The Three/Four bridge and solving quartic equations
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Solving polynomial equations, or equivalently finding factorizations into linear factors, is a major theme in algebra. Here we show how the general quartic equation can be so factored if we have a prior technology for solving quadratic and cubic equations. The key idea goes back to Lagrange and his theory of resolvents. We implement this using the Three / Four bridge and Viete's understanding of coefficients of polynomials as symmetric functions of the zeroes.
00:10
hello
00:11
and welcome to six i'm norman wahlberger
00:14
and this is an elementary course in pure
00:16
mathematics
00:16
meant for a very general audience in our
00:20
last few videos we've been talking about
00:21
the 3-4
00:22
bridge nice connection between
00:26
three things and four things and i want
00:28
to extend that
00:30
now in the direction of algebra and show
00:32
you how that
00:33
3 4 bridge is actually intimately
00:35
connected with an important problem
00:37
historically in pure algebra which is
00:40
the problem of solving
00:42
quartic equations those are equations of
00:44
degree four
00:46
so this talk is a little bit tangential
00:48
to the main thrust of the course
00:49
but i do want to occasionally take a
00:52
little bit of detour and show you some
00:53
adjacent landscape
00:55
to what we're doing some of the things
00:57
involved in today's lecture are a little
00:58
bit more
01:00
challenging uh they may require a little
01:02
bit more sophistication so if you don't
01:04
understand all of it it's not really
01:06
crucially important that you do i just
01:08
want you to get an overall idea
01:09
but for those of you a little bit more
01:11
mature there will be even some
01:13
interesting
01:13
sort of challenges for you okay so we're
01:16
talking about
01:17
equations polynomial equations i like to
01:20
call them
01:20
poly numbers and let's start with the
01:23
degree two case the quadratic case very
01:25
familiar
01:26
and very important going back to ancient
01:29
babylonian times
01:31
so here are two factors linear factors
01:34
in a variable alpha although i prefer to
01:37
think of alpha as being a
01:39
a poly number but in any case here is c1
01:42
minus alpha
01:43
times c2 minus alpha and if we expand
01:45
that out using just the laws of algebra
01:48
we get c1 c2 the product there
01:51
and then the terms involving alpha are
01:54
minus c1 minus c2 so all together minus
01:58
c1 plus c2 times alpha and then there'll
02:01
be
02:01
a quadratic term plus alpha squared
02:05
so this is a quadratic polynomial or
02:08
polynomial
02:09
in alpha and this is its factorization
02:12
into a product of linear factors
02:17
and note that these two numbers c1 and
02:19
c2 are the
02:20
zeros of this expression in other words
02:24
if you
02:24
replaced alpha here with say c1 and you
02:27
worked out what you get
02:29
you'll see you get zero similarly if you
02:31
replace alpha with c2
02:32
you're going to get zero you can work
02:35
that out directly
02:36
from this but it's also immediately
02:38
obvious from here because this
02:39
factorization
02:41
makes it clear that if you replace alpha
02:42
with c1
02:44
you're going to get a zero in that
02:46
factor and so the whole
02:47
expression will be zero so let's note
02:50
that the product of
02:52
c1 and c2 is appearing here as the
02:54
constant term
02:55
and the sum c1 plus c2 is occurring
02:58
with a negative sign as the linear term
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now let's look at the corresponding
03:04
situation for three
03:07
factors so we have c1 minus alpha c2
03:10
minus alpha
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times c3 minus alpha when we expand all
03:15
of that out
03:15
we're going to get a cubic polynomial or
03:18
polynomial
03:19
in alpha and here's what it is okay so
03:22
the constant term is
03:24
c1 c2 c3 and then
03:27
the linear term the term involving just
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alpha
03:30
is a bit more complicated it has three
03:33
terms
03:34
c1 times c2 plus
03:37
c2 times c3 plus c1 times
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c3 so it's like kind of a symmetric
03:43
quadratic expression in the three
03:48
variables or numbers c1 c2 and c3
03:52
and then the coefficient of alpha
03:54
squared will be
03:55
c1 plus c2 plus c3
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and finally minus alpha cubed to finish
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note that i'm writing my polynomials in
04:06
increasing order
04:08
that is not entirely standard kind of
04:12
thing but i think it's a useful
04:14
shift to make there's pros and cons both
04:18
ways but this is a
04:19
i think overall superior way of thinking
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okay so now what about the degree 4 case
04:27
which is what we're interested in
04:28
because quartic equations
04:30
concern equations of degree 4. here we
04:33
have 4 factors c1 minus alpha c2 minus
04:35
alpha c3 minus alpha c4 minus alpha
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and here is the resulting expression
04:40
that you get when you expand
