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About this Lesson
 Type: Video Tutorial
 Length: 6:09
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 66 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: The Remainder Theorem (2 lessons, $2.97)
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 athttp://www.thinkwell.com/student/product/collegealgebra. 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 UTAustin 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, padic 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.
About this Author
 Thinkwell
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 Joined:
11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
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So now let's use this remainder theorem to actually help us evaluate polynomials. Now what's the idea here? Well, let me remind you that with the remainder theorem what we saw was if you have p(x) and you're going to divide it by x  c, we now know what the remainder is. We'll have some quotient times x  c and then plus the remainder, and the remainder we saw was just the polynomial evaluated at c, so it's just going to be p(c), and that's the remainder. So the cool thing is, suppose someone asked us the question, find p of c. That is, here's a polynomial, evaluate it at c. Well, one way of doing that is just taking the value of c and placing it in wherever you see an x and evaluate. But now we see there's a completely different way of doing that. Namely, let's just do the division, divide the polynomial by x  c, using synthetic division, and then see what the remainder is, because the remainder will exactly be the polynomial evaluated at c.
Let me show you this with an example. Let's suppose that p(x) = 5x³  6x²  28x  2. And suppose someone asked us what is p evaluated at 2? So just plug in 2 here. Well, you could just do that. You could put a 2 in here, a 2 in here, a 2 in here. So that's going to be 2³ times 5 minus 6 times 2²  28 times 2 minus 2. And you can actually compute that. But now let me show you another way of computing that exact same number. What I'm going to do is, notice that this will be the answer to the following question. What is the remainder when this thing is divided by x  2, which by the way, is just x + 2. So what's the remainder? The remainder will exactly be p evaluated at 2. So all I have to do now is compute the remainder using synthetic division. So here we go. This thing is just x + 2. So if I'm going to divide this into this polynomial, I have to switch the sign of that, so notice that now becomes back to the 2 that we started with, and I write down the coefficients here, a 5, a 6, a 28, and a 2, and I go through synthetic division, and at the end of the day the remainder that appears right here, that should equal p evaluated at 2, so that should be our answer.
Let's see what happens. I bring down the 5, multiply it by 2 and get a 10. Add here and get a 16. I multiply the 16 by 2 and I get a +32. I take +32 and add it to 28 and I get a net gain of 4. 4 times 2 is 8, and 8 + 2 is 10. And that's the remainder, so, in fact, that should equal p evaluated at 2. That is to say if you actually evaluate this thing at 2, that is 2³, which is 8, 8 times 5, which is 40, then subtract 6 times 4 minus 28 times 2 + 2. All that should equal 10, and you can see it's actually easier to do the synthetic division. The numbers here were pretty reasonable; there was no high powers, and I able to actually evaluate a polynomial. What was the secret? The secret was to go from this idea, evaluating a polynomial, to thinking about it as a remainder. So I divided by x minus that number, which became x + 2, and then I did my synthetic division. So I switched the sign, did a synthetic division very carefully, and the remainder turns out to be the value of a function at 2. And you can try it. Take 2, plug it in here, and you'll see exactly 10.
Let's try another one. Suppose the polynomial someone gives us is 3x²  1, and the question is find p evaluated at 2. Well, this one is so easy we can actually do it by hand, so let me actually do it. What would I get? I would see 3 times 2²  1, and that would equal 3 times 4, which is 12, minus 1, which is 11. So there's the answer. So you're saying, "Okay, then why bother doing it another way?" Well, this is just a very, very simple example, and now I want to show you that, in fact, you get the same answer by doing it this other way. The other way would be to say, `Okay, let's actually divide the polynomial by x  2." I want to find the value at 2, so I look at x  2. So if I divide it by x  2, that means synthetic division. I switch the sign of that to give me back the 2. Here I see a 3. There's no x term, so I write a zero. There's a 1 here. Now, watch how easy this is going to beeven easier than this. I bring down the 3. 2 times 3 is 6. I add 6. 6 times 2 is 12. 12  1 = 11. It's actually easier than even doing this, because there was no squaring. It was just one multiplication and a couple of additions. And you can see the exact same answer.
So the point is, if you want to evaluate a really complicated polynomial at a particular point, you should do a different problem, namely find the remainder when x minus that point is divided into the polynomial. Use a little synthetic division, and the remainder turns out to be the polynomial evaluated at that pointa really neat fact, and actually very useful when you want to evaluate some really complicated polynomials at particular points. So it's cool. The remainder theorem turns out to save us. Try these and see if you can be saved too.
Polynomial and Rational Functions
The Remainder Theorem
More on the Remainder Theorem Page [1 of 1]
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