So as an example, prove that a real polynomial of odd degree has a real root.
So again, pause the video. It's good to actually try this.
The first step in actually attacking this problem is
really getting a handle on what
kind of proof technique to use, and I find it easiest to actually do this using a proof by contradiction.
So if we actually assume that a real polynomial of odd degree does not have a [real] root,
then we actually know from the previous theorem what all the factors must be.
The previous theorem tells us that all the factors must be quadratic.
If all the factors are quadratic, then
the total degree must be even. It's got to be 2 times the number of quadratic factors,
and that's the contradiction since we have a polynomial of odd degree.
So let's just actually write this up. So assume towards a contradiction that
p of x is a real polynomial of odd degree without a [real] root.
By the Factor Theorem, it can't have any real linear factors,
so something to think about here, let me pause at the sentence.
The Factor Theorem what does it really tell you? It tells you…
it's sort of like a change in dictionary.
[Real] roots of a polynomial are the same -
or correspond to linear factors of a polynomial, and vice versa.
So you have this duality and this change in dictionary is kind of what's happening here.
This is a very common theme throughout algebra
that you often look at something in a different light.
By changing the name, it seems like you're not doing anything,
but by looking at it in a different viewpoint, you actually
can get more information.
So we can't have a real linear factor, so by Real Factors of Real Polynomials,
we see that all the factors must be quadratic.
So what is the degree of this? Well this is 2 times k,
and this is something that's supposed to have odd degree.
2k is always even, that's a contradiction.
So hence the polynomial must have a real root,
and that's it.
So again, if these types of questions interest you,
one of my favorite courses in undergrad was a course in Galois Theory that I took.
So if you like dealing with polynomials and talking about roots in different fields and things like that,
I recommend highly a course in Galois Theory.
Okay.
That being said, let's do one more final example.
So something that I don't do in class a lot of is actually
trying to find square roots of complex numbers,
which might come in handy if you're trying to solve
for factors of a quadratic polynomial that are not nice.
So as an example,
I'm going to - let's talk about this one: find the square root of 5 plus 2i.
Now I've used this symbol here which is actually a little bit…
a little bit bad.
I mean we kind of understand what this means, this really means find the values z
such that z squared is equal to 5 plus 2i.
So, again, pause the video and actually try to solve that.
Hopefully you've actually tried, given this a shot,
and it's actually not bad when you reword it the way I've worded it, right?
Find number z squared - find number z so that z squared is equal to 5 plus 2i.
How do we…
so how do we do that? Well we can convert 5 plus 2i to polar form,
and then actually take… and then use the Complex
nth Roots Theorem.
So as I said, we're seeking values such that z squared is equal to 5 plus 2i.
There's a couple ways to do this, you can actually write z as x plus y i,
plug it through, and try to solve it,
or you can actually write this in polar form and then
try to…and then just use CNRT.
And that's what we're going to use here.
So I'm going to write 5 plus 2i in polar form.
So the modulus of 5 plus 2i is 5 squared plus 2 squared, that's 29,
and take the square root of that, so square root of 29.
And what's my angle here? My angle is not nice, it's going to be y over x.
This lives in the first quadrant, so arctan of 2 over 5 is
the correct answer here.
It's not arctan 2 over 5 plus pi.
If you don't understand that you should go back a couple videos and look at
when we talked about converting from polar to standard form and vice versa.
So thus, once we know this is the answer, so this is like
29 to the 1/2, so I take another 1/2, and that will give me the 4th root of 29,
times e to the i theta over 2, that's going to be a solution to this, right?
If I square this, I get exactly this number, which is what I wanted.
And the other solution is given by CNRT.
The other solution is you just add even multiples of 2 pi over
n, where n is the degree that we're looking for.
Here it's 2 so 2 pi over 2 is just pi.
So we add an i pi here.
So we're adding pi to theta over 2, which is the same as adding i pi to the whole thing
Either way you can use brackets or do it like this, it’s fine.
Either way, we see our two answers are plus minus the 4th root of 29,
e to the i theta over 2.
You could try to convert back to standard form. I don't
think the answer is going to be nice, maybe it is I'm not sure.
I think it'll be fairly ugly.
Since we didn't tell you which form to put it in,
leave it in this form.
Oftentimes, you'll know the angle theta and it'll be one of those
common angles like pi over 3 or pi over 6 and it's actually not too bad.
This one I just picked a number out of my hat
and went with it. But hopefully
this gives you a little bit of an
idea. Hopefully it helps you to solve these types of problems.
So again, where's this most useful? It's most useful when you're
trying to use the quadratic formula on a complex polynomial.
And I think that's basically it. So,
you know, if you've watched the previous 11 videos
and you somehow made it to video 12,
thank you very much for listening. I think it's been - hopefully this has been
a good video series.
If you have any questions or comments, as always feel free to email me
cbruni@uwaterloo.ca,
and that's it.
Thank you very much. Good luck.