04:42
this all out and collect terms
04:45
with powers of alpha so coefficient of
04:48
alpha coefficient of alpha squared
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coefficient of alpha cubed
04:50
alpha fourth and this is the constant
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term
04:54
and when you see this you start to see
04:55
this sort of a pattern here here
04:57
the constant term is the product of all
04:59
four the coefficient of
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alpha with the minus sign is the sum of
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all the possible products of three
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individual
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elements from c1 c2 c3 c4
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the quadratic term with a plus sign now
05:13
is the sum of all possible
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pairs of things from four things
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and here then the linear term with a
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minus sign is just the sum of all those
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things so we're really talking about um
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sets from four objects so here are all
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the one sets
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from four here are all the two sets from
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four
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essentially here all the three sets for
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four and here all the four sets from
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four
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okay so that's your typical quartic
05:41
equation expanded out
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and so now we get to a fundamental
05:46
problem in algebra
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which we can state in a number of
05:49
slightly different but almost equivalent
05:51
ways
05:52
the first is how do you solve a
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polynomial equation of the form
05:57
p of alpha equals zero so you take one
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of these polynomial
06:01
expressions and you set it equal to zero
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and you're trying to find
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values for alpha which will actually
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make this thing zero
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a closely related problem which is how
06:12
to factor
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the polynomials p alpha
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so let me illustrate that so here is a
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specific case of a quartic
06:22
polynomial which we'll have a look at 1
06:24
minus alpha times 3 minus alpha times
06:26
8 minus alpha times 11 minus alpha now
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you can practice your high school
06:30
algebra technology and expand that out
06:34
and hopefully you will get this poly
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number
06:38
264 minus 409
06:41
alpha plus 167 alpha squared
06:44
minus 23 alpha cubed plus alpha to the
06:47
fourth
06:48
now let's just pause here what is this
06:50
number here this number is the product
06:53
of these various things so it's 24
06:56
times 11. what is this 409
07:00
it is the sum of all four products of
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three
07:04
of these four things 167 is the sum of
07:07
the products of all
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possible pairs of these things
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and the 23 is just the sum of the
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four things by itself okay so
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there's the factorization there is the
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expansion
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so the problem that we're referring to
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here is
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how do you find solutions to p of c
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equals zero
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how do you find numbers c so that when
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you replace the alpha with c
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you do get zero now in this case this is
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easy because
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we've cooked up this polynomial to have
07:39
these linear factors and that makes
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reading off the zeros completely
07:43
elementary the zeros are simply the
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numbers 1
07:46
3 8 and 11.
