16 - Conjugate Root Theorem, Part 1 (Roots of Polynomials, Solving Polynomials & Conjugate Roots) - Free Educational videos for Students in K-12

16 - Conjugate Root Theorem, Part 1 (Roots of Polynomials, Solving Polynomials & Conjugate Roots) - By Math and Science

Transcript


00:00	Hello . Welcome back to algebra . The title of
00:02	this lesson is called the Conjugate route theorem . This
00:05	is part one of two . So I said uh
00:07	we're going to be covering some kind of random theorems
00:10	that deal with polynomial . They're all really important .
00:13	None of them are very hard to understand but we
00:14	have to cover all of them . This one is
00:16	called the Conjugate route theorem . And actually you already
00:18	know what it is because I've told you through examples
00:21	, but we haven't formalized as a theorem . So
00:24	here are the basic ideas between these theories . There's
00:26	actually two theorems that we need to learn both of
00:28	which you already actually know . The first one is
00:30	really an untitled theorem . It says that every polynomial
00:34	with positive degree in has exactly in roots . So
00:37	that's something that we've been telling you from the very
00:39	beginning . That's just a theorem that's formalized as true
00:43	for every polynomial has a positive degree . The degree
00:46	of the polynomial tells us how many routes it has
00:49	. And you already know this . Think of a
00:51	line , right ? Uh Mx plus B . So
00:53	three X plus four for instance , that's the equation
00:56	of a line . What is the degree of that
00:57	equation ? The degree is one because that line has
01:01	X . To the first power . Right ? This
01:03	thing tells you that every polynomial and yes online is
01:06	a polynomial . It's just uh you know , it's
01:08	just a degree of one has exactly en route .
01:12	So it tells us that a line has a degree
01:14	one . It should only have one root . And
01:16	if you think about it , if the line goes
01:17	slanted like this , it can only cross the X
01:20	axis in one location . Of course there are special
01:22	lines that are horizontal . They don't have any routes
01:24	because they're special horizontal lines . But for every other
01:27	line other than a horizontal line , it will cross
01:29	the X axis in one location because it's a first
01:32	degree polynomial . Quadratic . Our second degree polynomial ,
01:35	right ? They cross into locations . We've done this
01:38	kind of thing before . But in general , you
01:40	all know that uh polynomial is going to either cross
01:44	like this or it's gonna go upside down across like
01:46	this . So they would have to real roots for
01:49	a quadratic , which has a degree of two .
01:51	Now , of course the polynomial could be flipped upside
01:54	down again as it goes upside down , it will
01:55	still have to roots . And of course this polynomial
01:57	might exist up here where there's no crossings on the
02:00	X axis . But that's okay . That just means
02:02	that there will be too complex roots . But no
02:04	matter how you slice it , every quadratic is going
02:07	to have a degree of two . And because of
02:09	that it will have two routes either to real or
02:11	too complex roots going up one more . You can
02:14	see that we have the cubic . So we talked
02:17	about this in the past uh , cubic is a
02:19	degree of three . It means X cubed plus some
02:22	other terms in general , a cubicle looks something like
02:24	this . So you can see +123 crossing points ,
02:28	right ? Uh And of course it might be flipped
02:30	upside down so it goes up and down the other
02:32	way . But it'll cross in three locations . If
02:34	you have a weirder cubic that maybe they don't have
02:37	three crossing points like this , it might cross like
02:39	this and then go down and maybe it doesn't cross
02:42	on the 2nd and 3rd time . In that case
02:44	you have one real root from this crossing and this
02:47	one since it doesn't cut through the actual access ,
02:49	those would be too complex roots , but you would
02:52	still have a total of three routes . Some real
02:54	some complex . We'll do one more just to make
02:58	sure you understand because it's been a while since we've
03:00	talked about this stuff . But 1/4 degree polynomial in
03:03	general will look something like this . So to go
03:05	down up , down , up , of course it
03:07	could be flipped upside down and so on and you
03:10	can see that there's four crossing points there . Now
03:12	of course you can have other variations on that .
