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College Algebra: Factor a Polynomials Given a Zero


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About this Lesson

  • Type: Video Tutorial
  • Length: 11:09
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 119 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 Factor 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 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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The zero of a polynomial are just all the solutions to f(x) = 0. So, basically, these are the x intercepts where a particular polynomial crosses the x-axis. You find that by taking the whole polynomial, setting it equal to 0 and solve. Those values of x that you find are sometimes called the 0's of a polynomial. It's just those values of x that if you plug into the polynomial, they produce 0. That's why they're called 0's of a polynomial. Now, it turns out that if you just know, in some cases, if you just know one 0 of a polynomial, then you'll be able to find all the 0's. I want to illustrate this technique and also, bring together the idea of the factor theorem and synthetic division and some other things. Try to tie ideas together here, by looking at some simple examples. Well, actually not so simple examples, but examples nonetheless.
Let's take a look at the following. Here's a polynomial, f(x) = x^3 - x^2 - 4x - 6. Okay, so there's the cubic polynomial. We're told that 3 is a 0 of this. That means if you plug in 3, this produces a 0. You can check that. I won't do that right now, but you could plug in 3 here and get 27, subtract off 9, subtract off another 12, subtract of another 6 and you'll see you'll get 0. Well, I guess I just did it. The point is, that is a 0, we're told that. So, x = 3 is a 0. Now, the question is, "Can we use that information to find all the other 0's of this cubic?" A cubic will have three 0's and I just know one of them. Can I find the other ones?
Well, to find a 0, technically, we set that equal to 0 and solve. But, I don't know how to factor a cubic. I know how to factor quadratics, but I don't know how to factor cubics that well. So, what could I do? Well, one method is to say, "Wait a minute, I already know one of the 0's." So, by the factor theorem, if x = 3 is a 0, that means that x - 3 must be a factor of this. That's the factor theorem. If f(3) = 0, that means, from the factor theorem, this is a factor of f(x). Well, that means this must divide into this evenly and I can see what remains and try to factor that. So, knowing that I have a 0, that leads me to actually find a factor. Once I have a factor, I can use synthetic division to divide out, see what I'm left with, take what I'm left with and then factor that.
If I take out an x - 3 factor here, I must be left with a quadratic. Because a quadratic times stuff with x gives me a cubic. Once I have a quadratic, I know how to find the 0's of that. So, this is going to work. Let's see what happens when I divide x - 3 into that. I'm going to use synthetic division. Remember how that goes, I flip the sign of this, which gives me back this 3. I make half a house. I represent all the coefficients in order, 1 - 1 - 4 - 6. I'm going to start my process. By the way, a little teeny check or balance that we can use to see if we're okay is, at the end of the day the remainder, which lives here--that remainder must be 0. Right? Because this is suppose to go in evenly. So, if this remainder turns out to be 1 or 2 or -3 or 4, I know something is very wrong, because this number should be 0. This should go in evenly since it's a factor
Bring down the 1. Look how easy this is once you get the hang of it. 3 times 1 is 3, add, you get 2. Multiply, you get 6. Add, you get 2. Multiply, you get 6, add, 0. Okay, so that's a good check. This is what's left over. How do I decode that? Well, this is the x^2 term, this is the x term, that's the constant term. So, what I see is, f(x) this polynomial here, equals--well x^2 + 2x + 2 all multiplied by x - 3. That's the factor we knew. So, now we see what we have to multiply. This times that, gives me f. So, if I want to find the 0's, I set this whole thing equal to 0 and solve. Well, if I have product equaling 0--so I have (x^2 +2x + 2)(x - 3) and I want to find out when that equals 0. That's the polynomial. That's f(x), we just saw that. So, what I do now is say, either that's 0, so x = 3--well that's the one we knew to begin with, or this is 0, well that's quadratic.
