18 - Descartes Rule of Signs, Part 1 (Find Roots of Polynomials) - Free Educational videos for Students in K-12

18 - Descartes Rule of Signs, Part 1 (Find Roots of Polynomials) - By Math and Science

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00:00	Hello . Welcome back to algebra . The title of
00:02	this lesson is called Deckert Rule of signs . Part
00:06	one . So the way you say this person's name
00:08	is Renee this first name Renee de cart and he
00:11	invented a rule that helps us figure out the possible
00:15	number of routes that a polynomial has . And so
00:18	just to kind of let you know where this the
00:19	name of this person comes from is extremely famous and
00:22	invented lots and lots of things when it comes to
00:25	also algebra and also higher level math . In fact
00:28	when you think of the xy plane , the X
00:30	Y plane , we used for everything from plotting functions
00:32	, plotting lines , plotting out circles and ellipses later
00:35	on , the xy plane is actually called the Cartesian
00:38	plane from the cart . The cart Cartesian , that's
00:42	where it comes from . So he invented lots of
00:44	very useful things and when working on these polynomial is
00:48	now you know that we have polynomial of first order
00:51	is a line . We know how many routes online
00:53	has , We know how to solve pretty much all
00:56	lines to figure out their roots . We know how
00:58	to solve quadratic those are degree of two , we
01:01	know how to factor them . And if we can't
01:03	factor them , we can always use the quadratic formula
01:05	for routes of three and four and higher things get
01:08	a little bit more difficult . We can sometimes do
01:10	it by guessing one of the roots and then using
01:13	division to find the other two routes . But when
01:16	the polynomial is get higher and higher , like let's
01:18	say 1/6 order polynomial or 1/7 order polynomial , there's
01:22	not a good clear way to just figure out the
01:24	roots . I mean now we can use computers to
01:26	do it , but I'm talking about by hand To
01:28	actually find the exact solution to a 10th order polynomial
01:31	is not something that we know how to do in
01:33	algebra , but we do have this concept called d
01:36	cards rule of signs and it allows us to predict
01:39	the possible number and the possible types of routes that
01:43	a polynomial of any order can have . In other
01:46	words , how many , how many positive roots will
01:51	the thing have ? In other words , 1234 How
01:53	many positive roots will it have ? How many negative
01:55	roots will the thing have . And then because we
01:58	know the total number of routes of any polynomial ,
02:01	then we can oftentimes figure out how many complex roots
02:04	. So we may not be able to actually find
02:07	the actual value of the roots . But using the
02:09	card rule of signs , we can figure out how
02:11	many routes there are and the different types . So
02:13	often the problem will say here's a polynomial . Find
02:16	the nature of the roots . And what they want
02:18	you to tell them is they want you to tell
02:20	them how many positive roots , how many negative roots
02:22	? How many complex roots does the polynomial have ?
02:24	Even if you can't actually solve them . So in
02:27	order to do this um we have to learn about
02:30	something called variation inside because it's basically what the cart
02:35	rule of signs is based on . So this thing
02:37	is called variation in side and it's just much ,
02:45	much easier to explain it with an example , it's
02:47	actually really , really simple . So let's take the
02:50	first one . Let's say we have 1/6 order polynomial
02:52	that goes X to the sixth power minus two X
02:55	. To the fourth power minus five X . To
02:59	the second power plus three X minus six . Now
03:03	the first thing you need to do is you need
03:05	to arrange the polynomial in order to figure out the
03:07	variation inside . You have to arrange the polynomial in
03:10	decreasing degree of exo six . There's no 5/5 power
03:15	term than four . There's no third power term .
