Hello everyone. Welcome to Week 11 part 1 of Carmen's Core Concepts,
my name is Carmen Bruni,
and in these videos we talk about the
current week in Math 135.
This time I've decided to break it up into two parts,
part 1 and part 2, and I've decided to make part 1
a more theoretical part where we talk more about the
proofs related to the theorems that we did this week,
and in part 2 we talk about
the applications of the theorems.
So in theory, the two parts are independent,
however you should probably watch part 1 before you watch part 2
unless you feel very comfortable with the theory, then you can go ahead straight to part 2,
either way is fine.
With that being said, let's begin.
So polynomial statements.
So let f and g be non-zero polynomials over a field F
such that they are not additive inverses, and we'll see why we need that condition in a second,
then we get the following for free. So we get
the degree of f plus g is less than or equal to the max of [the] degree of f and the degree of g.
The degree of f times g is the sum of the degrees of f and g,
and if f divides g and g divides f,
then f is equal to c g for some
element of our field, F.
I want to talk about each of these a little bit individually.
Why do I need f and g to not be additive inverses?
Well the first one, then we would get the degree of the 0 polynomial
In this course, we say that the [degree of the] 0 polynomial is undefined.
Some textbooks, and it's not unreasonable to do so, but
it does require a little bit of understanding,
define the degree of the 0 polynomial to be negative infinity,
and it’s so that statements like this actually hold for free, right?
Negative infinity should, morally speaking, be less than any integers, so
that's perfectly fine,
but we're not going to do that. We just say the degree of the 0 polynomial is undefined.
That being said, it's always good to double check some of these theorems.
So why is this a less than or equal to statement? Well
the leading terms of f and g could cancel each other out,
that's possible, so you actually might get the strictly less than case, okay?
Usually though, as long as those two
terms don't cancel each other out, this will be an equality,
that's something you can check. These aren’t hard to prove, but I'm not going to prove them here.
We proved the last one in class, I mentioned the second one in class,
and the first one is actually not too hard either.
Just write out f and g in terms of polynomials,
add them up, and then argue why it must be less than or equal to the -
and then argue what the degree of the sum must be.
That's I think - oh and the last one, let me talk a little about the last one.
The second one is also fairly clear once you do it. You
just multiply the leading two terms and that'll be the sum.
You never get 0 when you multiply the two terms because you're over a field,
so you can't get cancellation so this is actually okay,
but this does require that we have a field.
f divides g and g divides f, the way I like to talk about this one -
or what I’d like to talk about this one
is I'd like to relate back to the integer case.
So if I have a divides b and b divides a, then what do we know?
As integers…
we think about this then we say, “Okay well we know that
a could be equal to b,” that’s a possibility.
and the case that a lot of us forget is that a could also be equal to the negation of b.
So realistically, a is equal to plus or minus b,
and it turns out that plus or minus 1 happens
to be the only two elements that are invertible in the integers,
and that actually matters a little bit. You'll see where that comes up in the proof,
but I won't talk about that anymore, but this is what I like to - or
what I'm trying to get at here is the analogy here, right? So we have these polynomials,
we have this notation for integers, how do these two things relate?
Well here we don't get just plus or minus 1, we get any invertible element in our field.
Turns out that 0 won't work because if f is 0 and g is 0,
we didn't define 0 dividing 0 as a polynomial, but that's not a big deal.
Oh we can't have them as 0 because f and g are non-zero, either way is fine.
But yeah, so something to think about here, right?
When you see a new problem in an unfamiliar situation,
reduce back to a situation that is familiar to you,
and try to reason out what the correct statement and the correct proof should be there.
It’s a good trick in mathematics. If the problem is too hard, get rid of it and solve an easier one.
This works well for, you know, assignments and when you're
doing work on napkins and [stuff] like that, but you know
be careful on a test, right? Don't just change the problem,
make it easier, and then solve the easier problem.
It is a good tactic to get a feeling for what the problem's saying.
Alright, Remainder Theorem. So we're going to start with our one by one theorem-crunching here,
and then in the next…like I said in the next part, we'll talk about the applications.
So what does the Remainder Theorem say? Well
it gives us a way to divide polynomials when we have a very specific form;
if we just care about the remainder. So if we're dividing a polynomial by
a linear term x minus c, so this specific form x minus c,
then the remainder must be f at c.
This will become important for us when we talk about the Factor Theorem.
It turns out that we actually divide by these types of polynomials more than you might imagine.
That being said, pause the video, see if you can give this proof a shot. It's not too bad,
none of these are too, too bad. Some of them are a little bit more involved, but
the proof technique part of it is not too bad.
So hopefully after you've paused it, you've given this a shot, you've tried it out,
how do we go through this?
Well this is the Remainder Theorem, so we need to talk about a remainder,
so we should actually just divide f of x by this x minus c
using the Division Algorithm for Polynomials.
That gives us some quotient and some remainder,
and then we need to argue whether [the] remainder is f at c.
Well if we're dividing by a linear polynomial, we know the remainder must be constant,
and the question is, “Well what constant is it,”
and that's what we can try to figure out.
So using the Division Algorithm gives us these polynomials q and x…
or q of x and r of x.
They turn out to be unique when we require that r of x is 0
or the degree of r of x is less than the degree of x minus c,
and the degree of x minus c, we know that, that's 1.
So what does that mean? Well that means that the degree of r of x is 0, or r of x is 0,
but the only degree 0 polynomials are constants,
so therefore r of x is equal to some constant k for some k inside the field F,
and here k can be 0 which also encompasses the first remainder case.
So how do we determine what k is? Well we have this relationship between f, and
q, and r, and x minus c,
so let's actually substitute a value in for x.
And what value are we going to substitute in?
We're going to substitute in the value x is equal to c.
That makes sense, c is inside my field, this is perfectly reasonable,
and when I do that, the middle term goes away
and I'm left with f at c is equal to r at c,
but r at c is just k, right, r at c is a constant polynomial.
So k is equal to f at c and we know that r of x
must be the constant polynomial f at c, and we're done.
That's the Remainder Theorem, and it's
particularly useful for us to talk about this in the context of the Factor Theorem, right?
So what - or how does this help us with the Factor Theorem? Well it says okay well,
I know what the remainder is and I want to know that x minus c is a factor,
well then I want a 0 remainder.
So therefore I would want f at c to be 0, and if I
get f at c to be 0, then I know that c is a root
and all of this kind of ties together to the Factor Theorem.
So suppose that f of x is a
polynomial over the field F
and c is some constant in F, then the polynomial x minus c is a
factor of f of x if and only if f at c is 0,
that is, c is a root of f of x.
I've kind of already given you the proof, but if you
don't see it yet, you should pause the video and try it,
and hopefully you've come back after trying it.
Let's see the proof. It's actually pretty quick.
So when is x minus c a factor of f of x? Well if I
divide by x minus c,
if it’s a factor, then the remainder must be 0.
That's just by the Division Algorithm for Polynomials, right,
the quotient and the remainder must be unique so if I divide by x minus c and it's a factor,
then it must have a 0 remainder.
And when does this have a 0 remainder? Well
the Remainder Theorem says that r of x is actually equal to f at c,
but r of x is 0, so therefore f at c is 0,
and this goes in both directions, right? So the remainder is always f at c
and f at c is 0 if and only if r of x is 0
because they're equal to each other and that's actually it, it's actually pretty quick.
I've written it out pretty compactly and pretty quickly, but
the idea I think is pretty clear.
And this gives us the Factor Theorem.
This is useful for finding roots which we'll see in the next video.