WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about the fundamental theorem of algebra.
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This lesson is not going to be like a standard lesson.
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So normally, we see some interesting ideas, and then we look at how numbers interact with those ideas.
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And then, we start working out examples.
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For this lesson, though, we are going to just see interesting ideas; there will be hardly any numbers, and there will be no examples whatsoever.
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Why? The ideas we are about to see are from advanced mathematics, and they are more
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about showing fundamental truths than just giving us a bunch of numbers and exercises to work with.
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Furthermore, since they don't really involve numbers, most teachers and textbooks can't and won't be able to test this stuff.
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And instead, they only briefly mention it.
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So, why are you still watching this lesson, if you are never going to be graded on it?
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It is because this stuff is cool--at least, I think so!
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You are about to see some amazing results from really high-level mathematics, and you are going to be able to understand it.
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You are actually going to be able to understand this stuff from what you have just seen in the previous lessons on polynomials.
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It is a really cool chance to have a culmination of understanding, and be able to get something pretty advanced from a level that is not super difficult to understand.
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If you have the time, humor me and watch this--you might find it interesting.
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I think it is pretty cool, and I hope you do, too; thanks.
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All right, previously we have seen that, while some polynomials seem to have no roots, they can actually have hidden roots, in a way, in the complex numbers.
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For example, if we look at x² + 1 = 0, there are no solutions to that.
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x² + 1 has no roots in the reals; if we look at its graph over here on the reals,
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we see that it never manages to cross the x-axis, so it has no roots.
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But if we switch into the complex numbers, we find roots--we find that i and -i work.
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For example, i² + 1 becomes -1 (i, the imaginary number, squared, becomes -1) + 1; so we get 0.
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Look, that is a solution; the same thing would happen if we took -i and squared that.
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So, this realization that there are roots, even though at first it doesn't seem like there are roots,
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might give us this sneaking suspicion that all polynomials have roots, if we allow for the complex numbers--
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that there is some root out there that we might not be able to find using the real numbers,
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but that it is out there when we expand our search to the complex numbers--at least we would have this suspicion.
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It turns out that our suspicion was correct: this idea is expressed by the **fundamental theorem of algebra**.
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It is such an important, fundamental idea about how polynomials work that it gets this name, "fundamental theorem of algebra."
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And it says that every polynomial of degree n > 0 has at least one root in the complex numbers.
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Notice that it is "at least"; so it could have more; and it also says "complex numbers."
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So, every polynomial that isn't just a constant--that has a degree of 1 or greater--has at least one root.
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We are guaranteed a root in the complex numbers, if we expand our search to the complex numbers.
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This seems like a pretty simple idea, probably; but it is actually very difficult to prove.
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The proof requires advanced mathematics, and so we won't be able to see it here in this course.
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But trust that it is true; and it can be proved if you get to a high enough level and have enough background.
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So, at this point, we might say, "Well, where are all the imaginary roots?"
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Considering the fundamental theorem of algebra, why haven't we seen more imaginary roots?
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If we are guaranteed what that says, why is it that x² - 4 = 0...why do we get only the roots of 2 and -2?
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Those don't use imaginary numbers; it seems that we are just fine without imaginary numbers; we can find roots just fine.
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So, what is up with the fundamental theorem of algebra?
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Well, remember: it didn't say imaginary numbers; it said complex numbers.
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It is the complex numbers; now, by definition, a complex number is of the form a + bi.
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So notice, if we felt like it, we could say b = 0; and if b = 0, then the complex numbers are all real,
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because we will be left with just the a portion; and a is just a real number.
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So, we see from this that the reals are a subset of the complex.
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The reals are contained in the complex numbers.
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The complex numbers have all of the real numbers inside of them, because we can just cut off that imaginary component by setting b to 0.
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So, the above solutions of x = 2 and -2...they are complex numbers, because they fit in the a portion.
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They fit in the real portion of a complex number, so it is still a complex number.
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We can show all real numbers as complex numbers; so that is what the fundamental theorem of algebra is telling us.
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It is telling us that the roots are going to be complex numbers.
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But since it is a complex number, it could just be a real number, just like all integers are contained in the real numbers.
