Hello everyone. Welcome to Week 12 of Carmen's Core Concepts, my name is Carmen Bruni. In this video series, we've been talking about Math 135 on a week-by-week basis. This is the final week, Week 12. This is a shorter video. Most of the content was in the Week 11 videos.
But there's one final topic that I'd like to talk about and that is given by the following table of contents, and that's factoring real polynomials over the real numbers. So that is our question of interest, and we'll do that with a couple of theorems, an example using these theorems, and a little bit of bonus content here is square roots of complex numbers.
Theorem (RQF): Let $f(x) \in \mathbb{R}[x]$. If $c\in \mathbb{C}-\mathbb{R}$ and $f(c)=0$, then there exists a $g(x) \in \mathbb{R}[x]$ such that $g(x)$ is a real quadratic factor of $f(x)$.
Okay so, again take a minute and try to prove this, and what you're going to possibly realize is that there are sort of two parts here. One part’s very subtle, which we'll talk about, and one part’s sort of clear.
So if you're trying to prove this, again, we know from the previous videos that if $c$ is a root of a real polynomial, then its conjugate is also a root. So if we use the fact that we have those two roots, we can actually pair them up and we're going to see that we get a quadratic real factor. So take:
$$\begin{align*} g(x)&=(x-c)(x-\bar{c})\\\\ &=x^2-(c+\bar{c})x+c\bar{c}\\\\ &=x^2-2\Re(c)x+|c|^2 \in \mathbb{R}[x] \end{align*}$$So here's $g(x)$. This is a real polynomial, it has $c$ and $\bar{c}$ as a root, and we claim that $g(x)$ is a factor of $f(x)$.
Now here's where the subtle part comes in, and it turns out that it can't happen, but you do have to think about it a little bit. What we talked about is we already know that $x-c$ and $x-\bar{c}$, we already know that these are both factors of $f(x)$ over the complex numbers, that we know, right? We saw that before.
The problem now is if I take $f(x)$ and I divide it by $x-c$ and $x-\bar{c}$, I know I get some quotient, let’s call it $q(x)$, plus some $0$ remainder, right, because they both divide it.
So at this point in our mind, we're thinking of that quotient, $q(x)$, as being a complex polynomial. The question now becomes well why when I divide it by this polynomial, do I get a real polynomial with $0$ remainder over the reals? It's not clear that quotient actually is a real polynomial.
It turns out that it is, which we'll see in a minute, but at this point it's not clear just because these two things are roots, and when I divide them over the complex numbers that I get some complex polynomial, it's not clear that if I divide by the product of these things, which is a real polynomial, that I get a real polynomial factor with no remainder. Again it's true, we'll prove it specifically in this case, you can do things a little bit more general than this, but let's just focus on this case for now.
So last thing we have to do is show that $g(x)$ is a factor, how do we show that it’s a factor? We're going to use the Division Algorithm, and reach a contradiction by assuming that it's not a factor.
So by the Division Algorithm there are polynomials $q(x)$ and $r(x)$, and these are unique, such that
$$f(x)=g(x)q(x)+r(x)$$$r(x)$ is either $0$ or $\deg(r(x)) < 2$.
So we're going to assume towards a contradiction that it's not the $0$ polynomial, then the degree is either $0$ or $1$. That is $r(x)$ is either linear or constant.
When we plug in $x=c$ into this equation, what do we get?
$$0=f(c)=g(c)q(c)+r(c)=r(c)$$So if $r(x)$ is constant, then we know it's the $0$ constant because it's $0$ at some point. Again this is actually also a little bit subtle, which I'm not going to talk about here, but this is constant at some complex point.
So therefore this is the $0$ polynomial as a complex polynomial, so why is it the $0$ polynomial’s a real polynomial? I'll let you think about that as an exercise, but I don’t want to talk about it too much.
And if $r(x)$ was a linear polynomial, let's say $r(x)=ax+b$, then if we plug in $c$ we get
$$r(c)=ac+b=0 \Rightarrow c=\frac{-b}{a}$$Therefore, $c$ must be a real number, it's $\displaystyle\frac{-b}{a}$, and that also is a contradiction. So if it's linear, then we see that our root must have been real, and if it's a constant polynomial, then we know that it must be the $0$ polynomial. So in either case, we see that the remainder must be $0$, and hence $g(x) \mid f(x)$, so $g(x)$ is a real factor of $f(x). \quad \blacksquare$
Okay good, so that satisfies the Real Quadratic Factors question.