03:15	Where for instance , let me do this one in
03:16	purple , you might go down and then up and
03:19	then down and then up again , something like this
03:21	. So you would have won two real crossing points
03:24	. These guys didn't really make it up to cut
03:26	through . So those other two would be imaginary or
03:29	complex roots . But either way 1/4 degree polynomial is
03:32	still going to give you four routes . So just
03:35	to wrap it up in a bow , this would
03:36	be like an X squared . I'm not saying this
03:39	is a graph of X squared , I'm saying this
03:40	is a second degree polynomial . This would be like
03:43	for a cubic type polynomial with all the terms after
03:46	it . This would be for like 1/4 degree polynomial
03:48	. And again here you have in the red curve
03:50	, you have the four real crossing points . Here's
03:52	another example where you have to reel and two imaginary
03:55	crossing points . And then of course the first one
03:57	that I opened up with was the concept of a
03:59	line . Right ? And so for a line in
04:01	general they can go either direction , but they're only
04:03	gonna have one crossing 10.1 route right here . So
04:05	this would be like an X . Why is equal
04:08	to X . Type of type of deal ? Of
04:09	course , that's not the exact curve . It's just
04:11	generally a first degree , second degree , third degree
04:14	, fourth degree like this . Now , the second
04:18	theory that I want to talk about is what we
04:19	call the conjugate route theorem . And that is a
04:21	theory that you actually know also what it says .
04:24	We've already talked about it , but we haven't formalized
04:26	that . It's actually a theorem that applies to all
04:27	polynomial . But notice it says with real coefficients .
04:31	Every polynomial we've ever done in this class is a
04:33	real number as a coefficient in front of the X
04:36	terms , right ? You know , three uh X
04:38	squared or minus four X squared or something . Those
04:41	are all real numbers . If you have a polynomial
04:43	and you can have polynomial is with imaginary coefficient ,
04:47	then this contrary route theorem doesn't apply . But for
04:49	every polynomial we have ever dealt with , they always
04:51	have real coefficients . And what it's saying is if
04:54	you have an imaginary or complex roots , A plus
04:57	B i is is just a general way of saying
04:59	any complex number like three plus four I . Then
05:02	if you know this route then you know that the
05:04	partner route is called the conjugate route , which would
05:07	be uh where the imaginary part has the opposite sign
05:11	. And you know that that's going to be a
05:12	route as well . So for instance if you know
05:15	that three plus two eyes , a route of of
05:18	a polynomial then by the conjugate roots here , um
05:21	you automatically know that three minus two I is also
05:25	a route , Right ? And it goes the other
05:27	way , if you know that negative 1 -5 I
05:31	is a route , then you automatically know that negative
05:34	one plus five I is also a route . These
05:36	are called conjugate pairs . The only difference between them
05:39	is the imaginary part has the opposite sign . Everything
05:42	else is exactly the same . See the opposite sign
05:44	here , opposite sign here . So the bottom line
05:46	is roots of equations . When it comes to the
05:49	complex roots , they always come in pairs . You'll
05:51	never ever ever ever solve a polynomial equation and get
05:56	a route that's complex without also getting the partner route
05:59	to conjugate of it . It's just always gonna happen
06:02	unless this polynomial that you're dealing with does not even
06:05	have real coefficients . If it has complex coefficients or
06:08	imaginary coefficients , which we've actually never dealt with .
06:11	But if you if you had that then of course
06:13	this theory doesn't apply . But for every problem in
06:16	this class we're going to basically put our hats on
06:18	this because it's always going to be true . So
06:21	the types of problems we're going to do here is
06:23	the following , for instance , find a cubic equation
06:31	with um integral coefficients , integral coefficients with the following
06:40	routes . So the root for this problem is negative
06:46	one and 5 . I've I want you to find
06:49	the cubic equation . Now , you might look at
06:51	this and say , well , didn't we do this
06:52	kind of problem before ? Didn't he already given me
06:54	the roots and that I had to find the equation
06:56	. That's true . I did . But that was
06:58	different because if you look here , I'm asking you
07:00	to find a cubic equation that means a third degree
07:03	polynomial . But I only gave you two routes .
07:06	So you need to then no . From the conjugate
07:08	route theorem , what the other route is , There's
07:10	no way to really figure it out without knowing what
07:13	the other route is . And because you know that
07:16	it's a cubic , that's what we're asked to find
07:19	, then you know that it's an X . To
07:21	the third power polynomial . That's what that tells you
07:24	, right . And because of that , you know
07:26	that there has to be three routes because of the
07:29	first hearing that we already learned and you already know
07:30	what two of them are and one of them is
07:32	imaginary . So that means that the five I root
07:35	produces a partner route , which is negative five I
07:39	. Is also a route right now , if you
07:44	look at the wording here , it says A plus
07:46	B . I . And then a minus B .