03:18	Uh then two , then there's no there's one power
03:21	term right here . And then the zero power of
03:23	the constant term is on the end . So you
03:24	have to write it in decreasing order and then you
03:27	look for changes inside . So you see how this
03:30	first term is positive . There's an invisible positive here
03:33	, but the next term is negative . So this
03:35	means there is a one variation and sign . But
03:39	when we go from the second term to the third
03:41	term , see this one is negative , but the
03:42	next term is also negative . So this is not
03:45	a variation and signed between those terms , but this
03:48	one is negative and this one is positive . So
03:49	this is number two . I'm numbering them number one
03:52	. First variation inside . Second variation inside this one
03:55	goes from positive to negative . So it's also a
03:58	variation inside . So there are actually three variations inside
04:01	in this polynomial , three variations of sign or inside
04:09	. And then one more thing I want to note
04:10	. I'm not gonna write it down for every problem
04:12	. But you see how we didn't have 1/5 order
04:14	term of 1/5 power term . We didn't have a
04:16	third power term . So to figure out this variation
04:19	and sign , you're gonna ignore all of these missing
04:23	terms . Like I'm just gonna write it down missing
04:28	terms for instance zero , there's a zero X to
04:31	the fifth power in here because it's missing . There's
04:33	a zero X . To the third power in here
04:35	. That's missing . So we're ignoring those were not
04:37	because zero really doesn't have a sign . Zero is
04:40	neither positive nor negative . So we just ignore those
04:43	. We use We look for the variation inside .
04:45	Now . I haven't talent told you yet what day
04:47	cart rule of sign is , but you have to
04:50	figure out how many variations and sign you have for
04:53	a polynomial in order to use it . So that's
04:56	why we're doing that first . All right . Let's
04:57	do one more little practice with variation inside . Before
05:00	. I actually tell you what did cart rule of
05:02	fine . Really is the next polynomial is let's say
05:04	we have 1/5 order polynomial X to the fifth plus
05:09	X . To the fourth . Power minus three X
05:12	. To the second power plus four X plus six
05:16	. So this is 1/5 order . We know that
05:18	this is the sixth order . This polynomial has six
05:21	routes and this is 1/5 order polynomial . It has
05:23	five total roots . Uh But we're now looking for
05:27	the variation and sign positive positive . There's no variation
05:30	and sign here positive , negative . There's a variation
05:33	and sign right there negative to positive . There's variation
05:36	and sign right there positive to positive . There's no
05:39	variation and sign here . So , what I actually
05:41	have here is two variations inside . All right .
05:49	So , it's important . Now , we're gonna kind
05:51	of uh put it all together here . We're gonna
05:53	write day card rule . Sign down . It looks
05:56	very complicated when I write it all down , you're
05:57	gonna be like , whoa , this looks so complicated
05:59	. It's not complicated . Trust me , just let
06:00	me get through it . I have to write a
06:02	few things down , but it's actually one of the
06:04	simplest theorems that we have in algebra . But the
06:07	key to it is understanding how many variations in sign
06:09	you have in the polynomial that you're given . Because
06:12	that's how the thing is written . That's how the
06:13	theorem is written . The most important thing I want
06:16	to say before I write it down is it only
06:19	tells us the number of possible positive roots and the
06:23	number of possible negative roots . In other words ,
06:25	the cart rule of sign doesn't give you a concrete
06:29	answer . Uh it's hard to describe without a problem
06:32	and we're gonna do a problem . So don't sweat
06:34	it , we're gonna get there . But it doesn't
06:35	tell you , hey , you have five positive roots
06:37	and to negative roots , it doesn't tell you ,
06:39	hey , we have one positive route and three negative
06:42	. It doesn't tell you those things . It tells
06:44	you you might have one or two positive roots and
06:47	one or three negative roots . It gives you possibilities
06:50	. It doesn't give you concrete things . So you
06:52	might say , well who cares about that ? Trust
06:55	me in higher level math and trying to solve a
06:57	higher level polynomial that has its uses . So let's
06:59	go ahead and write it down and it's called as
07:02	we have said Deckert . That's how you spell it
07:05	day cart rule of signs . Alright . Uh I'll
07:15	be 100% honest with you when you get in the
07:17	higher level math , you might use this a little
07:19	bit , but for everyday use and algebra , it's
07:20	not something you use very much . But as far
07:22	as the theory of algebra , the theory of roots
07:25	of polynomial zits , it's centrally critical to that .
