10 - The Remainder Theorem of Synthetic Division & Polynomial Long Division - Part 1 - Free Educational videos for Students in K-12

10 - The Remainder Theorem of Synthetic Division & Polynomial Long Division - Part 1 - By Math and Science

Transcript


00:00	Hello . Welcome back to algebra . We're going to
00:02	conquer the topic called the remainder theorem of polynomial in
00:06	algebra . This is part one of two . So
00:08	we're covering what we call the remainder theorem . Now
00:11	keep in mind that we're going to understand and learn
00:13	this theorem in this lesson . But in the next
00:15	couple of lessons will conquer something called the factor theorem
00:18	. The remainder theorem and the factor theorem kind of
00:20	go together like peanut butter and jelly . They they
00:23	very much are cousins of one another . But I
00:25	don't want to put them in the same lesson because
00:27	it will make it too long and too cumbersome .
00:28	So just keep in mind that you are going to
00:31	understand the remainder theorem in this lesson . But when
00:33	we get to the factor theorem , I'll be referencing
00:35	this theorem and it'll come together and even make more
00:38	sense once we can put both of them together .
00:41	Now this theorem frequently gives students problems . It doesn't
00:44	make a lot of sense the first time you read
00:45	it . So what I'm gonna do is I'm gonna
00:47	cut to the chase . I'm gonna write the theorem
00:49	down . You won't understand how it's true or why
00:52	it's true , but I'll explain what it does .
00:54	Then we'll do a quick example to show you how
00:56	to use it . That's what I really want you
00:58	to be able to do . It's very , very
00:59	simple to use the remainder theory of polynomial . And
01:03	then what I'll do is I'll I'm not gonna prove
01:05	the theorem , but I'm gonna show you why it
01:07	works . I want you to know why it works
01:09	. And then we'll wrap up the lesson with another
01:11	example . So we're gonna get a couple of examples
01:13	here , we're gonna get some a little bit of
01:14	theory , a little bit of background about why it
01:16	works . And then in the next lesson , will
01:19	crank up the complexity a little bit more . So
01:21	here we're talking about the remainder theorem . So you
01:24	might think it has something to do with the remainders
01:26	and you'll be right . So the remainder zero ,
01:31	that's what the th means remained your theory . Alright
01:34	, so here's my best guess at my best way
01:36	of writing it down in a way that's going to
01:38	be to the point if the polynomial , that's what
01:43	Polly means . P vex So I'm just naming any
01:46	polynomial P of X . Can be any polynomial you
01:48	want any degree , right ? Is if that polynomial
01:51	is divided , buy some quantity X minus C .
01:58	Now this could be X minus one , X minus
02:00	three . X minus four . Also it could be
02:02	because C could be positive or negative , it could
02:04	be X plus five X plus 10 . You know
02:06	, just like we were doing for synthetic division ,
02:08	if it's divided by anything like this , then the
02:12	following thing is true , then the remainder . That's
02:17	why it's called the remainder theorem . When you do
02:21	that division , the remainder left over . When you
02:22	do that division is the value P evaluated at seat
02:30	. Now I'm gonna let that sink in for a
02:31	little bit because it doesn't make a lot of sense
02:34	. We normally think about remainders as being something to
02:37	do with division because that's that's what it is ,
02:39	right ? But now I'm telling you that this remainder
02:41	is kind of connected to the value PFC . What
02:44	am I trying to say here ? I'm trying to
02:46	say that we know how to evaluate polynomial . When
02:49	I say PFC , that means I'm putting some number
02:51	into the polynomial . I'm calculating what it can do
02:54	. So let's say I want to calculate this value
02:56	of this polynomial when X is equal to two .
02:59	Of course I can do it manually . I can
03:01	just stick it in in the value and calculate the
03:03	polynomial right ? But what I'm also saying is there's
03:05	another way to find the value of that polynomial evaluated
03:09	at that number . And that way to do it
03:12	is to divide the polynomial by by that same number
03:15	basically X minus that number and then crank through it
03:18	all and figure out what the remainder is at the
03:20	end . The remainder is gonna be the value P
03:22	evaluated at sea . So for an example , I
03:26	always like to give examples that make everything easier .
