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College Algebra: Partial Fractions


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About this Lesson

  • Type: Video Tutorial
  • Length: 12:23
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 132 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

College Algebra Review (30 lessons, $59.40)

In this lesson, you will learn about partial fractions and the partial fraction decomposition of a rational function. If you can use partial fractions, you can break up fractions with complex denominators into component parts. To break something into partial fractions, you’ll start by factoring the denominator. In the end, you’ll be able to come up with the partial fraction decomposition of a complicated rational fraction. Understanding this technique is invaluable in calculus.

This lesson is perfect for review for a CLEP test, mid-term, final, summer school, or personal growth!

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.

About this Author

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You know in the advanced topics of calculus partial fractions really do play an important role so it's kind of fun to think about them now and to kind of see the pattern. The idea of partial fractions of course is that you take a fraction and what you want to do is you want to write it as the sum of simpler fractions in a way, kind of again this theme of divide and conquer. So for example if I take a look at this fraction (4x+1) all divided by x times the quantity (x+1)2 well we have already seen how to deal with things like that but now I have to repeat it linear factor right x+1 is linear and I have got it appearing twice so it's a cool little trick we can do to kind of break this apart. I know a cool trick, the secret, the secret is since it's a linear factor but it's squared we are going to repeat that linear factor once so let me show you how we are going to consider. We are going to consider this as some constant divided by x plus some constant divided by x+1. And here is the sneaky part, to get that squared in there we are going to introduce another constant divided by the quantity (x+1)2. Might be tempted just to do this, avoid that temptation. Put in the actual linear factor and then asset return with the linear factors squared and you are going to be good to go. Everything else is now going to be the same as in the other examples that we have considered. So in particular I have got to get a common denominator and then kind of watch what happens to the coefficients and the coefficient on the x is going to equal 4, the constant coefficient is going to be 1 and then we are good to go. So what do I do here? Here I have to multiply the numerator and denominator by (x+1)2 to get the common bottom. Here I have got to multiply the numerator and denominator by x times the quantity x+1 because I need an extra factor of x+1 to top it of to be a square and then I need an x over x as well. Here all I have to do is multiply the numerator and denominator by x and I am good to go. If I were to write that out which I am going to do right now it would look like this. Check it out, a(x+1)2 all over x(x+1)2 that’s just a/x + b(x(x+1)) all over x(x+1)2 really? Yeah because if you were to simplify you just get the b over x+1 so that’s perfect. And then finally last but definitely not least is my cx all divided by x(x+1)2. So there you have it boom, now what do I have to do I have to untangle all of this. I have the common denominator so I am going to write it all x(x+1)2 but now I have got to do foiling and distributing “a” and the thing you know. So I am going to do it and I am going to do it my way. So if I square this out that’s going to be x2 +2x +1 to multiply it all by a so I see ax2+2ax+a, so I did all that in just one step. Now here I have got to distribute the bx to the x and the bx to the 1 so I see a plus that’s this plus right here plus bx2+bx. And then don’t forget I have to add the plus cx and there I have it so that’s how we are great. Now of course you want to combine like terms so we want to combine like terms because that is the mantra that follows us through algebra. So let's do the x2 terms, I see x2 here and here and here so I have ax2 + bx2 if I factor out the x2 I see (a+b)x2. So in effect if you want you may like to keep track, you can like check them off. Chekhov was a famous writer and he inspired this method. Now what about the xs? Well here is an x by itself, here is an x by itself, here is an x by itself I can factor out that x and what do I see? I see 2a+b+c and that’s all times x so check, check, check. Anything unchecked? Yes this term right here and that’s it, it's a constant term just plus a. So I stick that on and that’s all over the common bottom, notice I keep writing the common bottom and you should too. Now here is the exciting point because you see all of these are equal signs, which means that if I scan this up then what I see is that this original thing is identical to this. This function equals that function and that of course gives me the rules and these are the important things, these are the rules of the road. Because I know that whatever is multiplying