07:48	Are all also route . So another way of thinking
07:50	about it is if you know that one of the
07:52	roots is five I . That's basically the complex number
07:55	zero plus five . That's what that means . No
07:57	real part at all . But the imaginary parts five
08:00	I . And that means the conjugate of it is
08:02	zero minus five I . And if you just strip
08:04	the zero away , then this is what I wrote
08:06	down here . So for cons you get numbers ,
08:08	if it's a if it's purely imaginary , you just
08:10	stick a negative sign on the front or if it
08:12	was already negative , you flip it back to positive
08:15	. If it's complex then you just retained the real
08:18	part and flip the imaginary part . So it's all
08:19	the same thing . But the bottom line is we
08:22	now know that since five I is a route then
08:26	we know that negative five I is also a route
08:27	. And that's because of this conjugal route theorem .
08:30	And so since you now know it , all three
08:32	routes are you can construct it the polynomial the same
08:35	way we did before . For this route it will
08:36	be X plus one because the factor will be x
08:40	minus a negative one . We've done that many times
08:42	before . And then for this route it's going to
08:45	be X -5 i . And then for the route
08:48	that we now know uh additional the context root ,
08:52	there has to be X plus five by and that
08:54	has to equal zero . If you think about it
08:56	. If I give you this and I ask you
08:58	what are the roots of this equation ? You set
09:00	this equal to zero , solve for X . You
09:02	get negative one . Set this to zero . Move
09:04	this over , you get five , I set this
09:06	to zero . Move it over . You get the
09:07	negative five I . So you know that these factors
09:09	are then correct . Now what you have to do
09:12	is perform the multiplication . To figure out what the
09:16	parents polynomial is . You have to multiply this out
09:18	. So we'll save the X plus one for last
09:22	and we'll multiply this X times X is X squared
09:25	. This is going to be negative five I X
09:29	. Then this one will be positive 5ix . Okay
09:34	, we multiply these these guys here negative five I
09:37	positive five . I will give you a negative because
09:39	the negative times positive 25 . And then the items
09:42	the eyes I squared . Easy to make mistakes with
09:45	complex numbers , imaginary numbers like that . So we
09:47	have to know how to deal with that . And
09:49	then on the next line will say , well we'll
09:51	take the X . Square term and we'll add these
09:53	together . But these at 20 negative five I .
09:56	X . Positive five I X . And then this
09:58	is negative 25 . But then I know that I
10:01	squared is negative one . So I'm gonna write it
10:03	like this . And in the next line , I
10:04	will simplify it . It's kind of its final form
10:07	Uh , x squared . This becomes plus 25 .
10:11	Just All right , so far so good . So
10:14	the way I want to do this , when you
10:15	have very complex multiplication is just multiply two of them
10:18	. When you get the answer multiply that by the
10:21	next one . So then you have to multiply X
10:23	times X squared is X cubed , then X times
10:27	the 25 is 25 X . Then one times X
10:30	squared is X squared , one times 25 is 25
10:34	. And now we collect terms , we have X
10:36	cubed this X squared is the next lowest term .
10:40	And then the 25 X . And then the 25
10:45	and you circle this . So execute plus X squared
10:49	plus 25 X plus 25 is equal to zero .
10:52	And if you were to plot this , we know
10:54	already that one of the roots is uh purely real
10:58	, it's a real root . But two of these
11:00	routes are complex , their congregants of one another .
11:03	So going back to our original sketches , we know
11:05	in general that cubic are gonna look like this ,
11:07	they're gonna look like a little s that goes down
11:09	, but in this case it might look more like
11:11	the purple curve where you have one real crossing point
11:14	, and then the other two never really made it
11:16	down to cross . So they're they're representing the imaginary
11:18	or complex roots . So if you re scale it
11:21	, this would be like a negative one , that's
11:22	the real root here . These represent the plus and
11:25	minus five . I and I've done an entire lesson
11:28	on what do complex and imaginary roots means . So
11:31	if you want to see more information about that ,
11:33	you can go dig up that lesson that I've done
11:36	. So um this basically is very similar to the
11:40	previous problems we've done , but we're just using the
11:43	conjugate route theorem and the the number of routes that
11:46	we know now is a theorem as well . So
11:49	we also want to find a cubic right with integral
11:53	coefficients . Notice we have integral coefficients . That's the
11:56	other thing we know . And then I want to
11:58	say one more thing before I move on . I
12:00	mentioned in previous lessons , this equation does have these
12:04	three routes that we've talked about , but it is
12:06	not the only equation that has those roots . If
12:08	I multiply the left hand side and the right hand
12:11	side by any number I want , let's say I
12:12	multiplied by five , then I'll get five x cubed
12:16	plus five X squared plus 125 x plus 125 I'll
12:22	get a larger , much higher coefficients right equals zero
12:26	. Because five times zero on the right is still
12:27	gonna be zero . Then I will get a different
12:30	polynomial that will graft differently than than this one .