03:28	Let's say we have some polynomial can be any polynomial
03:32	you want But I'm gonna pick one here . Two
03:34	X cubed minus seven X squared plus five x minus
03:38	one . So here's a polynomial . You all know
03:41	that I can put numbers into this polynomial and I
03:43	can calculate the answers . That's what we have been
03:45	doing forever . So I can put the number one
03:47	in here and I can calculate the value . I
03:49	can put the number 17 in here and calculate the
03:51	value . But you also have to agree that because
03:53	it's cube terms and X and X square terms and
03:56	so on that , it might be a little cumbersome
03:59	to evaluate that polynomial by sticking the number in there
04:02	. Because let's say I put the number 11 and
04:04	I'll have to take 11 cube then multiplied by two
04:07	, then 11 square , then multiply by negative seven
04:10	, then 11 times five and all this . And
04:12	I have to add all the terms together . So
04:13	there's a lot of multiplication is going on because of
04:16	the exponents . And then I have to add everything
04:18	together . So if I want to evaluate the value
04:20	of this polynomial and a number I can do it
04:22	. Of course I can . But I have to
04:24	crank through all of those exponents and and do it
04:27	. What this remainder theory is telling you telling you
04:30	that if you want to find , I can find
04:36	any value I want . But let's just pick one
04:38	p evaluated at three . So if I wanted to
04:40	do that I would stick three in here in Cuba
04:42	and three in here and square it three and so
04:44	on and multiply and add everything . But another way
04:47	to find it , then The value p . of
04:52	three that I'm seeking to find is the remainder left
04:59	over . When we divide This polynomial . That in
05:07	question divided by X -3 . I need to let
05:11	you sink this have the sink in because really the
05:15	black text is kind of like what you might see
05:17	in a book . But really everything below the black
05:20	text is really what I want you to think about
05:21	because it's much easier say you have some polynomial ,
05:24	you want to figure out what the value of it
05:25	is that X is equal to three . Of course
05:27	I can put the number three in there and calculated
05:29	. But there is an alternative way to find out
05:31	what the value is here . And that alternative way
05:34	is to take this polynomial and divide it by X
05:37	minus whatever I'm evaluating and at But I don't care
05:41	about the value of the division . I only care
05:43	about the remainder . Just that single number at the
05:46	very very end . That number is the one I
05:48	circle and that's gonna be the value P . Evaluated
05:50	at sea . I'm p evaluated at the # three
05:54	. So , let's give a concrete example . I
05:56	want to take this polynomial here . I want to
05:58	divide I want to figure out what P evaluated at
06:01	three is All right ? So , I'm telling you
06:04	that I take this polynomial which is too negative 75
06:08	negative one . I'm gonna do synthetic division here .
06:10	Right ? I'm telling you I have to divide it
06:12	. Of course I could do the long division .
06:13	There's no problem with that . But we already know
06:15	that synthetic division is much much easier to do when
06:19	we're dividing by X minus the number or X plus
06:21	the number . Remember I told you that there would
06:23	be other uses for synthetic division and here's here's one
06:26	of them . So , we're dividing by something that
06:29	allows us to use synthetic division . We're just gonna
06:31	do synthetic division because it's easier to do . So
06:35	, I'm gonna do this here . And I need
06:36	to divide by x minus three . We know when
06:38	we do synthetic division , that I have to put
06:40	the opposite sign here . The negative three becomes a
06:43	positive three . We're gonna do this division . So
06:45	I draw my horizontal line . I dragged my two
06:48	down three times two is six , add these together
06:51	, get negative 13 times negative ones , negative three
06:54	, add these together . Get a 23 times two
06:56	is six and add these together and you get a
06:59	five . I don't care about any of these numbers
07:02	or any of these numbers or any of these numbers
07:04	or any of these numbers . The only thing I
07:06	care about for the factor for the remainder theorem is
07:09	this number , the remainder was equal to five .
