| 11:55 | . And then the dividing catches of course the fraction | |
| 11:58 | so 1/2 1/3 1/6 . But we have to have | |
| 12:01 | the plus and minuses in front . 1/2 . Gotta | |
| 12:05 | have plus or minus sorry plus or minus 1/3 plus | |
| 12:08 | or minus 1/6 . So now we've exhausted all the | |
| 12:13 | possibilities of 1/1 1/2 1/3 1/6 . Then we switched | |
| 12:17 | to this 1 3/1 . 3/2 . 3/3 . 3/6 | |
| 12:21 | . And they continue on in the list here . | |
| 12:23 | So we have plus or minus don't forget that 3/1 | |
| 12:26 | plus or minus 3/2 plus or minus 3/3 Plus or | |
| 12:32 | -3/6 . You can see there's a lot of uh | |
| 12:37 | a lot of numbers there but we can simplify this | |
| 12:39 | list quite a bit . So what you do is | |
| 12:41 | you first write it all down , then we need | |
| 12:43 | to go one more step and simplify we simplify it | |
| 12:49 | . Like this . Plus or minus 1/1 is just | |
| 12:51 | plus or minus one plus or minus one half . | |
| 12:55 | I can't do anything more with that . So we're | |
| 12:56 | just gonna leave it like this plus or minus one | |
| 12:58 | third . That's in the list . We can't do | |
| 13:01 | much with that . Plus or minus 1/6 can't do | |
| 13:05 | much with that but 3/1 Is just plus or -3 | |
| 13:09 | . Three have you can't do much of what's for | |
| 13:11 | that ? So we have 3/2ves but then we have | |
| 13:13 | plus or minus 3/3 . But notice that becomes one | |
| 13:16 | , 3/3 is one . And then over here we | |
| 13:19 | have plus or minus 3/6 but we know that reduces | |
| 13:21 | to one half . All right . So now what | |
| 13:24 | we've done is we've simplified our list but then we | |
| 13:26 | noticed right away that we actually have some duplicates in | |
| 13:29 | the list . This is a duplicate of that number | |
| 13:33 | one . We don't need duplicates in the list either | |
| 13:35 | it's a root or it's not a root . And | |
| 13:37 | then this guy The 1/2 is also due . So | |
| 13:41 | what I would do whenever I'm doing these problems is | |
| 13:44 | I then we say the possible rational . Mhm . | |
| 13:52 | Roots Can only be in this list . We can | |
| 13:56 | only have plus or -1 here . Plus or minus | |
| 14:00 | one half from there . We're not gonna count those | |
| 14:03 | twice . Then we have plus or minus one third | |
| 14:06 | . Plus or -1 6th . Plus or minus three | |
| 14:10 | from here . And then plus or minus three halves | |
| 14:13 | . I'm not going to put this one down again | |
| 14:14 | because it's already in the list and I'm not gonna | |
| 14:16 | put this one down again because it's already in the | |
| 14:17 | list . So you have +123456 But each one of | |
| 14:22 | them has positive or negative possibilities . So we actually | |
| 14:25 | have 12 Possibilities . So you have 12 possibilities for | |
| 14:35 | the rational roots . So that must mean that all | |
| 14:38 | other routes that this polynomial has . All other routes | |
| 14:45 | must be either irrational , Like square root of two | |
| 14:52 | or something like that , or complex . Which means | |
| 14:56 | like to I or three I or something like that | |
| 14:59 | . So I have 12 possibilities here . And I'm | |
| 15:02 | not saying that these are all of the roots . | |
| 15:03 | I'm just saying that the only routes that can any | |
| 15:07 | polynomial can have are either going to be complex numbers | |
| 15:10 | , which this doesn't help you with it all . | |
| 15:11 | Or they can be like radicals and irrational things like | |
| 15:15 | that . Pie , rational radical things like this , | |
| 15:17 | which this isn't gonna help you for . But any | |
| 15:19 | whole number roots , any fractional roots , positive negative | |
| 15:23 | numbers , fractions whole numbers , things like that . | |
| 15:25 | They have to be in this list of 12 numbers | |
| 15:27 | . 12 numbers . How do you actually figure out | |
| 15:30 | which of these things actually are the roots ? Obviously | |
| 15:33 | the equation is only a third order equation that can | |
| 15:35 | only be three routes total . How can I find | |
| 15:38 | out which of these ? If any of them actually | |
| 15:40 | are a root of that ? Well , unfortunately . | |
| 15:43 | Mhm . You have to try then . Oh and | |
| 15:51 | that is not fun , right ? Because that means | |