12:35	Let's say . Maybe it goes much higher and so
12:37	on and so forth . But it will still have
12:39	the same routes , right ? So you can take
12:41	the answers you get here . You can multiply that
12:42	by anything you want uh to generate another polynomial .
12:47	That new polynomial will have the same routes as the
12:49	one this one has but it will be shaped differently
12:51	. So if I multiply any of these curves by
12:53	a number , it might stretch the curve but they'll
12:55	still have the same crossing points there . So the
12:58	reason I bring it up is because this is saying
13:00	find a cubic with integral coefficients . In this case
13:04	the polynomial already arrived . I already had integral coefficients
13:07	, no fractions in other words . So I didn't
13:10	have to mess around with it . But if I
13:11	needed to multiply that polynomial by two , I could
13:14	do that or by five or by whatever I wanted
13:16	to . So here I'm still trying to find a
13:19	cubic and here I have roots that are given to
13:23	me if I can spell the word roots of negative
13:27	two and the other route is negative one plus I
13:30	. So again , I'm told ahead of time as
13:33	part of this problem that I want a cubic .
13:35	Right ? And so because I want a cubic ,
13:38	I know that I want a polynomial with a third
13:40	degree highest power , Which means I need three routes
13:44	. But I've only been given two of the roots
13:46	and one of them was real . One of them
13:48	was complex . And because I know all of the
13:50	complex roots come in pairs , then I must also
13:53	then have negative 1 -1 is a route because this
13:59	is a complex route . I know that the conjugate
14:01	of it must also be this would be 123 routes
14:04	altogether from my cubic that I'm trying to find .
14:07	So then in order to actually find it , I
14:09	have to write the factored form . This route is
14:12	going to lead to a term X-plus two . This
14:15	term is going to lead to because it's basically x
14:18	minus the minus two . That's why it's X plus
14:20	two . This one is going to become , this
14:22	is how you write it down , X minus negative
14:24	one plus I . And then this last one here
14:28	will be x minus negative one minus I . Now
14:32	the reason these are written this way is because if
14:35	I get , if I were to give you this
14:36	and say what are the roots , what would you
14:38	do ? You would set this to zero and solve
14:40	it and find that that's negative to you would set
14:42	this 1-0 . You would take this whole imaginary complex
14:45	term , move it to the other side by addition
14:48	. And you would find that negative one plus I
14:49	as a route . And then you would do the
14:51	same thing here and you would find this is a
14:52	route . So the bottom line is whatever you're given
14:55	, you just say x minus that route . That
14:57	is one of the factors . And that comes from
14:59	the factor theorem that we've already learned . So now
15:02	that that's out of the way , we have to
15:04	multiply all this stuff out . But before we can
15:06	multiply we have to fix what's on the inside here
15:09	. So we have X plus two and this becomes
15:12	x minus times minus , gives us positive one negative
15:16	times eyes negative I and then we have x minus
15:20	. Times minus is positive . One minus times minus
15:23	is positive . I All right , so now we
15:26	have um two very large complex terms and then we
15:29	have another one here . So what we wanna do
15:32	is multiply Only these two terms together . First whatever
15:37	we get as an answer will then multiply by the
15:39	X-plus two there . So X times X gives me
15:43	X squared X Times one is X X times I
15:47	is I at times X . Then I move my
15:49	finger here . One times X is X one times
15:52	one is 11 times eyes I move my finger here
15:57	negative items . This is uh I guess I should
16:00	take away that positive sign their uh negative items .
16:03	X is negative . I X negative I times one
16:06	is negative I negative I times I is don't forget
16:09	be careful , negative I square . Do not try
16:13	to do too many things at a time . The
16:15	negative times positive means I have a negative sign .
16:17	The I times the eye gives me I squared .
16:19	Yes , I know that you know that this I
16:21	squared will eventually become a negative one . But to
16:23	keep it clean , write it like that initially .
16:25	Then we're going to clean up our terms . Um
16:28	And I think the way I'm gonna clean on my
16:30	terms . Yeah , like this X plus two .
16:34	And on the inside here I look for X squared
16:37	. Alright , I got an X squared term .