07:12	And what that means is that the polynomial evaluated at
07:16	the number three is actually equal to five . It's
07:19	crazy , right ? It doesn't seem like that would
07:20	be true , but it is true . Let's go
07:22	verify this . The polynomial is given us two x
07:26	cubed minus seven X squared plus five x minus one
07:29	. So let's evaluate P evaluated at three . So
07:33	it's two X cubed . So we'll say 23 cubed
07:37	minus seven X squared 73 squared plus five X ,
07:43	Which is five times 3 and then -1 . All
07:47	right . So what we have here is two times
07:50	three cube is 27 . Uh And then three squared
07:54	is 99 times seven is going to be 63 .
07:57	And then we have five jumps , three is 15
08:00	-1 . And if you go crank through this two
08:02	times 27 , you subtract the 63 and add these
08:04	numbers and so on . You're going to get five
08:08	, which is exactly what we predicted . So why
08:10	do we care about the factor theorem or the remainder
08:13	? I keep saying factor theorem because we're gonna be
08:15	learning about the factor theorem . It's very closely related
08:17	to this . The reason we care is because when
08:20	we're evaluating polynomial nowadays with computers , it doesn't matter
08:23	because a computer , you know , you can you
08:25	can calculate anything you want , but if you were
08:28	doing this by hand then you would have to calculate
08:31	three times three times three and then three times three
08:34	here and then three times five here . And you
08:36	have to multiply and add everything now . That's just
08:38	for the number three . What if I were calculating
08:40	the polynomial evaluated at the number 127 ? Right then
08:45	I would have to cube 127 and I have to
08:47	square 127 and multiply 127 times five . And multiplying
08:51	ad to get the value of what it is .
08:53	It's going to actually be easier to do the synthetic
08:55	division here because I don't have any cubes anywhere notice
08:58	I haven't cube anything . I just dropped . Multiply
09:01	drop , multiply drop , multiply drop all I care
09:03	about the last number . So the remainder theorem is
09:07	very often used , especially when you're doing things by
09:09	hand . But there's also another use for it because
09:11	we're going to tie it into what we call the
09:13	factor theorem Later on . For right now , I
09:15	only want you to know when you want to find
09:17	the value of a polynomial at a number just divide
09:20	by X minus that number and then the remainder that
09:23	you get is going to be the value that you
09:25	seek . That's really all I'm trying to say .
09:26	And so that's what the black text is saying .
09:28	If the polynomial P of X is divided by this
09:31	number , then the remainder . When you do that
09:33	is the value of P evaluated at that number .
09:36	Okay . Now we want to turn our attention to
09:38	why does it work ? I'm not going to do
09:40	a rigorous proof of it . So we're gonna say
09:43	why does the remainder theorem work ? Why does it
09:52	work ? I think uh even though we're not gonna
09:54	do a rigorous mathematical proof of it , we're not
09:56	gonna do a 10 page proof of why the remainder
09:58	theorem works . I think it's instructive for us to
10:01	decompose a little bit more . What what's really happening
10:04	here ? So notice that it all hinges back to
10:07	this division process . Were saying that when we take
10:08	the polynomial we divide by this , something magical happens
10:11	where the remainder becomes important , basically . So let's
10:15	kind of explore that a little bit . If we
10:17	take this polynomial whatever it is , I'm gonna generalize
10:20	it here , right ? And we divide by something
10:23	on the outside . It goes out here , right
10:26	? And the answer that you get , we call
10:28	it Q . Of X . Why ? Because we
10:31	have names for all of these things . This thing
10:33	is called the quotient . Okay . This thing on
10:38	the inside is called the dividend . Normally I don't
10:42	care about labelling things , but it's gonna help us
10:45	here in a second . This thing on the outside
10:47	this whole term is called the divisor . That's what
10:50	I'm dividing by now . When I go through this
10:52	process , you know , I'm gonna multiply going to
10:55	subtract and all this stuff . I'm gonna go blah
10:57	blah blah all the way to the end . Eventually
10:59	I'm gonna get some remainder . And this is called
11:02	the remainder . We've done long division enough to know
11:06	that you're going to get some kind of remainder at
11:08	the end . Right ? So let's go ahead and
11:10	go through this process . I've already done it with
11:13	synthetic division . We know the process ends with the
11:15	remainder of five . But let's just go through it
11:18	because it won't take long . And it's going to
11:20	help me explain to you why this remainder theorem actually
11:23	works . So for this polynomial we had two X
11:26	cubed minus seven X squared plus five X minus one
11:32	. And we were dividing it by x minus three
11:36	. Remember we said well we take the polynomial divide
11:38	by X -3 . That's gonna give us the value
11:40	. The remainder will be the value P evaluated at
11:42	three . So let's go ahead and do this real
11:44	quick , verify what the remainder is and then we
11:46	can crank through it . So X . Times what
11:49	gives me two X cubed . It's got to be
11:50	a two X squared there multiply down there . I'm
11:53	gonna get a two X cubed . Multiply down here
11:56	will be negative six X squared . Going a little
11:58	fast because we've done division many many times before .