| 15:56 | you have to take the number one and stick it | |
| 15:57 | into this polynomial . See if it equals zero . | |
| 15:59 | Take negative one . Put it in this polynomial . | |
| 16:01 | See if it equals zero , then I have plus | |
| 16:03 | or minus one half . So I put one half | |
| 16:04 | in here in cuba and all this stuff and see | |
| 16:06 | if it equals zero , negative one half . Put | |
| 16:08 | it in here . See if it equals zero . | |
| 16:09 | And I could go through all 12 of them . | |
| 16:11 | I might figure out because it's the third order polynomial | |
| 16:13 | . I might find that all three of the roots | |
| 16:15 | actually are in this list . I'm done . But | |
| 16:18 | maybe I only find out that one of them is | |
| 16:20 | in this list or something like this because you could | |
| 16:23 | have complex roots as well . Which this doesn't help | |
| 16:25 | you . It only helps you with the rational roots | |
| 16:27 | . All right . Um then you might say , | |
| 16:29 | well if it doesn't help you with all of the | |
| 16:31 | roots , then why do we even learn this thing | |
| 16:33 | and that for that ? I need to do a | |
| 16:34 | real example problem to show you how you can use | |
| 16:37 | this to help . Even if it doesn't give you | |
| 16:39 | all of the answers that can still be very useful | |
| 16:41 | . So this was a larger equation because you had | |
| 16:44 | big coefficients here , especially this one to give you | |
| 16:47 | a long list of numbers . The smaller the numbers | |
| 16:49 | are going to have , the smaller the number of | |
| 16:52 | factors and the easier the problem is going to be | |
| 16:54 | . So let's crawl before we can walk now that | |
| 16:56 | we know how to figure out what the possibilities are | |
| 16:58 | . Let's do a real problem . The problem basically | |
| 17:00 | says I want to solve and your problem will say | |
| 17:04 | something like this , solve the following polynomial X cubed | |
| 17:07 | , uh minus seven X plus six . And you | |
| 17:10 | know , obviously we're finding the roots so we're gonna | |
| 17:12 | set it equal to zero . So you want to | |
| 17:15 | you want to solve this now ? You know that | |
| 17:16 | there's three routes . So you know that there can | |
| 17:18 | be some rational roots in there . There can be | |
| 17:19 | some irrational roots in there . There can be some | |
| 17:22 | complex roots in there . But you know , you | |
| 17:24 | could do a cart rule of signs if you wanted | |
| 17:26 | to and it would tell you how many positive and | |
| 17:28 | negative rights you have . We're not gonna do that | |
| 17:30 | . We already learned that skill . Now we're gonna | |
| 17:32 | go and try to figure out and see if we | |
| 17:33 | can find out what routes are actually satisfied . This | |
| 17:37 | guy's , which means when it says solving it means | |
| 17:39 | find all the routes . All right , So the | |
| 17:43 | way we're going to do this is the the factors | |
| 17:46 | of this thing are plus or minus , the coefficient | |
| 17:49 | in front . Plus or minus one . That's the | |
| 17:50 | only factors of one is the number one itself . | |
| 17:53 | Okay . You might say why is negative one factor | |
| 17:55 | ? Well , negative one times negative one would give | |
| 17:58 | me positive one . So it's a factor if you | |
| 18:00 | can find anything at all to multiply by it to | |
| 18:02 | give you you know what you have . So , | |
| 18:04 | the factors of this , you're always gonna have plus | |
| 18:06 | or minus any time you write it down is one | |
| 18:08 | . Now the factors of this guy , we just | |
| 18:10 | wrote the list down . So we're very familiar . | |
| 18:12 | It's plus or minus one , plus or minus to | |
| 18:15 | plus or minus three and plus or minus six . | |
| 18:18 | Those are the only factors there's two times four is | |
| 18:20 | eight , there's eight of them . Right ? But | |
| 18:23 | in order to figure out what the rational root possibilities | |
| 18:26 | are the possible rational roots . The rats routes you | |
| 18:33 | can call it rat roots but whatever it's called , | |
| 18:35 | rational roots is going to be these factors divided by | |
| 18:38 | these factors which exactly what we did . We did | |
| 18:40 | essentially double check H over K . So these were | |