So, now I've got to solve x^2 + 2x +2 = 0. Well, how do I solve this? Well, I can try to factor. Try, try, try, doesn't work. So, I'll use the quadratic formula. X = -b plus or minus the square root of b^2, that's 4, minus 4ac. So, that's going to be 8 all over 2a. Now, what is that? Well, that's -2 plus or minus the square root of -4. Ouch. That's going to be an imaginary number. That's going to be a complex number. The square root of -4 is actually 2i. So, this equals -2 plus or minus 2i, all divided by 2. I can factor out that 2 on the top and cancel with this and I see -1 plus or minus i.
So, we see the other two roots of the cubic equation are actually imaginary numbers. That's okay. That means, by the way, that graphically this cubic is only going to cross through the x-axis only once, at 3 and then never again. So, it may look something like this. That could be how the cubic goes. It crosses here at 3, but never crosses again. It's sort of a cubic. It has three little wings. The other roots turn out to be imaginary. So, what are the 0's? There's that one real root, 3, then there's -1 + i and then there's -1 - i. So, look, I started off with this really complicated looking cubic. I wanted to find the 0's of it. Set it equal to 0 and solve. Didn't know how to do it. Since I was given one of the 0's, I was able to convert that to a factor, divide out. Then I have quadratic. Once I've got a quadratic, I'm home free.
Let's try one quick example to illustrate this a little bit more. f(x) = x^3 - 2x^2 - 5x + 6. I tell you that x = 1 is a 0 of this. Now, I want you to find all the 0's. So, what do I do? Well, if that's a 0, that means that f evaluated at 1 equals 0. You can check. By putting a 1 here I see 1 - 2 is -1, -1 and -5 is -6, and 6 is 0. It checks. This really is a root. Well, by the factor theorem, if this equals 0, that means x - 1 is a factor. So, if this is factor, then what does that mean? If I divide it through into this, it should go in evenly, no remainder and I can then see what's left.
So, let's now do a synthetic division. I flip the sign, 1, make a house. Write down the coefficients, 1, -2, -5, 6. Don't forget to write down all the coefficients. If there wasn't an x^2 term in there, we'd put a placeholder of 0 right in here. But, since there is one, we don't have to worry about it. Then, bring this down and multiply. Now remember, the remainder is going to be here. If this is really a factor, the remainder here should be 0. Let's see if we're okay. 1 times 1, 1, add, -1, 1 times 1 is 1, -1, add -6, 1 time 1 is 6, -6, add 0. Looks great. This is what's left over. That's the x^2, that's the x part, that's the constant part. So, what I see is, f(x), this whole complicated thing is just x - 1 times this thing. So, what I see here f(x) = x - 1. There's the factor and when I divide it out, I see an x^2 - x - 6. You see how I read off the coefficients right there?
Now, all I've got to do is see if I can factor that. So, you can try to factor. I think we'll have better luck here, I hope. X, x, opposite signs, plus, minus--how about 3 and 2? Looks good to me. What are the 0's? Set it equal to 0 and solve. I see either this is 0, that means, so here are the 0's. So, 0's of f(x), they're x = 1--that's the one we were told, in fact. So, we found it again, that's good. This equals 0. If x = -2 and this equals 0 if x = 3. So, here we see there are three real 0's of this cubic. So, what we see is that here they are, 1, -2, 3.
Just as a little side note. What does this look like graphically? Remember the other one graphically? It went like this. It crossed through and then it went and looped up here. This one would be a little different, because I have -2, crosses the x-axis there, 1 and then 2, 3. So, it crosses here and here. So, this one has to look something like this. You see it crosses at -2, it crosses at 1 and it crosses at 3. So, this one sort of zigzags right through--it crosses three times. Whereas the other one we saw, I'll just remind you, had the feature that it only crossed--what value was it? I forgot now. Was it 1 or something? I think it was at-- oh no, it was at 3.
See, the other one had that little bend still, but the bend didn't quite make it down far enough to hit the axis again. So, it only had one real root, two imaginary. This one has three real roots. A lot of possibilities, a lot of things could happen and it's really neat that using the factor theorem, I can actually report all the 0's just knowing one on a cubic.
Have some fun with these. I actually think these are pretty cool. Just be careful with the synthetic division. Enjoy.
Polynomial and Rational Functions
The Factor Theorem
Factoring a Polynomial Given a Zero Page [2 of 2]

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