12:01	We subtract both of these we get zero here seven
12:04	minus and negative 67 I'm sorry negative seven basically plus
12:07	six negative seven minus and minus six is negative seven
12:10	plus six . And that's gonna be negative one X
12:13	squared . So negative one X square . Then I
12:16	take and drag my next thing down like this X
12:19	. Times what gives me negative X . Squared .
12:21	It's gonna be negative X . Multiply here gives me
12:24	negative X squared negative X . Times . This gives
12:26	me positive three X . And I need to subtract
12:30	these two . These subtract to give me 05 minus
12:33	three gives me two X . Then I have to
12:36	drag my next guy down which is -1 X .
12:40	Times something gives me two X . It has to
12:42	be a to multiply . It's gonna give me a
12:44	two X . Two times the negative three gives me
12:46	negative six . I have to subtract them negative one
12:50	minus a negative six means negative one plus six .
12:53	Which gives me a five . And that's exactly what
12:55	we said . The remainder had to be equal to
12:56	five . I just wanted to do it again so
12:58	you can see I'm not doing any funny business with
13:00	the synthetic division . I'm taking the polynomial . I'm
13:02	dividing by X minus three the remainder . I'm getting
13:05	us five . And we've already shown you that when
13:07	you plug the value of three and you get a
13:09	five . Now what we want to do is explore
13:11	a little bit more closely why it falls out that
13:14	way . And does it work for all numbers ?
13:15	Right . All right . So what we want to
13:18	do ? Because we want to talk about if we
13:21	do this division here , right ? Or in general
13:24	, any kind of division here , how do we
13:26	check the answer ? Right ? If we wanted to
13:28	check the answer here , we've done that many times
13:30	before . When I taught you division , I'll taught
13:32	you how to check it . I said , well
13:34	, what you do is you take what you get
13:36	at the top , you multiply while it's out here
13:38	, and then whatever you get there as a result
13:41	, you add the remainder and then what you should
13:43	get back is what's underneath . So , another way
13:46	of saying is , is that is you take the
13:47	divisor , You multiply it by the quotient , which
13:50	is what the answer that you got was . Uh
13:53	and you're gonna get something but you have to add
13:55	the remainder back into it to get what's underneath .
13:57	This is exactly the same thing as dividing , you
14:00	know , uh dividing um 19 by three or something
14:05	like this . And you're gonna get a five here
14:07	, that's 15 . You're gonna subtract and you're gonna
14:10	get a four , you're gonna get a remainder of
14:11	four , right ? So what we're basically saying is
14:14	that in order to check this division , you take
14:16	what's in front the three , You multiply it by
14:19	the five and you have to add the remainder back
14:21	in . So you're gonna get 15 plus four .
14:24	And if you get 19 if you get what's underneath
14:26	here , then the division was correct . So we're
14:28	doing the same thing here , we take the divisor
14:30	. We multiply by the quotient . We add the
14:32	remainder and what we get should be equal to what's
14:35	under here the dividend . So if you're going to
14:38	generalize that , the way you say it is ,
14:41	you say the polynomial P of x is going to
14:47	be equal to the Q of X . That you
14:49	get for the answer that close in their times the
14:53	divisor x minus C plus the remainder . All I've
14:56	said is this polynomial here is going to be equal
14:59	to the divisor times the quotient , plus the remainder
15:02	were just checking our work and it has to equal
15:04	what's under here . So for this case it's this
15:06	times this plus the remainder has got to equal this
15:09	. All right . So let's generate let's kind of
15:11	change it a little bit into the problem that we
15:14	actually have . So what we're basically saying is this
15:17	polynomial P of X , which we know what it
15:20	is . Is given in our problem statement is going
15:22	to be equal to Q of x Times X -3
15:27	because we divided by X -3 plus the remainder .