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another way of asking this question is
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given this
07:54
polynomial here how do you reconstruct
07:58
its factorization and that was how do
08:00
you rewrite it as a product of linear
08:02
factors
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it's pretty well equivalent
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now this is a really fundamental problem
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okay it's uh it's motivated so much
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of work in in algebra even from from
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ancient times
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of course it's this is a unnatural kind
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of problem because i've already shown
08:22
you what the
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zeros are by having this factorization
08:25
to start with but if you just write down
08:27
some polynomial like this of degree 4
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and ask
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you know can you factor it it's actually
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a very subtle thing
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so it's important to realize that in
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general this may not be
08:40
possible in other words
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to go backwards from a polynomial like
08:45
this to a factorization
08:47
is not always possible in fact generally
08:49
speaking it's
08:51
to be expected that you're not going to
08:53
be able to do this you may be able to do
08:55
it approximately and that's an entirely
08:57
different
08:58
question and a very interesting question
08:59
but in terms of an exact
09:02
factorization it's typically not
09:04
possible
09:05
and you can see that at least sort of
09:08
you know
09:08
kind of heuristic for that why that is
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you see in this case here we started
09:13
with these four numbers and they were
09:14
relatively small you know around the
09:16
size
09:16
- but the coefficients of the
09:19
polynomial that we got when we expanded
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these four factors out involves some
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rather bigger numbers
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and that typically happens because
09:27
you're taking
09:28
products of these guys or sums of
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products of some subsets of them
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and that's telling you that this map
09:38
that starts with a set of four
09:42
numbers and that gives you this
09:45
polynomial is a is a map that's going to
09:47
really increase
09:49
the numbers involved so even if your
09:51
numbers are reasonably small
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the numbers in the coefficients of the
09:54
polynomial that result are bigger
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and if you think about it for a while
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that suggests to you that in fact you're
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not going to be able to easily reverse
10:02
this
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procedure because you know there's lots
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of bigger numbers and there's not so
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many
10:08
smaller numbers especially if say we're
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working over the integers which is a
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kind of a natural first approximation
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so you can't expect that
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typical polynomial factors like this but
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if it does
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the question is how do you have some
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technology to be able to recover this
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factorization
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this is a game you could play with your
10:29
friend you could cook up some
10:32
linear factors maybe four of them and
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you multiply them together and you give
10:35
your friend this big polynomial and you
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ask
10:38
him or her to recover the factorization
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this is the way i want you to think
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about what this fundamental
10:46
problem really amounts to okay it's
10:47
recovering a factorization of a
10:50
polynomial which has been created
10:52
by multiplying linear factors
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and this is a quite a challenging
10:59
problem especially
11:01
since we can ask the same question over
11:04
a rather
11:04
general number system so in this
11:07
discussion what we'd like to do is have
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a technology that doesn't
11:11
depend on which particular number system
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that you're you know using right a kind
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of a universal technology
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that's a a big challenge but if you want
11:22
to stick with something specific
11:23
the default number system is the
11:25
rational number system so what i want to
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do today
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is explain how the quartic case the
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degree
11:32
4 case actually follows from
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understanding the degree
11:36
2 and the degree 3 case so we're going
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to make this assumption that
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we can somehow solve
11:44
or factor degree 2 and degree 3 poly
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numbers
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if they can be factored so maybe that
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involves
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a formula like the quadratic formula
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works for
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the usual sort of quadratic equations
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the
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cubic formula much more complicated only
12:02
discovered in the 16th century
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but we're going to not worry about you
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know how to solve quadratic
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and cubic equations maybe we just have a
12:12
computer that is able to
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factor polynomials of these degrees if
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that factorization is possible
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and we're going to try to use those
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techniques to build a method for
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solving or factoring degree four
12:28
equations okay it's a beautiful fact
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that this can be done
12:32
that we can bootstrap from the degree
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two and degree three to get to the
12:36
degree
12:37
equation okay so here is the general
12:40
quartic degree 4 equation
12:44
with factors c1 minus alpha c2 minus
12:46
alpha c3 minus alpha c4 minus alpha
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we've already written that down in the
12:50
first slide
12:51
we saw it was a polynomial in alpha and
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let's call the coefficients
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e4 e3 e2 e1
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and 1 with this alternating aspect so
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we'll say e4
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minus e3 alpha plus e2 alpha squared
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minus e1 alpha cubed plus alpha to the
13:09
4.
