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Lecture Comments (3)

2 answers

Last reply by:
Wed Jul 31, 2013 3:24 AM

Post by StudentD on July 30, 2013

what if the degree of the polynomial is not an integer, say 1/2? would it be 1? I think I  might've missed it somewhere.

Fundamental Theorem of Algebra

  • This lesson will not be like a standard lesson: there will be hardly any numbers, and no examples at all. This is because we're learning some interesting ideas from advanced math.
  • The Fundamental Theorem of Algebra says, "Every polynomial of degree n > 0 has at least one root in the complex numbers."
  • Notice that "in the complex numbers" does not mean the root can't be real. An entirely real number is contained in the complex numbers because they contain both real and imaginary numbers.
  • The Fundamental Theorem of Algebra allows us to create the n roots theorem: "A polynomial of degree n has exactly n roots in the complex numbers (these roots are not necessarily distinct)." Equivalently, we can also write this as: "For any polynomial P(x) of degree n > 0, there exist n complex numbers z1, z2,..., zn (not necessarily distinct) and a constant number a such that
    P(x) = a  (x − z1)(x−z2) …(x−zn).
  • Some important things to note about the above theorem:
    • While a polynomial of degree n has n roots, they are not necessarily all distinct. We can have roots that "repeat" multiple times. This idea is called multiplicity.
    • The above theorem is only true if we're allowing complex numbers. If we only stick to the real numbers, there are many polynomials that cannot be broken up into linear factors.
    • While it was not explicitly stated, the theorem holds whether the coefficients of the polynomial are real or complex numbers.
    • The Fundamental Theorem of Algebra and the n roots theorem are existence theorems. However, they do not tell us how to find these roots. All they do is tell us that they're out there somewhere. They guarantee existence, but that's all.

Fundamental Theorem of Algebra

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:05
  • Idea: Hidden Roots 1:16
    • Roots in Complex Form
    • All Polynomials Have Roots
  • Fundamental Theorem of Algebra 2:21
  • Where Are All the Imaginary Roots, Then? 3:17
    • All Roots are Complex
    • Real Numbers are a Subset of Complex Numbers
  • The n Roots Theorem 5:01
    • For Any Polynomial, Its Degree is Equal to the Number of Roots
    • Equivalent Statement
  • Comments: Multiplicity 6:29
    • Non-Distinct Roots
    • Denoting Multiplicity
  • Comments: Complex Numbers Necessary 7:41
  • Comments: Complex Coefficients Allowed 8:55
  • Comments: Existence Theorem 9:59
  • Proof Sketch of n Roots Theorem 10:45
    • First Root
    • Second Root
    • Continuation to Find all Roots

Transcription: Fundamental Theorem of Algebra

Hi--welcome back to

Today, we are going to talk about the fundamental theorem of algebra.0002

This lesson is not going to be like a standard lesson.0006

So normally, we see some interesting ideas, and then we look at how numbers interact with those ideas.0008

And then, we start working out examples.0013

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.0015

Why? The ideas we are about to see are from advanced mathematics, and they are more0022

about showing fundamental truths than just giving us a bunch of numbers and exercises to work with.0027

Furthermore, since they don't really involve numbers, most teachers and textbooks can't and won't be able to test this stuff.0032

And instead, they only briefly mention it.0041

So, why are you still watching this lesson, if you are never going to be graded on it?0043

It is because this stuff is cool--at least, I think so!0047

You are about to see some amazing results from really high-level mathematics, and you are going to be able to understand it.0050

You are actually going to be able to understand this stuff from what you have just seen in the previous lessons on polynomials.0055

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.0060

If you have the time, humor me and watch this--you might find it interesting.0070

I think it is pretty cool, and I hope you do, too; thanks.0073

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.0077

For example, if we look at x2 + 1 = 0, there are no solutions to that.0087

x2 + 1 has no roots in the reals; if we look at its graph over here on the reals, 0092

we see that it never manages to cross the x-axis, so it has no roots.0097

But if we switch into the complex numbers, we find roots--we find that i and -i work.0102

For example, i2 + 1 becomes -1 (i, the imaginary number, squared, becomes -1) + 1; so we get 0.0108

Look, that is a solution; the same thing would happen if we took -i and squared that.0117