15:31	But what I'm trying to say is this checking of
15:34	the work here works for for for the polynomial whatever
15:38	you've divided by When you multiply these guys and add
15:41	the remainder , it must be what is you must
15:43	recover what is uh under here . And this is
15:47	equivalent . What we're saying is the polynomial is actually
15:50	equal to this thing , whatever it is . So
15:52	that means I can put whatever value of X into
15:55	both sides that I want to . So what if
15:58	I want to continue , I will say , well
16:00	what if I want to evaluate the polynomial at the
16:03	value of three ? That's what I want to do
16:05	. So that means that I'll take the quotient that
16:08	I get as an answer and then I'm gonna plug
16:11	the value of three in here , right ? Um
16:16	And I'm gonna have to add my remainder . Notice
16:18	what's happened here , 3 -3 is zero . I
16:20	guess what I'm trying to say is when you divide
16:23	by this , you're dividing by a very special thing
16:27	and you're dividing by X -3 because eventually I want
16:30	to plug in a value of three there . So
16:32	what is going to happen is when I plug a
16:34	value of three in there , this goes to zero
16:36	. So P evaluated at three is equal to zero
16:40	plus the remainder , which means the value evaluated at
16:43	three of this polynomial is equal to the remainder ,
16:46	which is equal to five . So , the the
16:49	uh polynomial evaluated at three is equal to five .
16:54	All right . And this is not a proof ,
16:56	but this is showing you that it works and why
16:58	it works . The reason why it works is because
17:00	when you do division what you have is the answer
17:03	that you get at the top times your divisor and
17:06	then you add your remainder in . But if you
17:08	divide this polynomial by a very special number , A
17:11	number such that if I am trying to evaluate it
17:14	, X is equal to three . I'm gonna put
17:15	X -3 here so that I will get a zero
17:18	here . Then whenever that happens the remainder is what
17:21	is equal to what I'm trying to find . And
17:23	it always works out that way . If I'm trying
17:25	to evaluate the polynomial X is equal to 10 ,
17:28	then I want to divide by x minus 10 .
17:31	Why ? Because when I put the 10 in here
17:33	which would be like 10 minus 10 Be divided .
17:36	He basically divided by X -10 . So this equation
17:39	will be having X -10 year But I'm gonna put
17:41	the value of 10 and here which drops the whole
17:43	term out . So that's why it's not so much
17:45	of a coincidence that whenever it's not a coincidence that
17:49	you're want to evaluate the polynomial X is equal to
17:52	three and you divide by X -3 . It's not
17:55	a coincidence at all . It's because when you put
17:57	a value of three in here , then when you
18:00	kind of quote unquote , check the polynomial , the
18:04	this term drops away . So whatever you're trying to
18:07	evaluate at , you divide by X minus that and
18:11	then the remainder that you have left over is always
18:13	going to be equal to the number that you seek
18:15	. That's why it works . All right . So
18:17	I want to do one more problem . I know
18:19	it's a little bit fuzzy . I mean even to
18:21	me I know what the sterile means , but it's
18:23	a little bit weird . But the reason it works
18:24	is because of the way division works . When we
18:27	divide by things , then in order to check them
18:30	, we have to backwards multiply and and and plug
18:32	in and add the remainder back in . So if
18:34	we divide by the by the very special thing that
18:37	we that we we we know we're going to get
18:39	it's gonna drop away to zero . If we basically
18:42	divide by X minus the number that we're trying to
18:46	evaluate at the term drops away and then the answer
18:49	is what the remainder is . So it's kind of
18:50	going a backdoor way of figuring out what that polynomial
18:53	is equal to . So just give one more example
18:56	, we're gonna use the remainder theorem to evaluate the
18:59	following polynomial P of x X Cubed -2 , x
19:04	squared -5 X -7 . And we're gonna evaluate ,
19:10	we want to find P of four . We want
19:14	to find the value of P is equal divorce .
