20 - The Rational Root Theorem, Part 1 (Rational Roots of Polynomials) - Free Educational videos for Students in K-12 | Lumos Learning

## 20 - The Rational Root Theorem, Part 1 (Rational Roots of Polynomials) - Free Educational videos for Students in k-12

#### 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 .
03:44 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
15:11 Or they can be like radicals and irrational things like
15:15 that . Pie , rational radical things like this ,
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 ,
16:31 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
23:16 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
26:11 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
28:19 . 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
29:30 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 .
30:19 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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