| 18:42 | called the H factors . These were called the K | |
| 18:45 | factors H over K . So but you know , | |
| 18:47 | we're getting a little bit into solving the problem . | |
| 18:49 | You don't have to write H and K down all | |
| 18:50 | the time . You just list the factors . List | |
| 18:52 | the factors and then we say that we're gonna have | |
| 18:54 | plus or minus 1/1 mm . Then we can have | |
| 18:58 | plus or -2/1 . Okay , Plus or -3/1 . | |
| 19:04 | Plus or -6 of them one ? Because this only | |
| 19:06 | had really two factors anyway . So you're taking each | |
| 19:09 | of these and dividing by this . So the list | |
| 19:10 | is much shorter than it was in the previous problem | |
| 19:13 | . Now we can simplify this list by just looking | |
| 19:17 | at it and saying we can simplify it by plus | |
| 19:19 | or minus 1/1 being one plus or minus to plus | |
| 19:23 | or minus three plus or minus six . So here | |
| 19:25 | we have a situation . This is kind of nice | |
| 19:27 | because we have a third order polynomial . We know | |
| 19:29 | we're gonna have three roots . Total sum can be | |
| 19:31 | complex of course , or irrational . We don't know | |
| 19:34 | yet but we know some could be rational which means | |
| 19:36 | their numbers or fractions but we know that if it | |
| 19:39 | has any rational roots at all , it has to | |
| 19:41 | be in this list . So there's four times to | |
| 19:44 | eight possibilities here . So really if you had lots | |
| 19:48 | of time on your hands or a computer on your | |
| 19:50 | hands , what you would do is take every one | |
| 19:52 | of these things and put it into the equation and | |
| 19:54 | see if it equals zero . That's gonna totally find | |
| 19:57 | out if all of them are rational . It's going | |
| 20:02 | to figure out all of the rational roots because you're | |
| 20:04 | gonna substitute every last one of them . But you | |
| 20:06 | can see that that strategy is going to get really | |
| 20:08 | cumbersome as we get larger problems and you have to | |
| 20:10 | start substituting all these fractional ones in there . So | |
| 20:13 | what I want you to do , what most teachers | |
| 20:15 | are , what's most books will tell you to do | |
| 20:17 | is don't try to substitute every one of these things | |
| 20:19 | in there . Just pick the easy one . The | |
| 20:21 | easy one is going to be one usually not always | |
| 20:24 | , but you actually always , you're gonna have plus | |
| 20:26 | or minus one is a possible rational root because every | |
| 20:29 | number has a factor of one . So you're always | |
| 20:32 | going to have plus or -1 , Right ? And | |
| 20:34 | you can of course substitute plus or -2 . You | |
| 20:37 | can do that . There's nothing wrong with that . | |
| 20:38 | But always try to substitute plus or -1 . You | |
| 20:41 | might get lucky , right ? If you get lucky | |
| 20:43 | , then I'm gonna show you how to handle that | |
| 20:45 | . But let's go ahead and see what happens uh | |
| 20:49 | and see if we can avoid plugging in every last | |
| 20:51 | one of these things . Because for every time we | |
| 20:53 | do it , we're gonna have to square CUBA and | |
| 20:55 | all this other stuff to figure out if it's a | |
| 20:56 | route or not . So we're gonna plug in this | |
| 21:00 | polynomial evaluated at the number one because plus or minus | |
| 21:03 | one is the lowest route . Now one cubed is | |
| 21:06 | , we're going to stick in there minus seven times | |
| 21:08 | one right here , plus six . One cube is | |
| 21:12 | one minus seven from this and then plus six . | |
| 21:15 | So the one plus six is seven minus +27 is | |
| 21:18 | zero . So it is a route . So you | |
| 21:21 | have figured out that the number one is actually a | |
| 21:23 | route , so X is equal to one is a | |
| 21:27 | route . Again you got lucky , but reality is | |
| 21:30 | we choose the problem so that in the beginning you | |
| 21:32 | find out that uh that that is a room . | |
| 21:34 | Now we have all we have one of them down | |
| 21:36 | , we only have two left . So maybe we'll | |
| 21:38 | get lucky again . Let's plug in negative one . | |
| 21:40 | Maybe it'll be a route as well , -1 . | |
| 21:43 | So we put negative ones cubit minus seven , negative | |
| 21:48 | one plus six . Now this negative one cube evaluates | |