16:38	I'm gonna put a little dot under that to tell
16:41	me I've looked at it . I have an X
16:42	term here and an X term here . X term
16:45	here in X term here . So that's gonna be
16:46	two X . Mhm . Right . But then I
16:51	have an I . X . Term here and a
16:53	negative I . X . Term here . So these
16:56	guys go together and basically add to zero . So
16:58	I've taken care of this one . This one this
17:00	one this one I have a one and I have
17:03	an eye and a negative I so these guys are
17:04	gonna go together And make zero . So I still
17:08	have a plus one right there . So carry that
17:09	through . I have a plus one and now everything
17:12	is accounted for except for this , I have a
17:14	minus and then the I squared is negative one .
17:16	That's when you finally deal with that at the end
17:19	. And then it becomes very simple because what you
17:22	have is X squared plus two X . And then
17:25	you have one . But this becomes a positive one
17:28	. I'm sorry . Yeah , positive one negative time
17:29	. Seconds positive one plus the one gives you two
17:33	. Yes . So let me double check myself .
17:35	I have X squared plus two X plus two .
17:36	And that's correct . It's very easy when you're multiplying
17:39	complex numbers and imaginary numbers to screw up the signs
17:41	because if you're trying to do I squared is negative
17:44	one . And you have a negative times negative and
17:47	all that , it's very difficult to keep it straight
17:49	. And that's why I'm showing you how I typically
17:51	do it . All right . Next thing we need
17:53	to do is multiply these together . X times X
17:56	gives me X . I'm sorry . X times X
17:58	squared gives us X cubed X times the two X
18:01	is two X squared X times the two is two
18:04	X . And then we have two X squared And
18:09	then we have four x . And then we have
18:12	four and then we collect terms again . All right
18:17	. So we have X cubed two X squared ,
18:21	two x squared . That's a four X squared .
18:24	Put a dot there and a dot there . Two
18:25	X and four X six X . And then we
18:29	have a four right here . And I'll double check
18:32	myself X cubed plus four X squared plus six X
18:35	plus four . That equals zero . And again we
18:37	check because the problem said , find a cubic equation
18:40	with integral coefficients with these routes . All of the
18:43	coefficients are basically there's no fractions are integral coefficients .
18:47	So I don't need to multiply this equation by anything
18:50	to clear out any fractions . I've already have integral
18:53	coefficients . So I'm good . But also keep in
18:55	mind this is just one of many uh Equations and
18:58	one of an infinite family that has the same routes
19:01	. If I multiply this equation by four or by
19:03	five or by 10 , I will get new coefficients
19:06	. The curve will look different but it will still
19:08	cross in the same locations . So I will have
19:11	if I did that , I would have a similar
19:13	equation that's related to this one that has the same
19:15	routes . Mm . Okay , last question it says
19:22	the following , find the remaining route uh and verify
19:36	, verify it . So what I'm gonna do is
19:37	I'm gonna give you some routes , you're gonna have
19:38	to figure out what route is left and go from
19:42	there . So the equation we have is x cubed
19:45	minus three , X squared plus four , X -12
19:50	equals zero . And the routes that I'm giving you
19:54	here R three and 2 I so the question says
19:58	find the remaining route and verify it . So what
20:00	you're supposed to do here is look at the equation
20:03	. You see that it has a degree of three
20:05	that tells you it's a cubic equation that tells you
20:07	there's only three routes of this equation . Some could
20:10	be real , some could be complex . That's the
20:12	first thing . And we've learned . The second thing
20:14	is I'm only giving you two of these routes .
20:16	So you know there must be a third one .
20:18	You know that you have a two I . Is
20:20	one of the roots and you know that imaginary or
20:22	complex roots always come in pairs . Conjugate pairs .
20:25	So if you know that this is a route then
20:27	you also then know that all of this stuff kind
20:30	of implies that negative two I the conjugate of the
20:32	two I is also a route . And they're asking
20:37	you to verify that there's a couple of ways to
20:39	verify it . We've actually used synthetic division and things
20:42	like this in the past . But the cleanest way
20:44	to verify this is a route is to put it
20:46	in here . So negative two . I cubed minus
20:50	three . Negative two I squared from this term four
20:56	negative two . I this term minus 12 . And
20:59	we're asking ourselves does this equal to zero ? If
21:01	it is equal to zero then this is a root
21:03	. And we've done everything correctly . And so it's
21:06	really easy to make mistakes here because notice we have
21:08	negative two . I the whole thing is cubed .