19:16	So this means in terms of our theorem , C
19:19	is equal to four . What do I mean by
19:20	sea ? The theorem says if a polynomial P of
19:24	X is divided by x minus C , then the
19:27	remainder is the value P of C . So I
19:30	kind of bring the letter C . And they're not
19:32	to confuse you just because it's kind of a generalized
19:34	thing . And that's how the theorems are written in
19:36	your books . But the bottom line is if I'm
19:38	trying to evaluate this thing at P is equal to
19:42	four , all I have to do is find the
19:43	remainder when I take P of X and I divided
19:46	by x minus four . Right ? So I divide
19:50	by X -4 because there is a positive for in
19:52	there , I divide by X -C . In terms
19:54	of the theorem . So I'm divided by X -4
19:56	right there . So I can say that uh the
20:01	remainder I get when I do that is going to
20:03	be equal to this value here . Right ? So
20:05	now I want to divide it . Now I can
20:06	do long division , I can take this thing and
20:08	I can divide by x minus four . That would
20:10	work fine . But I like synthetic division , that's
20:12	actually easier to do . So I'm gonna do synthetic
20:14	division one , X cube minus two , X squared
20:18	minus five X minus seven . And I'm gonna divide
20:20	that by X -4 . But for long debate for
20:23	synthetic division , I have to switch the sign ,
20:25	make it for . So then I'm gonna go down
20:29	and drop a one and then I do it four
20:31	times one is four , add these together , I'm
20:33	gonna get it to two times four is eight ,
20:36	Add these together , I'll get a 33 times four
20:39	is 12 , and I'll add these together and I
20:41	will get a five . And the only number that
20:43	I care about is the very last one . The
20:45	remainder . So the remainder is equal to five .
20:50	So that means because of the serum that this polynomial
20:54	evaluated at the number four is equal to five .
20:57	You don't have to think about it . You don't
20:59	have to go through the whole , this is ,
21:00	I'm going to check this polynomial by doing this and
21:03	substance . You don't have to do any of that
21:04	. The purpose of this was just to show you
21:07	why it works . It works because if I take
21:09	a polynomial and divide by something , I'm gonna have
21:11	this format with a remainder to check it . The
21:14	original polynomial is always going to be equal to the
21:18	answer . You get times what you divided by plus
21:23	the remainder . So if you divide by something very
21:26	closely related to what you're trying to evaluate here ,
21:29	in this case we're doing um X is equal to
21:31	three . We wanted to evaluate P is equal to
21:34	P . Of three . What we're trying to do
21:36	then the polynomial was equal to the quotient times the
21:41	divisor plus this . If we plug put a value
21:44	of three in here , the three minus three makes
21:46	it drop out . So that's why we're always dividing
21:48	by something closely related to what we're trying to evaluate
21:51	by . So the bottom line is You don't have
21:54	to think about it . You don't have to go
21:56	through the proof of why it works every time .
21:58	But I do like you to know where things come
21:59	from . What you do is give you a polynomial
22:02	. They say fine p evaluated at four you say
22:05	, Okay , I'm gonna divide by X -4 .
22:07	Now . You can do that with long division as
22:09	we did before . Or you can do it with
22:10	synthetic division . We do a synthetic division . We
22:12	have to change the sign of this number . We
22:15	do this in the synthetic division process , we get
22:17	a remainder of five . So five . So we
22:19	know that P4 is equal to five . The remainder
22:22	of a polynomial is always equal to that polynomial divided
22:26	by um When it's divided by X minus the number
22:30	, so the better way to think about it is
22:32	generally how you'll see it in your textbook . If
22:35	a polynomial is divided by x minus something , then
22:38	the remainder . When you do that division is the
22:40	value P evaluated at sea . These are always going
22:42	to be the same number . If you want P
22:44	evaluated at 34 , divide by X -34 . Find
22:48	the remainder . If you want P evaluated at seven
22:51	, divide by x minus seven . Find the remainder
22:53	if you want P at negative two because they can
22:56	be negative too . If you want to be evaluated
22:58	at negative two , then divide by X plus two
23:01	. Because you can have x minus of minus two
23:03	, divide by X plus two and then find that
23:05	remainder . So make sure you understand this and do
23:08	all of these problems yourself , Follow me on to
23:10	the next lesson , we'll get a little more practice
23:12	with the remainder theorem in algebra .

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