| 21:52 | to negative one negative one times negative one times negative | |
| 21:55 | one this becomes a positive seven . This becomes a | |
| 21:58 | positive six . Right ? So you can think of | |
| 22:01 | the one minus the the seven or negative one plus | |
| 22:05 | the seven is six and then plus six is 12 | |
| 22:08 | . So no so we say that X . Is | |
| 22:12 | equal to -1 not a route . So you can | |
| 22:17 | see you're not gonna get lucky all the time , | |
| 22:18 | right ? It'll be nice if everything had a root | |
| 22:20 | of plus or minus one . Those are easy numbers | |
| 22:22 | to calculate putting a number one in there is like | |
| 22:24 | really really simple putting a number two really isn't that | |
| 22:27 | hard putting three and all that isn't that hard ? | |
| 22:29 | But what you're gonna run into isn't uh more complicated | |
| 22:32 | problems you're gonna have to put in one half And | |
| 22:35 | 1/6 and -16 . So you have two cubit and | |
| 22:38 | then add it and then all this stuff with fractions | |
| 22:40 | and so you're gonna have to do a lot of | |
| 22:41 | work . So what I want you to do is | |
| 22:44 | go ahead and try to substitute , keep going up | |
| 22:46 | your list until you find at least one of them | |
| 22:48 | . That is a route because once you identify one | |
| 22:52 | of them is being a route then we can use | |
| 22:54 | some tricks . I don't want to like to call | |
| 22:56 | it a trick but we can then find the remaining | |
| 22:59 | routes . By the way . I'm going to show | |
| 23:00 | you here in a second . We know that this | |
| 23:03 | guy is a route . So what does it actually | |
| 23:05 | mean ? Right , what does that actually mean ? | |
| 23:08 | It means that if the equation , I'm trying to | |
| 23:10 | decide if I want to even do this on this | |
| 23:11 | page or not ? Yeah , I think I am | |
| 23:14 | going to do it on this page . If x | |
| 23:16 | minus , if x equals one is a route , | |
| 23:19 | then that means this polynomial here can be written as | |
| 23:22 | a factored form of X -1 times some other factors | |
| 23:29 | of x . And that's going to equal what this | |
| 23:32 | polynomial is . We've we've done this before and I | |
| 23:35 | realized some of you may not have seen those other | |
| 23:37 | lessons . So I'm going to review it a little | |
| 23:38 | bit here . But the bottom line is we've actually | |
| 23:40 | done this before . If if this looks foreign to | |
| 23:42 | you , then you probably should go back and look | |
| 23:44 | at the factor theorem and the remainder theory and we've | |
| 23:46 | done the factor theorem and the remainder is there . | |
| 23:48 | And we did problems like this before . But anyway | |
| 23:50 | , this is the polynomial given to us . We | |
| 23:52 | know that if we knew all the routes , we | |
| 23:54 | should be able to write it in some kind of | |
| 23:56 | factored form like we do for binomial or uh for | |
| 23:59 | for quadratic equations all the time . One of the | |
| 24:02 | roots we actually already found . So we know that | |
| 24:04 | there has to be one parentheses as x minus one | |
| 24:07 | . Because if you set this thing equal to zero | |
| 24:09 | then you would solve for the one and that would | |
| 24:10 | be one of the roots , but that's not all | |
| 24:12 | of them . There has to be some other term | |
| 24:14 | multiplied , which I'm representing as Q Of X . | |
| 24:18 | It's some function of X . in other words it | |
| 24:20 | could be X -2 or X -3 as the other | |
| 24:23 | two routes here . I don't know what they are | |
| 24:25 | , but I know that there has to be something | |
| 24:26 | out here because all of them multiplied together has to | |
| 24:29 | equal the original cubic . Okay , so what this | |
| 24:33 | means is that To figure out what the other routes | |
| 24:36 | are ? We need to do Long Division , which | |
| 24:38 | I know you guys really love but that's why we | |
| 24:40 | did spend so much time on it . We need | |
| 24:42 | to take the original polynomial and divide by the X | |
| 24:46 | -1 . That's gonna give us whatever this is . | |
| 24:48 | And because this thing is a cubic equation and we | |
| 24:51 | already have one root when we divide it , we're | |
| 24:54 | gonna be left with Q . Of X . Which | |