13:10
and here we're just recalling what these
13:14
expressions are these in fact are called
13:16
the elementary symmetric
13:18
functions of the four quantities c1 c2
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c3 and c4 e1 is just the sum of them all
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that's the coefficient with a minus sign
13:27
of
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alpha cubed and then here is e2 the sum
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of all possible
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products of different ci's here's e3
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the sum of all possible triples products
13:39
of
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three of these things and e4 is the
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product of all four of them
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so these elementary symmetric functions
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uh probably introduced first and studied
13:50
by francois viete
13:51
in the 16th century great french
13:54
algebraist
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and and they play a very important role
13:58
in lots of branches of mathematics
14:00
but especially in this subject they are
14:02
kind of all important okay
14:04
so we've identified these four different
14:07
expressions in c1 c2 c3 c4
14:11
that are the coefficients of this
14:14
particular
14:15
poly number when you expand it out
14:19
okay so suppose we have one of these
14:21
quartic equations
14:23
and we're interested in solving it that
14:24
is recovering the factorization
14:27
if we don't know the initial
14:28
factorization and
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we're claiming that we can do this by
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using the
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three-four bridge and a certain kind of
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set of ideas
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roughly called resolvence go back to
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a lagrange it was a great 18th century
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french italian mathematician so what
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we're going to do
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is from our four quantities c1 c2 c3 c4
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we're going to build three quantities
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we're going to use exactly the 3 4
14:59
bridge to do that so let me remind you
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that if you have four things
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okay four things say forming a square
15:07
that you can
15:08
subdivide those four things into two
15:10
groups
15:12
of two to two two sets
15:15
in exactly three ways so you could take
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these
15:19
pairs like this pair and this pair or
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you could take
15:22
these pairs okay or you could take the
15:26
sort of diagonal pairs
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there's three possible combinations that
15:31
will do that
15:32
okay we're going to exploit that so
15:33
we're going to use that to
15:35
form these three new expressions which
15:38
we might call
15:39
a b and c so a is
15:42
c1 plus c2 times c3 plus c4
15:46
corresponding to the subdivision c1 and
15:49
c2
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in one part and c3 and c4 in another
15:53
and b is c1 plus c3 times c2 plus c4
15:58
corresponding to another subdivision and
16:00
c
16:01
is c1 plus c4 times c2 plus
16:06
c3
16:08
so if you have four say numbers
16:11
c1 c2 c3 c4 then a b
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and c are three auxiliary numbers that
16:17
you can create
16:18
from them and now here is the main
16:21
theorem that
16:22
if we create a poly number
16:26
with zeros a b and c in other words if
16:29
we take the three
16:30
factors a minus alpha b minus alpha and
16:33
c minus alpha
16:34
and expand that out that's going to be a
16:37
cubic
16:38
polynomial because we're starting with
16:39
three things
16:41
that we can express the coefficients of
16:44
that
16:44
cubic polynomial in terms
16:48
of the coefficients e1 e2
16:51
e3 e4 that we were looking at originally
16:55
coming from the degree 4 polynomial
16:57
which is the result of the linear
16:59
factors
17:00
corresponding to these four things and
17:03
the expression is quite involved
17:07
it's not something you could guess okay
17:10
so
17:10
here is a big looking
17:13
poly number of degree 3 in alpha
17:16
and the constant term is e1 times e2
17:19
times e3
17:20
minus e1 squared times e4 minus e3
17:24
squared the
17:27
linear coefficient with a minus sign is
17:30
e1 e3
17:31
plus e2 squared minus 4e4
17:35
and coefficient of alpha squared is 2e2
17:39
and there's a minus alpha cubed
17:43
this is a theorem that you could
17:46
demonstrate to yourself by just
17:48
expanding
17:49
okay so you replace the a b's and c's
17:52
with these expressions here in the c's
17:55
you expand that all out you get some big
17:57
expression involving
17:58
c's and the alphas each one of these
18:02
things is an expression in the c
18:04
some of them are quite long nevertheless
18:07
you could expand
18:08
all of this out as a polynomial in alpha
18:11
with coefficients depending on the seas
18:15
the claim is these two polynomials are
18:17
remarkably
18:18
the same so to get some kind of formula
18:22
like this
18:22
it's really useful to have some
18:26
familiarity with the theory of symmetric
18:29
functions or symmetric polynomials so
18:32
this is a little bit more advanced and
18:33
i'm not expecting you to
18:35
to you know to go through a proof of
18:37
this
18:38
i just want you to admire
18:41
the theorem and for those of you are
18:43
interested you could actually uh
18:44
try to see if you could prove this and
18:46
maybe even discover it
18:47
that's not impossible so in our example
18:51
remember we were looking at the
18:52
particular example with
18:54
roots one 3 8 and 11. we had this
18:57
poly number of degree 4 and that's
19:00
telling us that the
19:02
elementary functions were e there's e4
19:05
there's e3 there's e2 and there's e1
19:07
okay so then the claim is from the
19:10
theorem that if we take
19:11
these three quantities a minus alpha b
19:13
minus alpha and c minus alpha
19:15
we're going to get this thing here
19:18
so this is what you get when you take
19:21
this one here okay that's the constant
19:23
term that's
19:24
e1 times e2 times e3 so that's e1 times
19:27
e2 times e3
19:29
minus e1 squared times e4
19:32
minus e3 squared okay you calculated
19:36
that all out using these values of e
19:38
and you're going to get this number 1
19:40
264 032
19:44
and this expression here evaluates to
19:45
this coefficient
19:47
and this thing here 2e2 is just
19:52
it's just two times this thing here so
19:53
that's 334.
