So the theorem here, Rational Roots Theorem, if we have a polynomial of degree n
that's integral, so with integer coefficients,
and we have a rational root, r is equal to s over t in lowest terms,
then the numerator of that rational root must divide the constant term,
that's what this is saying, s divides the constant term,
and the denominator of that rational root must divide the leading coefficient.
I won't do an example of this theorem here, that will be in part 2
so if you want to check that out,
here I'm just going to do the proof. Again, pause the video here.
This is a proof that you should be able to do, it's not too bad.
The only trick would be - the only trick while doing this proof is to actually get - once you
do step 1, actually get an integer
object.
So hopefully you’ve paused the video, come back,
when I said step 1, what I really meant was well
when you plug in the rational root, clearly that's something that you should do.
You have a root you know if you plug it in you get 0,
that sounds like a reasonable first step in this problem,
but then once you do that, you get like
a n times s over t to the power of n
and it's like well okay I have these rational numbers, the sum
of these rational numbers is an integer, it's 0.
How do I deal with that?
And the correct way is to actually get rid of the fraction part.
How do we do that? Well we multiply by the denominator, which is t to the n.
So if you look at all these terms, so I've only written a couple of terms
out here but we also have like
a n minus 1, times s to the n minus 1 over t to the n minus 1,
so if I multiply that by t to the n, I'm just going to get a t term,
and I'm left with the numerator which stays untouched, s to the n minus 1.
And we keep going all the way down.
So what do I do at this point? Well now I know I look at this and I have
s divides a, and I want to show that s divides a naught.
All of these terms depend on s.
So…by the way somebody in class said, “Well you should use DIC,”
and I said, “Well you can, but DIC is really only for two terms,
so you have to use some sort of like DIC plus induction type thing.”
I don't really want to do that so I'm just going to write it out
as a factor like this.
By the way, I'm using a “dot dot dot” argument here, which is kind of
contradicting what I said earlier. If you really want to be like
super, super formal with this, you can write this as a summation.
The reason why I don't is because I think that this is a little bit easier to understand and cope with,
but if you want, write it as a summation, the proof will pan through.
So I'm going to rewrite this by factoring out the s
and bringing all of the first…
how many terms is this? n minus 1 terms to the other side.
So I'm going to leave a naught times 2 to the n on one side,
you know flip the sides because it's a little easier to see.
Negative s will factor out of all these terms, and I'm left with this integer.
Okay?
Therefore s divides a naught t to the n.
We're almost done, we want to show s divides a naught, we got this t to the n here
that's bothering us,
and the trick now to solving this is to recall oh well
s over t was a root in lowest terms,
and because it’s a root in lowest terms, the gcd of s and t must be 1.
So therefore the gcd of s and t to the n must also be 1 by, let's say,
GCDPF that's perfectly reasonable, right?
If s and t don't share a prime factor, then surely s and t to the n don’t share a prime factor,
so they must be co-prime.
And if s and t to the n are co-prime, then we can use Coprimeness and Divisibility
and say hey well s doesn't divide t to the n,
but s divides this product, so therefore s must divide
a naught.
And I only proved the first part. Why? Well
the proof is similar for t divides a to the n.
Take the last…
n minus 1 terms, factor out a t,
bring it to the other side, it's the same argument.
So here “similarly” actually is valid as a proof technique.
If you need to see it, write it out. If you want some practice, you should write out that proof.
That's a very easy proof to write out the other direction,
but I won't do it here.
So the Rational Roots Theorem gives us one way to help to factor,
and the last theorem that we're going to talk about,
I believe unless I'm mistaken, is the Conjugate Roots Theorem,
and that will also help us try to factor polynomials.
It's a really, really cute theorem, it's actually not hard to prove either
so I do encourage you to pause the video and try it.
If c is a complex root of a real polynomial,
then c bar is a root of p of x.
So if c is - so if we take a polynomial with real coefficients
and we try to factor it over C and we know that c is a root,
then we also know that c bar is a root of p of x
It's kind of neat, and the proof really isn't that hard.
You really should pause the video, make sure that you can do this proof.
It turns out all you have to do to prove this is actually write p of x
as a polynomial a to the n x to the n
plus dot dot dot plus a1 x plus a0,
plug in c, and then take the conjugates of both sides.
So here we go, that's exactly what I'm doing here. So
write p of x as a polynomial with p at c equals 0,
then let's look at p at c bar. Well if I take p at c bar, I'm going to just
plug in c bar into x for every possible x.
Since my coefficients are real
and by Properties of Conjugates, I can…
so if I take the conjugate of c to the power of n, that's the same as taking
c the n and the conjugate of that, those are the same.
Since a n is real,
it's the same as a n bar. So I'm writing this as all of this.
Now I'm adding a bunch of terms that I've taken the conjugate of,
so it's the same as adding all the terms then taking the conjugate.
So the sum of the conjugates is the conjugate of the sum,
that's by PCJ again,
and this just becomes…this is just the polynomial now when I plug in c.
So it’s p of c and I'm just taking the conjugate of it,
but p of c is 0, c was a root.
So the conjugate of 0 is still 0, we're done. So
c bar is also a root of a real polynomial when c is a root.
This is not true if your polynomial is not real, right?
If your polynomial has complex coefficients, this proof breaks down.
I can't just take the bar of each coefficient, that doesn't work.
But when it is real, you can do that.
So something…it's pretty cool. I know I think it's a really neat theorem,
and you can use this - so for example
if you can find a complex root, a strictly complex root,
then you actually know 2 roots,
and when you know 2 roots, then you can multiply, you know, their factors together and
divide out and get some new polynomial that’s degree 2 less.
By doing that, you reduce your computation and
you reduce your search space and so on and so forth. So these give you ways to factor
these polynomials.
I think that's all I want to talk about in this video.
I'm just trying to scroll and I don't think I have anything
else to talk about.
So again this was the theoretical video.
If you want to see applications of this, check out part 2 of Week 11.
Hopefully this gave you an idea of what we did this week.
Hopefully this helps you to try to factor
polynomials, gives you some ideas,
and hopefully it helps clear up some of the proofs that you might have been struggling with this week.
Okay that's all I have to say, thank you very much and all the best.