07:28	Okay . All right , so I'm going to kind
07:30	of simplify this . Um So let's let I'm gonna
07:34	do it in a shorthand way . In other words
07:36	, you're gonna see a much longer version of this
07:38	in your book . So I'm gonna say let P
07:41	. Of X be a polynomial uh with real coefficient
07:50	, that's pretty much every polynomial that we have ,
07:52	every coefficient of this . All the terms and this
07:55	polynomial . They're all real numbers . In other words
07:57	, you don't see any imaginary numbers are complex numbers
07:59	anywhere in here . All of the coefficients are just
08:02	numbers negative and positive numbers . All right . So
08:06	let's let some polynomial B A . P of X
08:08	have real coefficients . Alright , So there's two parts
08:11	to decorate rule of signs . All right . The
08:14	the number of positive , whoops , positive . I
08:20	think . I must feel positive . Sorry , positive
08:24	, positive . Yeah . Number of positive roots um
08:29	is equal to the number of variation inside . Now
08:44	. This would be easy if it would just stop
08:46	here . Right ? But it's not equal to the
08:47	number of variation inside . It's equal to the variation
08:49	of the number of signs of the polynomial variation in
08:53	signs of the polynomial or less . This is really
08:57	important , or less than this um by an even
09:04	number . All right . So the bottom line is
09:09	if you want to know how many routes , how
09:11	many positive roots of polynomial house . When I say
09:13	positive roots , I mean routes that are like 1354
09:17	to anything on the positive side , positive numbers that
09:20	are roots then day card rule assign tells you that
09:23	. But it doesn't give you a concrete answer because
09:25	if you find the number of the variation and signed
09:27	there , It's telling you that the number of positive
09:29	roots is equal to the however many variation and sign
09:31	you found . Or it can actually be less than
09:35	that number , but not just less than it can
09:37	be less than that number by an even number .
09:39	So for instance if you have five variations and sign
09:43	then the number of positive roots roots can be five
09:47	or three because that would be to be even number
09:51	less . So you can be three positive roots or
09:54	one positive route you see . So the variation and
09:57	sign tells you the maximum number of positive roots but
10:00	you could have less than that but not just less
10:01	than it can be less than that number by an
10:03	even number meaning by you have to count by twos
10:06	down basically that's what even numbers are when you count
10:08	by twos . So if you had five variation and
10:11	sign , you have five positive roots or three positive
10:13	roots or one positive roots . That cart rule of
10:15	mine doesn't tell you how many you have . It
10:17	tells you the possibilities . But you know for sure
10:20	that if you had a polynomial with five variation and
10:23	sign , you will definitely not have four positive roots
10:26	. Because you can't because if you had five variation
10:29	and sign , then it has to be that or
10:31	less by an even number . You can't before you
10:33	have to count by twos down . Right . You
10:36	can't have for instance zero uh positive real roots .
10:40	Because that it's not an even number less as well
10:43	. All right . Now there's a part two to
10:45	this you might say . Well this tells me how
10:46	many positive roots I have . But what about the
10:49	negative roots ? What about negative one being a route
10:51	? What about -17 being a route . What about
10:53	negative one half being a route . How many of
10:55	those roots do we have ? So it's very similar
10:58	to this . Um But you know very similar .
11:02	Let me just get it on the board . The
11:05	number of negative roots is um All right . Equal
11:15	to the number of variations in signs of And this
11:28	is critically important . Not the variation and sign of
11:31	the polynomial you started with but the variation and signs
11:34	of the polynomial replaced with negative X . You see
11:38	replaced with negative X . I'll show you how to
11:39	do that in a second and the same sort of
11:41	thing at the end or less . Then this bye
11:47	. And even number now . It kind of makes
11:51	sense that you have something to do with negative X
11:53	here because you're looking for negative roots . So what
11:55	you have is you have some polynomial . We know
11:57	how to figure out how many positive roots are possible
11:59	. Right ? How do we figure out how many
12:01	negative roots are possible ? You take your polynomial for
12:04	instance one of these here everywhere where you put an
12:07	X here . You stick a negative X in its
12:09	place . That's what this P of negative X means
12:11	. You know if you have P . Of one
12:14	, then you put the number one into the polynomial
12:16	. If you have a P . Of five ,
12:18	you put five everywhere where you see X in the
12:20	polynomial . If you have P . Of 17 ,
12:22	you calculate by putting a 17 everywhere where you see
12:24	X in the polynomial . If you have a P
12:26	of negative X , then everywhere where you see an
12:29	X . You just stick a negative X in there
12:31	. That's going to slightly change the polynomial because you
12:34	put a negative X in there , right ? And
12:35	you find the variation inside of that version . And
12:38	then however many variation of sign you have is the
12:41	number of negative roots . Or it can be less
12:43	than that by an even number . Sounds really complicated
12:46	because there's a lot of words on the board .