Now we have the final part which is how do real polynomials factor? And it turns out that real polynomials, they either factor as a product of linear factors, quadratic factors, or some combination of the two. And that's what this theorem says, Real Factors of Real Polynomials.
Theorem (RFRP): Let $f(x)=a_nx^n+...+a_1x+a_0 \in \mathbb{R}[x]$. Then $f(x)$ can be written as a product of real linear and real quadratic factors.
You should actually try this proof. It's a little bit subtle, but you can do it. And let me give you a hint, if you forget this proof, the way that we would start this proof is start by actually writing down all the possible roots, they're either strictly real, or they're strictly complex, and see if you can find some sort of relationship between factors and these roots.
So what's the idea here? This is more of a proof sketch than a real formal proof, but that's okay. So what do we do? We're going to take the real roots and the complex roots, and the complex roots come in pairs, one and its conjugate because this polynomial is real.
For each pair, what you're going to do is you're going to associate to it a quadratic factor, which we know from the previous theorem exists. So each pair of complex roots corresponds to a quadratic factor, say $q_i(x)$ and every individual real root corresponds to a real linear factor, say $g_i(x)$.
We put those two things together:
$$f(x)=cg_1(x)...g_k(x)q_1(x)...q_{l/2}(x) \quad \blacksquare$$It turns out that this constant, $c$, will be the leading term of our polynomial. And that's it, so that's the main idea here, okay? So we can find the roots by CPN, Complex Polynomials Have n Roots, and once we have these roots, we can pair them up and match them to factors using Real Quadratic Factors, and using the Factor Theorem.
And that's it, so that's Real Factors of Real Polynomials. There's a couple more examples that we can talk about right now, but this is the last theorem of this course. How do I factor real polynomials? Well it factors as linear and quadratic factors.
Prove that a real polynomial of odd degree has a real root.
So again, take a minute and give this a shot. 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 assume towards a contradiction that $p(x)$ 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's sort of like a change in dictionary. Real roots of a polynomial 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 we write:
$$p(x)=q_1(x)...q_k(x)\quad\text{for some quadratic factors}\,q_i(x)$$So what is the degree of this? Well this is $2k$, where $k$ is the number of quadratic factors, 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. $\quad \blacksquare$
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.
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.
Find $\sqrt{5+2i}$.
Now I've used this $\sqrt{\,\,\,}$ symbol here which is actually a little bit bad. I mean we kind of understand what this means, this really means find the values $z$ such that $z^2=5+2i$.
So, again, take a minute and actually try to solve that. It's actually not bad when you reword it the way I've worded it, right? Find number $z$ so that $z^2=5+2i$.
So how do we do that? Well we can convert $5+2i$ to polar form, and then use the Complex $n$th Roots Theorem.
So as I said, we're seeking values such that $z^2=5+2i$. There's a couple ways to do this, you can actually write $z=x+yi$, plug it through, and try to solve it, or you can actually write this in polar form 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 $|5+2i|=\sqrt{5^2+2^2}=\sqrt{29}$. And what's my angle here? My angle is not nice, it's going to be $\displaystyle\frac{y}{x}$. This lives in the first quadrant, so $\arctan\left(\displaystyle\frac{2}{5}\right)$ is the correct answer here. It's not $\arctan\left(\displaystyle\frac{2}{5}+\pi\right)$. 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, we have:
$$\sqrt{\sqrt{29}e^{i\theta}}=\sqrt[4]{29}e^{i\theta/2}$$That's going to be a solution to this, right? If I square $\sqrt[4]{29}e^{i\theta/2}$, I get exactly $\sqrt{29}e^{i\theta}$, which is what I wanted.
And the other solution is given by CNRT. The other solution is you just had even multiples of $\displaystyle\frac{2\pi}{n}$, where $n$ is the degree that we're looking for. Here it's $2$, so $\displaystyle\frac{2\pi}{2}=\pi$.
$$\text{Hence,}\,\sqrt[4]{29}e^{i\theta/2}\,\text{and}\,\sqrt[4]{29}e^{i\theta/2+i\pi}\,\text{are solutions.}\\\\ \text{Simplifying gives}\,\pm\sqrt[4]{29}e^{i\theta/2}$$So we add an $i\pi$ here. So we're adding $\pi$ to $\displaystyle\frac{\theta}{2}$, 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.
You could try to convert the solutions 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 $\displaystyle\frac{\pi}{3}$ or $\displaystyle\frac{\pi}{3}$ 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. 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.