21:11	So what happens is that cube applies to the sign
21:14	which is like a negative one out there . It
21:16	applies to the tube and it also applies to the
21:21	to the eye right ? So if you think about
21:23	it negative times negative gives positive times negative again would
21:26	give you negative two times two times two . Actually
21:29	gives you eight . And then the I . Q
21:31	. We're gonna save that for later because if we
21:33	try to do all the signs associated with the imaginary
21:36	number at the same time as everything else , we're
21:37	gonna make a huge mistake eventually . So negative two
21:40	times negative two times negative two gives you negative eight
21:43	. I cubits this term . Then you have the
21:46	-3 . Then on the inside here again negative two
21:49	times negative two is positive four items . I as
21:52	I square . We are not trying to another other
21:55	students again . We'll try to do all of this
21:57	and also multiplied by the negative three . All in
21:59	the same step . That's a huge mistake . You're
22:01	going to make an error eventually . Now this one
22:04	there's actually no exponent of the exponents one . So
22:06	I can multiply this negative eight . I And then
22:09	I have -12 . Is that equal to zero ?
22:12	So over here I have negative eight I cube .
22:15	I'm gonna write that as I squared times I why
22:19	am I doing that ? Because I know that I
22:21	squared is negative one . And in the next step
22:23	I'll replace it with negative one . If you try
22:26	to do items , items . I it often gets
22:28	confusing . What what's going to happen just as an
22:31	intermediate step , right ? As I squared times .
22:33	I that's equal to I cube . Right ? This
22:36	is going to be negative 12 . I squared again
22:39	. I'm not gonna do anything with the eyes yet
22:42	. And then I have negative eight . I and
22:44	I have -12 . Is that equal to zero ?
22:48	Okay . What we have here is negative eight .
22:50	The I squared . Now it can insert that .
22:51	That's equal to negative one . I still have an
22:53	eye there and then I have minus 12 . The
22:56	I squared I can insert is actually equal to negative
22:59	one . Then I have negative eight . I then
23:01	I have negative 12 and that is that equal to
23:04	zero . So now I'm at the home stretch this
23:06	becomes positive eight I this becomes positive 12 . This
23:10	is negative eight I and this is negative 12 .
23:13	Is that equal to zero ? What you can see
23:15	that the ai and the negative eight I subtracted zero
23:17	. The 12 and the negative 12 subtracted 00 equals
23:21	zero is a check . So what you have found
23:24	that a root or the remaining route is indeed negative
23:27	two I . Which is what you're asked to do
23:30	. The question said find the remaining route and verify
23:32	it . And we know from the conjugate route theorem
23:35	that if this is a root negative too , I
23:36	must also be a route . And so we put
23:38	it in here and you might say why do I
23:40	crank through all of that ? And show you every
23:42	little step to verify that the root is correct ?
23:45	Because to be honest with you , there's a lot
23:46	of places to make an error here . When you
23:48	have imaginary numbers , Most students will try to Cube
23:52	the -2 and then Cube the I at the same
23:54	time keeping track of all the signs in their head
23:56	and eventually make a sign error there . You can
23:59	also do the same thing here Trying to for instance
24:03	, I had it right here and I wrote it
24:04	as negative 12 I square . But a lot of
24:06	students will try to , in their mind say ,
24:08	well negative 12 and then this becomes a negative one
24:10	that becomes a positive and they'll try to write it
24:12	, do it all in their head and you might
24:14	get the problem right , but you eventually will make
24:17	a make a sign error . And so that's why
24:19	I'm showing you the right way to do it .
24:20	You need to spend a little bit of time writing
24:22	it out so that you make sure you get the
24:23	right answers . So that was the conjugate route theorem
24:26	, one of the most important theorems of algebra ,
24:28	because it tells us that all of these routes that
24:31	are conjugated always appear in pairs , and it's true
24:34	for every polynomial . Uh And so every polynomial ,
24:38	what's the what's the wording we used ? Uh if
24:41	a polynomial with real coefficients has a complex route ,
24:44	then the conjugate of that route is also a a
24:47	route . So if the polynomial has real roots ,
24:49	which every problem we've done here , then this is
24:51	true . If you're polynomial in some crazy class down
24:54	the road has imaginary coefficient , then this actually isn't
24:57	true anymore . You don't have to have conjugate roots
25:00	. All right , so make sure I understand this
25:02	. Solve all of these problems yourself . Follow me
25:04	on to the next lesson . We'll get a little
25:05	more practice with the conjugate roots here in algebra .

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