19:58
so we've started with
20:02
a quartic equation
20:05
and we've cooked up these new variables
20:09
and they
20:12
generate this something's called a
20:14
resolved kind of equation
20:15
which is a cubic and
20:18
we're assuming that we can factor cubics
20:21
if they're factorizable okay that was
20:24
one of our assumptions
20:26
so we can apply our technology for
20:29
solving cubics whatever that might be
20:31
maybe it's just using a computer or
20:32
maybe we have a formula it doesn't
20:33
matter
20:34
to factor this thing and it factors as
20:36
76 minus alpha
20:38
times 126 minus alpha times 132 minus
20:42
alpha
20:43
okay so it's at this stage here that
20:46
we've
20:47
we've used the 3 4 bridge to go from
20:49
four
20:51
variables to three variables go from a
20:54
quartic
20:54
to a cubic now we're assuming that we
20:57
know how to solve a cubic and then so
20:59
now we've solved that cubic mean so
21:01
we've made good progress to solving the
21:04
original
21:05
quartic so that means we can read off
21:07
the values of
21:08
capital a capital b and capital c
21:11
because we were able to factor that
21:12
cubic
21:13
so we can identify that these three
21:15
variables in some order are 76
21:17
126 and 132.
21:20
and by the symmetry of the situation it
21:22
doesn't really matter which ones we
21:24
assign to
21:25
which values so a which was c1 plus
21:28
c2 times c3 plus c4 we can set that
21:30
equal to 76
21:32
b which is c1 plus c3 times c2 plus c4
21:35
is say 126
21:37
and c is c1 plus c4 times c2 plus c3
21:40
equals 132
21:43
but we still want to find out what are
21:45
c1 c2
21:46
c3 c4 right that's what we want to do we
21:50
want to solve this
21:51
original quartic now we also know
21:55
that e1 which is the sum of the four
22:01
numbers c1 c2 c3 c4 is equal to 23
22:05
that was the coefficient of alpha or
22:07
minus alpha cubed in
22:09
in the polynomial so what we can do is
22:12
we can
22:12
use this additional equation and combine
22:16
it with each one of these
22:17
three to get additional quadratic
22:19
equations
22:21
and that's good because we are assuming
22:23
that we can solve quadratic equations
22:25
so let me show you how that goes let's
22:26
look at a if we look at the two
22:28
constituents here the c1 plus c2
22:31
and c3 plus c4 we know that their
22:33
product
22:34
is 76. we know that their sum
22:38
is 23 because of this equation
22:41
so these two quantities we know what the
22:43
product is we know what the sum is that
22:45
means we know
22:46
what the quadratic equation is that they
22:48
satisfy
22:49
remember we started off by talking about
22:51
the quadratic equation
22:52
having the form c1 c2 minus c1 plus c2
22:55
alpha plus alpha squared
22:58
so the the quadratic equation
23:01
factored is c1 plus c2 minus alpha c3
23:05
plus c4 minus
23:06
alpha that's going to have the form
23:07
here's the product
23:09
76 here's minus the sum and then plus
23:13
alpha squared
23:14
and because we can solve quadratic
23:17
equations we're assuming that
23:18
we know how to factor this thing and it
23:20
factors into 4 minus alpha
23:22
times 19 minus alpha
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and similarly for b you get this
23:28
equation
23:29
for the product of c 1 plus c 3 minus
23:32
alpha c 2 plus c 4 minus alpha there's
23:34
the product
23:35
minus the sum there's the factorization
23:38
9 minus alpha and 14 minus alpha
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and for c c1 plus c4 minus alpha times
23:45
c2 plus c3 minus alpha
23:47
is 132 minus 23 alpha plus alpha squared
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that's 11 minus alpha 12 minus alpha
23:55
what's that telling us that's telling us
23:57
that c1 plus c2 and c3 plus c4
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they are the zeros over here that means
24:05
they must be the zeros over here that's