So, this realization that there are roots, even though at first it doesn't seem like there are roots,0121

might give us this sneaking suspicion that all polynomials have roots, if we allow for the complex numbers--0126

that there is some root out there that we might not be able to find using the real numbers,0132

but that it is out there when we expand our search to the complex numbers--at least we would have this suspicion.0136

It turns out that our suspicion was correct: this idea is expressed by the fundamental theorem of algebra.0142

It is such an important, fundamental idea about how polynomials work that it gets this name, "fundamental theorem of algebra."0148

And it says that every polynomial of degree n > 0 has at least one root in the complex numbers.0156

Notice that it is "at least"; so it could have more; and it also says "complex numbers."0163

So, every polynomial that isn't just a constant--that has a degree of 1 or greater--has at least one root.0168

We are guaranteed a root in the complex numbers, if we expand our search to the complex numbers.0175

This seems like a pretty simple idea, probably; but it is actually very difficult to prove.0180

The proof requires advanced mathematics, and so we won't be able to see it here in this course.0186

But trust that it is true; and it can be proved if you get to a high enough level and have enough background.0191

So, at this point, we might say, "Well, where are all the imaginary roots?"0198

Considering the fundamental theorem of algebra, why haven't we seen more imaginary roots?0202

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?0207

Those don't use imaginary numbers; it seems that we are just fine without imaginary numbers; we can find roots just fine.0215

So, what is up with the fundamental theorem of algebra?0222

Well, remember: it didn't say imaginary numbers; it said complex numbers.0225

It is the complex numbers; now, by definition, a complex number is of the form a + bi.0231

So notice, if we felt like it, we could say b = 0; and if b = 0, then the complex numbers are all real,0239

because we will be left with just the a portion; and a is just a real number.0246

So, we see from this that the reals are a subset of the complex.0250

The reals are contained in the complex numbers.0254

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 0.0257

So, the above solutions of x = 2 and -2...they are complex numbers, because they fit in the a portion.0264

They fit in the real portion of a complex number, so it is still a complex number.0272

We can show all real numbers as complex numbers; so that is what the fundamental theorem of algebra is telling us.0277

It is telling us that the roots are going to be complex numbers.0283

But since it is a complex number, it could just be a real number, just like all integers are contained in the real numbers.0287

You can be a real number, but still be an integer; you can be a complex number, but still also be a real number.0295

All right, with the fundamental theorem of algebra learned, we can now go on to the n roots theorem,0301

which is also sometimes called the end zeroes theorem or the linear factorization theorem.0306

A polynomial of degree n has exactly n roots in the complex numbers.0311

And these roots are not necessarily distinct; so they are not necessarily going to be different from one another.0317

And we will talk more about all of these things.0322

Another way to say this is that, for any polynomial p(x) of degree n > 0, there exist n complex numbers,0324

z1, z2, up until zn (n complex numbers, and each one is going to be some z).0334

And they are also, once again, not necessarily distinct: z1 is not necessarily going to be different than z2.0339

We are just guaranteed that there will be a total of n numbers.0345

And a constant number a, such that the polynomial is equal to a(x - z1)(x - z2)(x - zn)--0348

that is a way of saying that we can break it into a bunch of linear factors.0357

Since x minus just some number is a linear factor, we can break any polynomial into n linear factors.0362

Impressively, we will actually be able to prove this theorem from what we currently know.0371

We are going to see a sketch of the proof after we discuss what the theorem means.0375

And it is pretty cool that we can actually see how such an advanced piece of mathematics0379

can be proven from fairly simple things, assuming we can assume the fundamental theorem of algebra.0384

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.0391

For example, if we consider f(x) = x5 - 5x4 + 10x3 - 10x2 + 5x - 1,0400

and we factor it, we will eventually figure out that that is (x - 1)5.0406

Now, that means that, when we look at it on the graph, it has just one root--just one x-intercept.0411

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.0419

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).0428

They are just all the same linear factor; this idea is called multiplicity--that we can have the same root show up multiple times.0437

We can have the same factor show up multiple times.0446

In this case, we would say that f has a root at x = 1 with multiplicity 5,0448

because the same linear factor, containing one root, shows up 5 times when we break it into its factors: multiplicity.0453

Complex numbers are necessary: it shows up in the statement of the theorem, but it is really important0463

to be aware that this n roots theorem thing is only true if we are allowing complex numbers.0467