12:48	But it's actually really easy . So let's go and
12:51	do the first one fully the full the full problem
12:55	. So the way that this problem might go ,
12:58	it would say list the nature of the roots positive
13:01	, negative or imaginary roots of the following polynomial .
13:04	The polynomial is P . Of x . Is X
13:08	to the 5th Plus X . to the 4th minus
13:13	three X squared . Sorry , this is 1/4 right
13:16	here . Three X word . Then we have four
13:20	X . And then we have plus six . Right
13:23	? So we're trying to find the roots and you
13:25	know that roots mean that you take P . Of
13:27	X and you set it equal to zero . So
13:28	we want to find how many and and what are
13:30	the locations of those routes ? So we look at
13:33	the variation and sign to figure out how many positive
13:36	roots we could possibly have . We have to figure
13:38	out the variation and sign of this polynomial that were
13:40	given . So this polynomial has a positive term ,
13:44	positive term . And notice , first of all we
13:46	wrote it in descending order of X . So that
13:48	has to be done . So there's no variation in
13:50	sign there . But there is a variation of sign
13:52	here . Then we have negative to positive here .
13:54	We have a variation and sign right here and positive
13:57	to positive there . So what we have for the
14:00	basic polynomial ? Yeah P of X . As we
14:03	have two variations inside . Right ? So what does
14:11	that mean ? It says the number of positive roots
14:14	is equal to the number of variation and signs or
14:17	less than that number by an even number . Which
14:20	is a fancy way of saying . You count by
14:22	twos downward . That's all it's saying . So since
14:25	we have two variations and sign what this thing is
14:28	really telling us is for the number of positive roots
14:34	. That's what the pos means means positive . It
14:36	can equal to positive roots or zero positive roots because
14:40	this is less than by an even number . So
14:42	you start with two and you count down by two
14:44	, you can't go down below zero anymore , you
14:47	can't have negative number of routes . So you basically
14:49	stop there . So we know we do not know
14:52	what the roots of this fifth order polynomial is ,
14:54	but we know that there's got to be five total
14:57	routes because across real or imaginary roots we have to
15:00	have that amount . We learned that a long time
15:03	ago right now we know that there could be two
15:06	positive roots or there might be zero positive roots .
15:09	We don't have any idea which we don't know .
15:11	Dick Heart rule sign doesn't tell us , but we
15:13	know that we can't have one . We cannot have
15:15	one positive route , we cannot have three positive roots
15:17	, We cannot have four positive roots . We cannot
15:19	have five positive roots . We can only have two
15:21	or zero . That's it . That's the only two
15:22	possibilities . So , it eliminates a lot of possibilities
15:28	. All right . So that's for the number of
15:30	positive roots . Now , for the number of negative
15:33	roots , it says we have to do the variation
15:36	and sign on a modified version of the polynomial .
15:39	So what we do is we say , well ,
15:40	let me take this polynomial and I'm gonna stick a
15:42	negative X everywhere . I see an X . But
15:45	you have to be careful because you're raising everything to
15:47	an exponent . You have to do it like this
15:49	negative X . You have to raise the whole thing
15:52	to the fifth power . You don't want to put
15:54	like negative X to the fifth , That's different .