24:06
4 and 19. so we have two possibilities
24:09
either c1 plus c2 equals 4 and c3 plus
24:12
c4 equals 19
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or it's the other way around c1 plus c2
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equals 19 and c3 plus c4 equals
24:20
- exactly the same way
24:23
the second equation tells us that c1
24:25
plus c3 equals
24:26
9 and c2 plus c4 equals 14
24:29
or it's the other way around and the
24:32
third equation tells us that
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c1 plus c4 is 11 and c2 plus c3 is 12
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or it's the other way around
24:43
now we have a very limited set of simple
24:46
equations
24:47
but there's various possibilities and so
24:48
a little bit of
24:50
thought now has to be you know applied
24:53
to deduce that in fact c1 equals 1
24:57
c2 equals 3 c3 equals 8 and c4 equals
25:00
11.
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so maybe you have to even go through
25:03
some possibilities in fact you could go
25:04
through all the possibilities and see
25:05
which ones actually work and which ones
25:07
don't
25:08
and you will find that's the solution so
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what have we done
25:11
we've started with a quartic equation
25:15
of degree 4. we've used the 3 4
25:19
bridge to create three auxiliary
25:24
kind of objects like resolvent kind of
25:28
objects
25:28
and they satisfy a cubic equation and
25:31
we've been able to use the fact that we
25:32
can solve cubic equations and also
25:35
quadratic equations to combine the
25:37
information
25:38
to actually find the solution to the
25:40
original quartic
25:42
so this lecture has been meant to mostly
25:45
demonstrate to you the importance of the
25:46
three four bridges not just an
25:48
abstract combinatorial thing it actually
25:50
has a practical important application in
25:53
an important chapter in algebra
25:57
and it connects with polynomials and
25:59
factorizations of polynomials and
26:01
elementary symmetric functions and
26:03
sort of nice other topics in high school
26:05
algebra
26:07
for those of you who like this kind of
26:08
thing i'm going to give you a little bit
26:10
more of a challenge here so there's
26:11
actually
26:12
another way we could have used the 3 4
26:14
bridge
26:15
so if we go back to the setting up a b
26:19
and c
26:19
another possibility would be to set up
26:22
things differently to introduce a d
26:24
e and f which are also formed from the
26:27
three ways of combining
26:30
pairs of two sets we could create d
26:34
to be c1 c2 plus c3 c4
26:38
and e to be c1 c3 plus c2 c4
26:42
and f is c1 c4 plus c2 c3
26:46
so these are new quadratic expressions
26:49
in the cis
26:51
which are also created from the 3 4
26:53
bridge but they have a slightly
26:55
different form
26:56
they're sums of products rather than
26:59
products of sums
27:01
okay and then you will be happy to check
27:04
this
27:04
relation that's obtained by just doing
27:08
the same kind of thing we did
27:09
previously and expressing the cubic
27:12
whose
27:12
zeros are d e and f in terms of the
27:15
coefficients of the original quartic
27:17
there's the expression there so i won't
27:20
go through it but you can
27:23
try to establish this or verify it
27:26
and then you can show how to use this to
27:28
solve the quartic equation in fact you
27:30
can even use the same chord equation
27:31
that i've used as an example
27:33
and see if you can get up the solution
27:36
from this point of view
27:38
so some interesting chapter of algebra
27:41
we're touching base with
27:42
some polynomial algebra it's a little
27:45
bit more advanced
27:46
don't worry if some of it was too
27:49
sophisticated
27:50
we'll return to combinatorial issues in
27:52
our next video
27:54
i'm norman wahlberger thanks for
28:00
listening
28:06
you