For example, f(x) = x3 + 1; we can break it into (x + 1)(x2 - x + 1).0473

So, (x + 1) has a solution in the reals, x = -1; but x2 - x + 1...that thing is an irreducible quadratic.0481

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.0491

But this theorem tells us that there must be an additional two roots, because we started at a degree of 3 in our polynomial.0501

So, since our polynomial had a degree of 3, we are guaranteed a total of 3 roots.0509

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.0513

So, those two roots must be complex.0521

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.0524

If we restrict ourselves to the reals, we won't necessarily be able to find n roots.0531

Complex coefficients are also allowed; it wasn't explicitly stated, but the n roots theorem is still true,0537

even if the coefficients of the polynomial are complex numbers.0543

For example, if we have this second-degree polynomial, this quadratic polynomial, (2 + i)x2 + -15ix + (-7 + 49i),0547

well, 2 + i, -15i, and -7 + 49i--those are all just complex numbers.0558

And it turns out that the n roots theorem is true here; we can break it into two linear factors,0563

because remember, we started with a degree of 2 on our polynomial.0570

So, we can break it into two linear factors, along with some constants at the front.0574

2 + i times x minus 3 - i times x - 7i--the degree of the polynomial is 2, and it has 2 roots;0579

or we can look at that, equivalent-ly, as being factored into two linear factors.0589

So, the theorem holds, whether the coefficients of the polynomial are real or complex numbers--pretty cool.0593

Existence theorem--this is a really important idea.0600

The fundamental theorem of algebra and the end roots theorem are existence theorems.0603

They guarantee the existence of roots; we know that there have to be n roots.0608

However, they don't tell us how to find the roots; all they do is tell us that they are out there somewhere.0614

They guarantee existence, but that is all; they don't tell us how to actually find what the roots are.0620

For example, we know that x4 + 1 = 0; since we have a degree of 4, it must have 4 roots.0626

But that doesn't actually put us any closer to figuring out what they are.0632

All we know is that they are out there somewhere; but we don't know what they are going to be from these theorems.0635

The theorem guarantees the existence, but it doesn't tell us how to actually get to them; that is an important point.0640

All right, we can now actually see a proof sketch of how the n roots theorem can be proven.0647

The fundamental theorem of algebra, like I said, requires some very advanced mathematics.0655

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.0660

So, starting with the fundamental theorem of algebra, which we will not be able to prove,0665

and the division algorithm, which we addressed earlier when we talked about polynomial division: the proof is within our graphs.0670

Start by considering some polynomial, p(x), with degree n > 0.0676

Now, the fundamental theorem of algebra, remember, said that there is at least one root.0682

So, by the fundamental theorem of algebra, we are guaranteed that there is at least one root.0686

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.0691

And we know that p(z1) = 0; it is a root, because if we plug z1 in, we get 0 out of the polynomial.0697

So, there is going to be at least one root to our polynomial, p; and we will call it z1.0705

Now, from earlier work, we know that, if we have a root, that is the same thing as saying (if there is some root0711

that causes our polynomial to give a 0) that there is a factor, (x - z1) somewhere inside of our polynomial.0718

So, by the division algorithm, we know that there exists some polynomial q1 (we will just call it q, for quotient;0727

and then we will need to give it a number, because there are going to be a bunch of these quotients coming up soon);0735

so we will call it q1(x), because it is the first time we have divided our initial polynomial.0739

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)).0744

We can pull out our factor, because we were guaranteed this factor,0753

because we know there is a root of z1, so there must be a factor of (x - z1).0757

So, we can pull that factor out; and we will be left with that factor, times some quotient.0762

So, we will have (x - z1), the factor that we know must be inside, because of the fundamental theorem of algebra;0766

and then, we pull it out through the division algorithm, and we must be left with some q1(x).0773

There can't be a remainder, because we know that that factor must be cleanly in there;0777

it must be able to divide out evenly; otherwise it couldn't be a factor.0781

So, we have that our initial polynomial is equal to (x - z1) times some other quotient polynomial, q1(x).0785

And since we pulled out a degree of 1 from a polynomial that initially started at n, we started at n;0793

and then we pull out 1; and we know that we are going to have a degree of n - 1 in our quotient polynomial.0798