15:57	You want to take the whole thing . And because
15:59	you're sticking it into there so you know you don't
16:01	want to do that , you want to put it
16:03	in and raise the whole thing and we'll calculate the
16:05	answer to this in the next step . Then we
16:07	have negative x rays to the fourth . Then we
16:10	have negative three negative x rays to the second power
16:15	. And let me go ahead and just take a
16:19	second and get this out of the way . We
16:21	don't want to do that , we decided we don't
16:22	want to do that . Then we have over here
16:24	plus four negative X . Um and then we have
16:30	plus six . So now what we have to do
16:32	is calculate kind of the new version . Now ,
16:36	you know whenever you raise negative numbers to an even
16:39	power , then the signs go basically make it positive
16:43	. When you raise negative one or negative numbers or
16:45	whatever to odd powers in the sign becomes negative because
16:48	negative one times negative one times negative one an odd
16:51	number of times makes it negative . So here we
16:54	have this negative term raised five times . So what
16:57	we're going to have is a negative X to the
16:58	fifth . So it worked out this way because the
17:01	negative one was multiplied by itself five times . Which
17:04	makes a negative in front . The X was multiplied
17:06	by itself five times . Makes X to the fifth
17:08	. But here this negative basically goes away . We
17:10	have a positive X to the fourth because the negative
17:13	one was multiplied 1234 times that kills it makes it
17:16	positive . Here we have a negative three but this
17:20	negative X squared . Makes it a positive X squared
17:23	. Right , We'll do the signs later over here
17:25	. Then we have four times negative X means negative
17:28	four X . And then we have positive six .
17:31	So really I put the parentheses around here , but
17:34	you know , you don't really need them . So
17:35	I don't need to rewrite the thing again . I
17:37	just want to make sure you understand how we're getting
17:38	the signs correct everywhere . And it's because of the
17:41	way we're raising the exponents of the negative powers here
17:45	. All right . So the next thing we're gonna
17:46	do , we catch up to myself and make sure
17:47	I've done everything correct . This is a variation of
17:50	sign going from negative to positive . This is a
17:52	variation of sign going from positive to negative . This
17:56	is no variation inside , but this one is a
17:58	variation inside . So there's your third variation inside .
18:02	So , what this means is kind of mirroring what
18:05	we did before we have three variations inside . Yes
18:13	. Right . And so then from that we can
18:15	figure out the number of negative roots Has to Equal
18:22	1 , 2 , 3 . It has to be
18:23	three . But according to the theorem the number of
18:26	negative roots is equal to the number of variation and
18:28	signs of this modified polynomial or less than that by
18:32	an even number , which means you count down by
18:34	two . So it can be three or one .
18:36	It can't be less than that because you can't subtract
18:38	two and still have a positive number or zero .
18:41	So you see what you have learned here is that
18:43	you know That the number of positive roots of this
18:45	parents polynomial can be two or 0 . The number
18:48	of negative roots of this parent polynomial can be three
18:51	or one . But you see you have a lot
18:53	of possibilities right ? Because you know that you have
18:57	five total roots total . So do you have to
19:01	positive and three negative , which would be five ?
19:04	Or do you have some other combination of them ?
19:06	Because it doesn't tell you how many negative and how
19:08	many positive exactly ? It just gives you options .
19:12	So we know for instance , we don't we cannot
19:14	have two negative roots , we cannot have four negative
19:17	roots for instance . So the best way to do
19:19	this is to write a table and we know that
19:22	there's five total roots of this polynomial . But don't
19:27	forget that for a polynomial in general you can have
19:30	positive roots , you can have negative roots and you
19:32	can have complex roots , right ? One plus two
19:35	, I one minus two . I . And we
19:37	know that those complex roots if there there are always
19:40	going to be in conjugate pairs because of the conjugate
19:42	route theorem that we've already learned . So the best
19:44	way to do this is to is to write a
19:46	table down . So we know we have five total
19:49	roots , and we know that we're gonna have some
19:51	number of positive roots , right ? We know that
19:56	we're gonna have some number of positive roots . We
19:58	know we're gonna have some number of negative roots ,
20:03	right ? And then we also know that we could
20:04	have some number of complex roots . Yes . And
20:10	when I say complex roots , I'm saying it could
20:12	be one eye to eye , pure imaginary , or
20:16	it could be one minus two , I one plus
20:18	two . Anything with an imaginary number is going to
20:19	go in this column . So what you have to
20:21	do is you have to take into account all the
20:23	possibilities here . So the cleanest way to do this
20:26	is you start with the positive roots , you could
20:29	have two routes , I could have two positive roots
20:32	, but if I lock this down , kind of
20:34	grab it and lock it down , I could have
20:36	two routes but I still have two possibilities of the
20:39	number of negative roots . Three or one . I
20:41	could have three or one negative roots . While I
20:44	while I lock this down saying I have to positive
20:46	route , so I could have three negative roots .