| 24:55 | will be quadratic . It has to be an X | |
| 24:57 | square quadratic because the entire thing is X . Q | |
| 25:02 | . And we have an X minus one multiply . | |
| 25:04 | So this has to be one degree lower because this | |
| 25:06 | is an X . And this will have to be | |
| 25:07 | an X squared term . And I guess I'm trying | |
| 25:10 | to tell you that once we do the division we | |
| 25:12 | expect to see a quadratic term and we know how | |
| 25:14 | to solve all quadratic equations . We can just use | |
| 25:16 | the quadratic formula , even if we can't factor it | |
| 25:19 | . So even though you only found one route , | |
| 25:21 | we can do division to find what's left over as | |
| 25:24 | the remaining factors . Once we have that we can | |
| 25:27 | always solve quadratic formulas and we can find the other | |
| 25:29 | two . And if those other routes are rational then | |
| 25:33 | they will have been in our list all along . | |
| 25:34 | We could of course continue cranking through this list , | |
| 25:36 | but this is going to take time as well . | |
| 25:39 | Okay , so what's the easiest way to do this | |
| 25:41 | division ? I think you will agree with me . | |
| 25:43 | The easiest way to do it is synthetic division . | |
| 25:45 | So whenever we write this down we're dividing uh by | |
| 25:49 | uh one X cube . There is no x squared | |
| 25:52 | term . So we have to uh this will remind | |
| 25:54 | myself this is the x squared term . So I | |
| 25:55 | have to put a zero there . We have a | |
| 25:57 | negative seven for the X . Term . And we | |
| 25:59 | have a six there and then I'm going to divide | |
| 26:02 | it by this term . But remember when we do | |
| 26:05 | synthetic division we change the sign of what we're dividing | |
| 26:09 | by . That allows us to do adding . Instead | |
| 26:11 | of subtracting , we covered all of this in synthetic | |
| 26:13 | division . So if you are looking at me cross | |
| 26:16 | eyed because you don't know what I'm doing , it | |
| 26:18 | just means you need to go back and learn synthetic | |
| 26:20 | division with me . So first step is dropped the | |
| 26:22 | one down . Then we multiply one times one is | |
| 26:24 | one . Then we add we get a one then | |
| 26:27 | one times one again this one then we add we | |
| 26:30 | get a negative 61 times negative six is negative six | |
| 26:34 | . We add and we get a zero . Now | |
| 26:36 | we always expect the remainder to be zero . When | |
| 26:38 | we divide by a factor , you know what I | |
| 26:40 | mean by that is if I take the number 45 | |
| 26:44 | okay , that's some number . I know that the | |
| 26:46 | number five is a factor of 45 . I know | |
| 26:49 | it divides evenly . So if I divide by five | |
| 26:53 | I know what I'm going to get is nine with | |
| 26:55 | a remainder of zero because it goes in an even | |
| 26:58 | number of times . Because when I divide by a | |
| 26:59 | factor , I expect to get something with no remainder | |
| 27:02 | at all left over . So what I'm saying is | |
| 27:04 | when I take the polynomial , I divide by a | |
| 27:07 | known factor . I know it's a factor because it's | |
| 27:08 | a route then what I left with should have no | |
| 27:11 | remainder just like this . And so what I've found | |
| 27:14 | is that what's left over here is X squared . | |
| 27:17 | This has to be one degree less than the cube | |
| 27:19 | . So it has to be x squared one X | |
| 27:21 | squared plus one x minus six . This is what | |
| 27:26 | is left . What we found out is that cube | |
| 27:28 | bags is equal to that . So let me just | |
| 27:32 | double check myself X squared plus six minus six . | |
| 27:34 | So what we're saying is that these two things multiplied | |
| 27:38 | together are what multiply to give us this guy . | |
| 27:41 | So let me write that on the last board here | |
| 27:46 | . What we have basically said is that x minus | |
| 27:49 | one Which was that known factor that we guessed ? | |
| 27:52 | And then we just figured out the answer for Q | |
| 27:54 | two Q of X . The division . We're saying | |
| 27:58 | that this has to equal X cubed -7 . x | |
| 28:02 | plus six . All right . Now we know how | |
| 28:06 | to solve these things because we're trying to set him | |
| 28:08 | equal to zero . Right ? How do we set | |