So, we can keep going with this procedure.0805

We did it once, and we pulled out a root, and then we broke our polynomial up.0807

We knew that there was a root, and then we were able to factor out that linear factor, because of that root.0811

So, if we have n - 1 > 0 (remember, n - 1 is the degree of our first quotient), we can use the fundamental theorem again.0817

The fundamental theorem says that as long as your degree is greater than 0, there is a root in there.0826

So, if that is the case, we know that there must be some root, z2, where,0831

when we plug it into our quotient (remember, since n - 1 > 0, we can use the fundamental theorem of algebra0835

to guarantee that there is a root z2, so we plug it into our quotient), q1(z2) is equal to 0.0841

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.0848

So, we have (x - z1)(q(x)) is the same thing as p(x).0856

So, if we plug in z2 into p(x), then we are going to have z2 plugged in for our x.0862

And so, we will get z2 - z1; and that is some number; but we are also going to have q1 plugging into z2 here.0869

So, we knew that q1 of z2 equals 0; so that means that the whole thing has to come to 0,0876

because we take out a 0 here, and 0 will knock out whatever else we have.0881

So, the whole thing comes to 0, which means that p(z2) has to equal 0--pretty cool.0884

Now, we can just use the division algorithm again.0893

We know that q1 of z2 equals 0;0895

so it must be the case that (x - z2) is a factor of q1(x), since it is able to be pulled out.0899

So, we can pull out a factor, (x - z2), from q1, which gives us,0906

once again, another quotient--our second quotient, so we will call it q2(x).0910

And q2 will be a polynomial of degree n - 2, because remember:0915

q1 started at n - 1; and then we are going to subtract 1, because we are pulling out a degree of 1.0919

So, we subtract 1; so we are going to get n - 2 as the degree of q2(x).0924

Now, this means that we can also express p(x) as (x - z1)(x - z2)(q2(x)).0931

Since q1(x) = (x - z2)(q2(x)), and we know that p(x)0939

equals (x - z1)(q1(x)), we take this right here, and we swap it in for q1(x) right here.0943

And we are going to get this thing right here, (x - z1)(x - z2) times our quotient 2, our second quotient of x.0951

And we just keep going with this method, until eventually, with the quotient polynomial, we get down to some qn0960

that is going to just be a constant; we will be stuck at degree 0.0967

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.0972

And so, we will have taken n steps down; we will have managed to pull out z1, z2, all the way up until zn.0977

Effectively, what we are doing is whittling down the degree of the polynomial we started with.0984

So, we are whittling down the degree: for every step we lower the degree by, we manage to get another root.0988

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.0994

For example, let's consider if p(x) is degree 5; we start at degree 5 with p(x), our initial polynomial.1001

By the fundamental theorem of algebra, we are guaranteed that it must have one root; z1, we will call it.1009

Then, we use the division algorithm to break it into q1(x).1016

Now, q1(x) is going to have degree 4; and so, by the fundamental theorem of algebra, we are guaranteed another root, z2.1022

And since z2 is the root of q1, and q1 is contained inside of our initial polynomial p,1031

z2 must also be a root for p(x); great.1038

Now, we use the division algorithm again; we can break this down,1042

and we are able to get to saying that q2(x) is able to come out of q1(x).1046

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, z3.1051

And since q2 is contained inside of q1, which is contained inside of p,1058

we know that z3 is, once again, a root for our initial polynomial that we started with.1062

We then break down q2(x); we get to q3(x).1067

q3(x) will now have a degree of 2, because we are stepping down once each time.1071

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.1076

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 z5.1083

We plug that in; and so, that is going to once again be yet another root.1094

And finally, we use the division algorithm one last time, and we get down to q5(x), which now has a degree of 0.1098

And since we have a degree of 0, we are not able to get a root out of it.1106

So, we have used up all of our roots; we have managed to pull out 5 roots, because we started at degree 5.1109

And each step, we are able to pull out one root as we subtract by 1.1115

And so, we are finally left with some constant polynomial.1121

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.1123

It is going to be a(x - z1)(x - z2)(x - z3)(x - z4)(x - z5)--pretty cool.1130

This is really, really deep stuff from advanced mathematics that we can actually be pretty good at understanding,1140

just from what we have managed to learn about polynomials so far--pretty impressive.1144

All right, we will see you at later--goodbye!1148