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 Week 11…
in the Week 11 videos.
Okay, Real Quadratic Factors.
What we'd like to - okay let's read this proposition first.
Let f at x be a real polynomial.
If c is a strictly complex root,
so remember this is set difference so c is a complex number not inside the real numbers,
and f at c is 0, so it's a root of our polynomial,
then there exists a polynomial g of x, real, such that
g of x is a real quadratic factor of f of x.
Okay so, again pause thevideo 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 clearer.
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 let's do that. Let's pair up c and its conjugate, c bar.
We know that both of these are roots by the Conjugate Roots Theorem,
that will come into play a little bit later.
If we multiply this out, we get
the following: so we get c plus its conjugate, that's just 2 times the real part,
and if we multiply c and its conjugate, we just get the modulus squared.
That's Properties of Modulus, and that's Properties of Conjugates there.
So good, okay, so here's g of x. This is a real polynomial, it has c and c bar as a root,
and we claim that g of x is a factor of f of 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 minus c and x minus c bar, we already know that these are both factors of f of x
over the complex numbers, that we know, right?
We saw that before.
The problem now is…
if I take f of x and I divide it by x minus c and x minus c bar
I know I get some quotient, let’s call it q of 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 of 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 of 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 of x and r of x,
and these are unique,
such that it satisfies this equation f is equal to g q plus r,
and r is either 0 or its degree is less than 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] of x is either linear or constant.
When we plug in x equals c into this equation, what do we get? We see that
well c was a root so we know that f at c is 0.
c is the root of our quadratic polynomial g of c,
so we know that's also 0. So we see that r of c must be 0.
So if r of 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, right?
So therefore this is the 0 polynomial as a complex polynomial,
so why is it that 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 of x was a linear polynomial, let's say r of x is equal to a x plus b,
then if we plug in c we get
a times c plus b equals 0, and so c must be a real number, it's minus b over 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 divides f, so g is a factor of f…
is a real factor of f.
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.
My acronym is wrong here, it should be RFRP, not RFPF.
Okay, let's see a proof of this.
I guess I should…before we read the proof, you should actually try this proof.
It's a little bit…
it's 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 a little bit
of a sketchy proof, but…
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,
and every individual real root corresponds to a real linear factor.
We put those two things together,
so here we have our quadratic factors and here our
real linear factors are given by g1 to g k,
multiply them together, and that gives us the f of x, up to some possible constant here,
which is the - it turns out that this constant 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.