| 28:10 | this thing equal to zero ? Basically we have X | |
| 28:11 | -1 . X squared plus x minus six equals zero | |
| 28:16 | . Well , we already have this one already here | |
| 28:19 | . Now let's try to factor this . If it | |
| 28:21 | doesn't factor , if this doesn't factor , we can | |
| 28:23 | still do the quadratic formula on it . But let's | |
| 28:24 | try to factor X and X . And then two | |
| 28:28 | times 3 is six . And if you work out | |
| 28:30 | the signs , this is the way that it works | |
| 28:32 | because you'll have X times X is X squared negative | |
| 28:36 | two times three is negative six . That's negative two | |
| 28:38 | X . That's positive three . X . That adds | |
| 28:41 | together to give you the positive X in the middle | |
| 28:44 | . So you see now we have figured out that | |
| 28:46 | from this , which we already knew X can be | |
| 28:49 | equal to one as a route from this . When | |
| 28:51 | we set it equal to zero , X can be | |
| 28:53 | equal to two as a route from this guy . | |
| 28:55 | When we set it equal to zero , X can | |
| 28:56 | be negative three is a route , so X is | |
| 28:58 | equal to one . X is equal to two . | |
| 29:00 | X is equal to negative three . Those are the | |
| 29:04 | three routes . We didn't have any complex roots in | |
| 29:06 | this case . That means this cubic polynomial cuts across | |
| 29:09 | the X axis in three locations and notice we have | |
| 29:12 | three rational roots . So 12 negative three . When | |
| 29:16 | we go back to our problem , our list did | |
| 29:19 | include 12 and negative three . So all of the | |
| 29:22 | rational roots that were actually solutions of the problem were | |
| 29:25 | in our list . So if we didn't do this | |
| 29:27 | division business , we would have had to go and | |
| 29:30 | continue on substituting things in plus one minus one plus | |
| 29:33 | two minus two plus three minus three . We eventually | |
| 29:36 | would have got them . All right . But the | |
| 29:38 | problem is you just never know when you're gonna get | |
| 29:40 | you're gonna get the answer . So you can be | |
| 29:42 | substituting in all day . You might substitute everything in | |
| 29:45 | that whole list and and get none of them to | |
| 29:48 | be in there because all the routes might be uh | |
| 29:50 | you know , um irrational or or something else . | |
| 29:53 | You don't have any idea what they are . But | |
| 29:55 | if the polynomial has any rational roots meaning routes that | |
| 29:59 | are whole numbers or fractions , then they're going to | |
| 30:01 | be in this list . So step number one , | |
| 30:03 | write your equation down , descending step number two , | |
| 30:06 | right ? The factors down here , uh end up | |
| 30:08 | here , then you write the possible rational roots as | |
| 30:11 | the divisions of all of these numbers , divided by | |
| 30:13 | all of these numbers . Simplify your list . And | |
| 30:16 | then what you have to do is start guessing . | |
| 30:17 | Start with the easy numbers , Find one . Once | |
| 30:19 | you find one route then you know that that route | |
| 30:22 | is a factor times more factors must equal the polynomial | |
| 30:26 | that you have , then my division . You can | |
| 30:28 | figure out what the other factors are . And then | |
| 30:31 | once we had this we were able to factor . | |
| 30:33 | If we were not able to factor , I could | |
| 30:35 | just set this quadratic equal to zero by itself and | |
| 30:39 | use the quadratic formula and figure out what the remaining | |
| 30:43 | two routes are . Maybe I get too complex routes | |
| 30:45 | out of it . You don't know until you do | |
| 30:46 | , but that's what you have to do . So | |
| 30:48 | you write your list , find at least one of | |
| 30:50 | these routes that work and then probably gonna end up | |
| 30:53 | using division to find the rest of them . So | |
| 30:55 | make sure you understand this . We have a few | |
| 30:57 | more lessons to get more practice with it , but | |
| 31:00 | it's all going to follow the same process . I | |
| 31:02 | want you to do this problem yourself , Make sure | |
| 31:03 | you understand and follow me on to the next lesson | |
| 31:05 | . We'll get more practice with the rational root theory | |
| 00:0-1 | . |
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