the x2 here has to be same thing multiplying this x2. Whatever is multiplying the x same thing here, whatever is multiplying the constants, so what do we see? What I see here a+b well there is no x2 term here, since there is no x2 term that means that a+b has to be the coefficient=0. And then what about this coefficient 2a+b+c well that has to be the coefficient multiplying the x term which has to equal 4. And finally the constant plus 1 has to be the same thing as the a and so a=1. So if I make a=1 2a+b+c=4 and a+b=0 these two things really are identical. Now of course this is real easy because this tells me that a=1 when I combined it with this I see that b is going to be the opposite of a so b=(-1) and then if I plug all that into here what do I see? I see that 2(1) which is 2+b which is (-1) plus c = 4 which means 1+c=4 and even I can do this c has to be 3 so therefore this implies that c has to equal 3. And now I found a, b and c and if I go back way back, way back when we were young and our whole mathematical lives were in front of us before we got all jaded to here I now see these values of a, b, c. I see that a has to equal 1 which means that this equals 1 over x. And what is b? Well b is going to be (-1) so I could write +(-1) or I can just write -1/x+1 either is correct. And then what about c? c is 3 so I have 3 over x+1 all squared and I have just written this as a sum of fractions, the partial fraction decomposition. Now I know you are saying, you are saying okay that’s enough. No it's never enough so I want to do one more for you and I think you are going to enjoy this one because this one is a little bit different and I just want to show you how to deal with this so we can viva. Now the first thing we tried to do of course is try to factor the denominator so if we factor the denominator what would we see? Factor the denominator we would see the following. Keep the numerator as it is, don’t touch the numerator. The only thing I see that I can do is factor out the common factor of x do you see anything else because I don’t. I factor out the x I see an x2+9 this can't be factored anymore over the real numbers and so that is called irreducible can't break it up anymore. This is not a linear factor. I don’t have a linear factor to a power like we saw in the previous example. This is really a quadratic factor. So how do we handle a quadratic factor if you want to write it as a partial quotient? What I do hear is I am going to break this thing up as a/x and instead of writing b see I just can't write because this is a quadratic so I got to put kind of 1 polynomial degree smaller which would in this case be a linear. So I got to put in a bx+c all over this. Now let's just think about this for a second. If you have a linear thing what's one degree small than a linear it's constant. What's one degree smaller than a quadratic it's linear and that’s the pattern here. Now everything else is the same we have to get a common denominator and we combined so I have to multiply top and bottom here by x2+9 and then I multiply top and bottom here by just x and so I see the common denominator which is awesome and when I combined what do I see? Well what I see is the following fantastic, fantastic looking thing. I love this because what I see is ax2+9a because I have to multiply that by a plus and then b(x2+cx) and it's all divided by the common denominator. And now I again combine like terms so what do I see? I see this equals (a+b)x2 and then I have got a plus cx+9a and that’s all over my common bottom. And what do I know? I know that this has to equal this so I know that this coefficient has to be 2, this coefficient has to be negative 1, this coefficient has to 9 and that gives me the ammo to get this going. In fact I see that 9a constant has to equal the constant and that immediately tells me that a=1. And then what do I know? I know that c this coefficient has to equal that so c=-1. And then how do I find out what b is? Well a+b is going to be the coefficient on x2 which is 2. I know that a is 1 that forces b to be 1. And so I see exactly what this coefficients are so I come back, I got to go back to my little grey shack – this equals well a is 1 so I just have 1/x plus and the b is remember 1 so it's just going to be x and the c is negative 1 minus 1 all over x2+9. And there I have the partial fraction decomposition for this crazy looking rational function. Now if you don’t believe me that’s fine, good for you if you are not trusting me, I wouldn’t trust you either but you can actually see if this is correct by getting a common denominator and combining everything. And when you combine the numerator I predict you are going to see 2x2-x+9 and the denominator will be the common denominator of x times x2+9 which of course is x3+9x. So you can even break up complicated looking rational functions that even have irreducible quadratics down there. All you are going to do is put a linear in this case bx+c in the numerator and you are good to go. Congratulations on conquering partial fraction so important, so great when you take calculus you are going to thank me and I will be there. I will see you soon.

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