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College Algebra: Finding a Polynomial's Real Zeros


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About this Lesson

  • Type: Video Tutorial
  • Length: 8:16
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 89 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Polynomial & Rational Functions (23 lessons, $35.64)
College Algebra: Zeros of Polynomials (5 lessons, $7.92)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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So if you want to find the real zeros of the polynomial, we now have a sense how to go about that. What I do is I take the polynomial and I set it equal to zero and solve for x and find those values for x, which make the polynomial zero. Those are the zeros. And how do you do that? Well, you try to factor the polynomial, so you have a product of a whole bunch of things equal zero, and then you know that if a product of a whole bunch of things equals zero, either one of them is zero or the other is zero or the other is zero, and again, you divide and conquer and you can solve those little teeny equations separately.
But I'm going to tell you the truth right now and maybe you didn't know this, or maybe you did know this, but the reality is, if you just write down a polynomial and you want to find its real zeros; that is really hard business--really, really hard. Now happily the kind of things that we're looking at together and the kind of things you'll probably do on your own, those are usually doable. But let me just tell you that it's not the case that every single polynomial can be easily factored and found the roots of. In fact, mathematicians are constantly wondering how can we do that because it's so hard.
But in particular cases it's not so bad to do. Let me show you some of those particular cases now and let's take a look at how you can find the zeros of some polynomials. So, for example, let's do a real easy one just to warm up here. Suppose f(x) = x(x - 4)(x - 1)(x + 7). Well, even though this is a fourth degree polynomial, if you multiplied all this out, which would take a lot of time, you'd see x^4 as the highest power, this is actually real easy to find the zeros of because it's already in factored form, so it's a piece of cake. If I want to find the zeros, I just set this equal to zero and what do I do? I say, well, this product equals zero, so therefore either this term equals zero, that term equals zero, this term equals zero, or that term equals zero, and I just go through now and solve each of those things. For example, if this equals zero, that means x = 0. So there's one zero. If this term were to equal zero, that would mean x has to be 4. So then we'd have x = 4 as another zero. If this term were to equal zero, that would mean x has to be 1, so then I would see x = 1. Well, there's another zero. Look at all these zeros. And then if x + 7 were to be zero, I'd see that x would have to be -7. So again, in some sense I'm sort of using that factor theorem, right? If I have this as a factor I know that 4 must be a zero. You see how the factor theorem plays in here? If this really is a zero, x - 1 should be a factor, and it is, it's right there. So you can see that connection between factor, zero, factor, zero, factor, zero, and so on. It's an important connection. But here it was so easy just to read them right off--0, 4, 1, -7, those are all the zeros of this polynomial. Why was it so easy? Well, they handed it to us on a silver platter because it was factored.
Okay, what about platters that aren't silver? So suppose I take a look at the following: f(x) = 2x³ - 7x² - 15x. And here I'm giving you a cubic and I want you to find all the zeros of this. So I want you to set that equal to zero and solve. Well, how in the world would you do that? Well, you look at that and you say, "Gosh, cubics are hard. How do you factor cubics?" But look. But just look at it and don't panic. I see there's an x here, an x here, and an x here, so I can just factor out that x and I'd be left with some quadratic. Well, quadratics, those are a little better to solve. So let's factor out that common factor of x and see what we're left with. Well, I'd have an x times... and what's left? I would have a 2x² - 7x - 15. So I just factored out the x here and now I'd set that equal to zero, and I want to solve. So one solution is apparent: x = 0. But what about here? Well, can I factor that? Let's try it.
So I'm going to make some big parenthesis here and see if we can make some progress. I'll put a 2x here and an x here. This tells me that we're going to have opposite signs, but I don't know how to put them down. Remember, you can't put them down cavalierly if you've got a 2 factor in front of 1. You've got to sort of jockey everything. The product of these two numbers has to be -15, but when you combine them with that 2, I've got to end up with -7. So what in the world should I do? How about 3 and 5? They're a product of 15. And can I take the 3 and the 5 and put them... Well, 5 and 2 would be 10, and then if I subtract off 3, I would get 7. So, in fact, it looks like I should put the 5 here, and the 3 here. Now, where's the negative sign? I want the final answer to be negative, so I want the big value to be negative. Let me put the negative here and the positive here and see if that works. This is 2x² and then I have a -10x + 3x, gives me a -7x. Good, and this is a -15. Great. I factored, therefore I'm done with business. X = 0 would be one zero, and then where does this equal zero? Well, if I set that equal to zero--2x + 3 = 0, that means that 2x = -3, which means that x = -3/2. So there's another zero, and then a final zero will be one this equals zero, which is just when x = 5. So this cubic actually has three real zeros: 0, -3/2, and 5. Easy to find it? I factored out a common factor and was able to factor the rest as a quadratic. Cool.
One last one just for fun. f(x) = x³ - 1 and I want to find zeros of this polynomial, so I want to set this equal to zero and solve. Now, I've got to factor that cubic. Well, there's no common factor of x there. That's not going to work, so that's not looking too good. So what should I do? Well, actually, I remember something from a long time ago. This actually is the difference of two perfect cubes. It's x³ and 1³, and in fact, we saw a long time ago, and if you want to review this--you might want to actually review this other lecture--you can click around me and actually review the lecture--the difference of two cubes can be factored in the following way. So we worked this out a while ago. There's an x - 1 factor and then there's an x² + x + 1 factor. So I'll let you review that if you want, but there it is. And now I can try to factor this. Well, let's try to factor that. First of all, let's just compute the discriminate of that. Now, the discriminate is -b, so it would be b², which would be 1² - 4 times ac, so the discriminate is -3. That means that if I were to solve this using the quadratic equation at the end of the day I'm going to have all sorts of junky, junky, junky stuff, plus or minus, and I'm going to have the square root of -3. So that negative in the square root, the fact that the discriminate is negative tells me these are going to be imaginary roots, imaginary zeros, because I'm going to have the square root of negative numbers.
So, in fact, this parabola is not going to cross the x-axis at all. This is going to contribute no real zeros, only some complex ones, and I'm asked to find the real zeros, so in fact, there will be no contribution from here. The only contribution will be x = 1. That's the only zero. And again, this is one of those deals where, in fact, what you have here is the following. You have a picture that looks like this. So let me try to draw this for you really fast. So you have a cubic that only crosses the x-axis at one point, and that's at 1. Even though potentially other ones can cross at three, this one only crosses at one. The other two roots are imaginary.
See if you can find the real zeros of these polynomials by factoring in some sort of clever way. Enjoy.
Polynomial and Rational Functions
Zeros of Polynomials
Finding the Real Zeroes for a Polynomial Page [2 of 2]

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