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You can be a real number, but still be an integer; you can be a complex number, but still also be a real number.
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All right, with the fundamental theorem of algebra learned, we can now go on to the n roots theorem,
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which is also sometimes called the end zeroes theorem or the linear factorization theorem.
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A polynomial of degree n has exactly n roots in the complex numbers.
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And these roots are not necessarily distinct; so they are not necessarily going to be different from one another.
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And we will talk more about all of these things.
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Another way to say this is that, for any polynomial p(x) of degree n > 0, there exist n complex numbers,
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z₁, z₂, up until z<font size="-6">n</font> (n complex numbers, and each one is going to be some z).
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And they are also, once again, not necessarily distinct: z₁ is not necessarily going to be different than z₂.
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We are just guaranteed that there will be a total of n numbers.
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And a constant number a, such that the polynomial is equal to a(x - z₁)(x - z₂)(x - z<font size="-6">n</font>)--
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that is a way of saying that we can break it into a bunch of linear factors.
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Since x minus just some number is a linear factor, we can break any polynomial into n linear factors.
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Impressively, we will actually be able to prove this theorem from what we currently know.
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We are going to see a sketch of the proof after we discuss what the theorem means.
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And it is pretty cool that we can actually see how such an advanced piece of mathematics
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can be proven from fairly simple things, assuming we can assume the fundamental theorem of algebra.
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The first comment is multiplicity: we need to notice that a polynomial of degree n has n roots, but that they are not necessarily going to be distinct.
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For example, if we consider f(x) = x⁵ - 5x⁴ + 10x³ - 10x² + 5x - 1,
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and we factor it, we will eventually figure out that that is (x - 1)⁵.
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Now, that means that, when we look at it on the graph, it has just one root--just one x-intercept.
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But in a way, it is showing up 5 times; so f only has one distinct root at x = 1; that root repeats 5 times.
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We have it to the fifth; and so we can break f into five linear factors, (x - 1)(x - 1)(x - 1)(x - 1)(x - 1).
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They are just all the same linear factor; this idea is called multiplicity--that we can have the same root show up multiple times.
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We can have the same factor show up multiple times.
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In this case, we would say that f has a root at x = 1 with multiplicity 5,
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because the same linear factor, containing one root, shows up 5 times when we break it into its factors: multiplicity.
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Complex numbers are necessary: it shows up in the statement of the theorem, but it is really important
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to be aware that this n roots theorem thing is only true if we are allowing complex numbers.
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For example, f(x) = x³ + 1; we can break it into (x + 1)(x² - x + 1).
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So, (x + 1) has a solution in the reals, x = -1; but x² - x + 1...that thing is an irreducible quadratic.
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It doesn't bring any real solutions; so if we restrict ourselves to the reals, f has only one root at x = -1, if we are stuck just in the reals.
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But this theorem tells us that there must be an additional two roots, because we started at a degree of 3 in our polynomial.
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So, since our polynomial had a degree of 3, we are guaranteed a total of 3 roots.
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And if we found that there is one root already, then we know that we can subtract one off; so there must be two roots left that we haven't found yet.
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So, those two roots must be complex.
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In other words, the theorem guarantees n roots (allowing for multiplicity, as we just talked about), but only if we are using the complex numbers.
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If we restrict ourselves to the reals, we won't necessarily be able to find n roots.
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Complex coefficients are also allowed; it wasn't explicitly stated, but the n roots theorem is still true,
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even if the coefficients of the polynomial are complex numbers.
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For example, if we have this second-degree polynomial, this quadratic polynomial, (2 + i)x² + -15ix + (-7 + 49i),
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well, 2 + i, -15i, and -7 + 49i--those are all just complex numbers.
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And it turns out that the n roots theorem is true here; we can break it into two linear factors,
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because remember, we started with a degree of 2 on our polynomial.
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So, we can break it into two linear factors, along with some constants at the front.
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2 + i times x minus 3 - i times x - 7i--the degree of the polynomial is 2, and it has 2 roots;
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or we can look at that, equivalent-ly, as being factored into two linear factors.
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So, the theorem holds, whether the coefficients of the polynomial are real or complex numbers--pretty cool.
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Existence theorem--this is a really important idea.