20:49	But if I have to positive and three negative ,
20:51	there's five total roots in this polynomial . Two plus
20:54	three is five . So in that case there would
20:55	actually be zero complex roots . Right ? If that
20:59	were true . Now if I'm locking this down this
21:01	to I've already done the three now , I can
21:03	say well what if I had the one negative roots
21:05	? So I would have in that case to positive
21:07	roots , one negative route that only gives me three
21:11	real roots . But I know there has to be
21:13	five . So there has to be in that case
21:16	too complex roots because they have to add up to
21:20	five . Right ? In this case it happens that
21:22	both were real and I didn't have any imaginary roots
21:24	or complex roots here in this line . If this
21:26	were the case , that's true , then I would
21:29	have to positive one negative too complex . So that
21:32	was if I held this constant , I'm done with
21:35	that , there's no other possibilities here . Right ?
21:37	Then I go to zero and said , well what
21:38	if I had zero positive roots and I could have
21:40	three or one negative roots ? So I could have
21:43	zero positive locking that down . I could have three
21:46	negative if those are if that's the case and that's
21:49	only three real roots , I would need again to
21:52	complex roots to make it equal five and rounding it
21:56	out . If I had zero positive and one negative
21:59	, Right , then that's only one . If you
22:02	add , there's only one real root . Now I
22:03	would have to have four um complex roots . So
22:07	let me double check myself to 30 to 1 to
22:10	032014 That's correct . So notice a couple of things
22:15	. The number of in the complex column here is
22:19	always going to either be zero meaning there's no complex
22:21	roots or it's going to be an even number .
22:23	So I had to complex roots or four . Why
22:25	do you think it has to be an even number
22:27	? It's because the conjugate route there and tells us
22:29	when we have complex routes they have to come in
22:31	pairs , you have to have one plus two I
22:33	and one minus two . I that's too you have
22:36	to have a negative one plus seven I . Negative
22:39	one minus seven I . That's too . So you
22:41	can never have an odd number of these complex routes
22:44	because they always have to come in pairs . And
22:46	so you can see that even though we don't know
22:49	what the roots of this equation are , We have
22:51	built a table . We know there's only really four
22:53	possibilities either 230 to 1 to 032 or 014 And
22:59	it's it's it's something that is useful in higher level
23:02	math . Sometimes when you're solving a very complicated problem
23:04	, maybe , you know in quantum mechanics or gravity
23:07	relativity theory or something , you have to solve a
23:09	large polynomial . Maybe we don't know how to solve
23:11	that polynomial . Maybe it's a polynomial so complex ,
23:14	believe me , they do exist , that we don't
23:16	know how to solve it . But by using something
23:18	like this , you can figure out what is possible
23:21	. So , you know that there's only four possibilities
23:23	. And you can investigate those four possibilities . You
23:25	know , for instance , that you can never have
23:27	, you know , something that's not in this table
23:31	as a possible solution set of this polynomial . That's
23:34	what Dick Art Rule of sign is all about .
23:36	So we've done this topic , I I think it's
23:38	a good idea if you do this one yourself just
23:39	to make sure you understand , but we do have
23:41	a few more problems . So follow me on to
23:43	the next lesson , we'll get a little more practice
23:44	with this day cart rule of signs .

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18 - Descartes Rule of Signs, Part 1 (Find Roots of Polynomials) is a free educational video by Math and Science.

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