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The fundamental theorem of algebra and the end roots theorem are **existence theorems**.
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They guarantee the existence of roots; we know that there have to be n roots.
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However, they don't tell us how to find the roots; all they do is tell us that they are out there somewhere.
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They guarantee existence, but that is all; they don't tell us how to actually find what the roots are.
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For example, we know that x⁴ + 1 = 0; since we have a degree of 4, it must have 4 roots.
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But that doesn't actually put us any closer to figuring out what they are.
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All we know is that they are out there somewhere; but we don't know what they are going to be from these theorems.
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The theorem guarantees the existence, but it doesn't tell us how to actually get to them; that is an important point.
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All right, we can now actually see a proof sketch of how the n roots theorem can be proven.
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The fundamental theorem of algebra, like I said, requires some very advanced mathematics.
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We won't be able to see it here; but we can actually understand how the n roots theorem is being built out of the fundamental theorem of algebra.
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So, starting with the fundamental theorem of algebra, which we will not be able to prove,
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and the division algorithm, which we addressed earlier when we talked about polynomial division: the proof is within our graphs.
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Start by considering some polynomial, p(x), with degree n > 0.
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Now, the fundamental theorem of algebra, remember, said that there is at least one root.
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So, by the fundamental theorem of algebra, we are guaranteed that there is at least one root.
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So, let's give that one root that we are absolutely guaranteed a name; and we will call it z₁, because it is our first root.
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And we know that p(z₁) = 0; it is a root, because if we plug z₁ in, we get 0 out of the polynomial.
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So, there is going to be at least one root to our polynomial, p; and we will call it z₁.
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Now, from earlier work, we know that, if we have a root, that is the same thing as saying (if there is some root
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that causes our polynomial to give a 0) that there is a factor, (x - z₁) somewhere inside of our polynomial.
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So, by the division algorithm, we know that there exists some polynomial q₁ (we will just call it q, for quotient;
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and then we will need to give it a number, because there are going to be a bunch of these quotients coming up soon);
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so we will call it q₁(x), because it is the first time we have divided our initial polynomial.
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And it is going to be degree n - 1, because we are going to pull out x - z₁ so that we will get (x - z₁)(q₁(x)).
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We can pull out our factor, because we were guaranteed this factor,
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because we know there is a root of z₁, so there must be a factor of (x - z₁).
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So, we can pull that factor out; and we will be left with that factor, times some quotient.
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So, we will have (x - z₁), the factor that we know must be inside, because of the fundamental theorem of algebra;
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and then, we pull it out through the division algorithm, and we must be left with some q₁(x).
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There can't be a remainder, because we know that that factor must be cleanly in there;
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it must be able to divide out evenly; otherwise it couldn't be a factor.
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So, we have that our initial polynomial is equal to (x - z₁) times some other quotient polynomial, q₁(x).
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And since we pulled out a degree of 1 from a polynomial that initially started at n, we started at n;
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and then we pull out 1; and we know that we are going to have a degree of n - 1 in our quotient polynomial.
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So, we can keep going with this procedure.
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We did it once, and we pulled out a root, and then we broke our polynomial up.
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We knew that there was a root, and then we were able to factor out that linear factor, because of that root.
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So, if we have n - 1 > 0 (remember, n - 1 is the degree of our first quotient), we can use the fundamental theorem again.
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The fundamental theorem says that as long as your degree is greater than 0, there is a root in there.
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So, if that is the case, we know that there must be some root, z₂, where,
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when we plug it into our quotient (remember, since n - 1 > 0, we can use the fundamental theorem of algebra
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to guarantee that there is a root z₂, so we plug it into our quotient), q₁(z₂) is equal to 0.
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Cool; but we also know that z₂ has to be a root of p(x), because the way that we got q₁(x) was by dividing it out.
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So, we have (x - z₁)(q(x)) is the same thing as p(x).
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So, if we plug in z₂ into p(x), then we are going to have z₂ plugged in for our x.
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And so, we will get z₂ - z₁; and that is some number; but we are also going to have q₁ plugging into z₂ here.
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So, we knew that q₁ of z₂ equals 0; so that means that the whole thing has to come to 0,
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because we take out a 0 here, and 0 will knock out whatever else we have.
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So, the whole thing comes to 0, which means that p(z₂) has to equal 0--pretty cool.
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Now, we can just use the division algorithm again.
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We know that q₁ of z₂ equals 0;
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so it must be the case that (x - z₂) is a factor of q₁(x), since it is able to be pulled out.
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So, we can pull out a factor, (x - z₂), from q₁, which gives us,
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once again, another quotient--our second quotient, so we will call it q₂(x).
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And q₂ will be a polynomial of degree n - 2, because remember:
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q₁ started at n - 1; and then we are going to subtract 1, because we are pulling out a degree of 1.
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So, we subtract 1; so we are going to get n - 2 as the degree of q₂(x).
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Now, this means that we can also express p(x) as (x - z₁)(x - z₂)(q₂(x)).
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Since q₁(x) = (x - z₂)(q₂(x)), and we know that p(x)
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equals (x - z₁)(q₁(x)), we take this right here, and we swap it in for q₁(x) right here.
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And we are going to get this thing right here, (x - z₁)(x - z₂) times our quotient 2, our second quotient of x.
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And we just keep going with this method, until eventually, with the quotient polynomial, we get down to some q<font size="-6">n</font>
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that is going to just be a constant; we will be stuck at degree 0.
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At this point, we will have a total of n roots, because we have pulled out one root each time we take a step down.
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And so, we will have taken n steps down; we will have managed to pull out z₁, z₂, all the way up until z<font size="-6">n</font>.
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Effectively, what we are doing is whittling down the degree of the polynomial we started with.
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So, we are whittling down the degree: for every step we lower the degree by, we manage to get another root.
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If we start at degree n, that gives us n steps to go down; so we have n steps to pull from--we manage to get n roots out of it.
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For example, let's consider if p(x) is degree 5; we start at degree 5 with p(x), our initial polynomial.
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By the fundamental theorem of algebra, we are guaranteed that it must have one root; z₁, we will call it.
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Then, we use the division algorithm to break it into q₁(x).
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Now, q₁(x) is going to have degree 4; and so, by the fundamental theorem of algebra, we are guaranteed another root, z₂.
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And since z₂ is the root of q₁, and q₁ is contained inside of our initial polynomial p,
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z₂ must also be a root for p(x); great.
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Now, we use the division algorithm again; we can break this down,
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and we are able to get to saying that q₂(x) is able to come out of q₁(x).
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And it is going to have a degree of 3; by the fundamental theorem of algebra, we are guaranteed that it must have a root, z₃.
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And since q₂ is contained inside of q₁, which is contained inside of p,
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we know that z₃ is, once again, a root for our initial polynomial that we started with.
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We then break down q₂(x); we get to q₃(x).
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q₃(x) will now have a degree of 2, because we are stepping down once each time.
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Since it has a degree greater than 0, we are guaranteed a root there, which is once again going to be a root of our initial polynomial.
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We break it down once again; we get to q₄(x); it has a degree of 1, so it is guaranteed to have a root of z₅.
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We plug that in; and so, that is going to once again be yet another root.
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And finally, we use the division algorithm one last time, and we get down to q₅(x), which now has a degree of 0.
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And since we have a degree of 0, we are not able to get a root out of it.
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So, we have used up all of our roots; we have managed to pull out 5 roots, because we started at degree 5.
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And each step, we are able to pull out one root as we subtract by 1.
00:18:40.700 --> 00:18:43.200
And so, we are finally left with some constant polynomial.
00:18:43.200 --> 00:18:50.400
And that is where our a, the a that multiplies the whole thing--that is going to be our fifth quotient polynomial at the very end.
00:18:50.400 --> 00:18:59.900
It is going to be a(x - z₁)(x - z₂)(x - z₃)(x - z₄)(x - z₅)--pretty cool.
00:18:59.900 --> 00:19:04.200
This is really, really deep stuff from advanced mathematics that we can actually be pretty good at understanding,
00:19:04.200 --> 00:19:07.800
just from what we have managed to learn about polynomials so far--pretty impressive.
00:19:07.800 --> 00:19:09.000
All right, we will see you at Educator.com later--goodbye!