In Exercises 9–42, write the partial fraction decomposition of ea...

Question

In Exercises 9–42, write the partial fraction decomposition of each rational expression. x^2/(x − 1)² (x + 1)

Accepted Answer

Hello. Today we're going to be performing partial fraction decomposition on the given expression. So the expression given to us is four Y squared Over. Why -2 squared Times Y Plus two. Now, before we start our partial fraction decomposition, we always want to check to make sure that our denominator is fully factored. The quantity y minus two can no longer be simplified. So that's in its simplest form and Y plus two cannot be factored. So it's in its simplest form. So our denominator is currently in its fully factored form. Now, let's go ahead and see what kind of partial fractions the factors produce. Why -2 is a linear quantity as the leading coefficient has an exponent of one, but why -2 is repeated two times. So this quantity why -2 squared is going to produce two partial fractions. It's going to give us a over y minus two plus B over y minus two squared and y plus two is also a linear quantity as the leading coefficient has an exponent of one but it's not repeated. So it's only going to produce one more partial fraction and that partial fraction is going to be C over Y plus two. Now we want to go ahead and solve our A. B and C. However, in order to do that, we want to go ahead and get rid of all of our denominators in order to get rid of our denominators, we can multiply by the lowest common denominator. And in this case here the L C D. Is going to be y minus two squared times Y plus two. So we're gonna want to multiply everything by the L. C. D. So let's just quickly write that out. We have four Y squared over y minus two squared times Y plus two times the L. C. D. Which is why minus two squared times Y plus two all over one. And that is equal to our first partial fraction which is a over y minus two times the L. C. D. Which is why minus two squared times Y plus 2/1 plus our second partial fraction which is B over y minus two squared times the L. C. D. Which is why minus two squared times Y plus 2/ Plus our 3rd partial fraction which is C over Y plus two Times Y -2 Squared times Y plus 2/1. Now multiplying the first quantum on the left hand side is going to completely get rid of the denominator for our first partial fraction. Multiplying this by the L. C. D. Is going to get rid of the denominator but still leave us with both quantities of y minus two and Y plus two, multiplying our second partial fraction by the L C. D. Is going to completely get rid of the denominator but leave us with y plus two in the numerator. And multiplying the last partial fraction is going to get rid of the denominator. Believe us with y minus two squared in the numerator. So simplifying this is going to give us four Y squared equal to a times y minus two Times Y-plus two plus B. Times Y-plus two Plus C Times Y -2 Squared. Now we need to go ahead and simplify the right hand side. In order to do that. We can first start by distributing A into y minus two, distributing B into Y plus two And expanding why -2 squared. And this is going to give us four. Y squared is equal to the quantity A y minus two A Times Y-plus two Plus B, Y Plus two, B plus C. Times Y squared -4 Y Plus four. And we want to continue simplifying this. We're going to foil A y minus two A. With y plus two. And we're going to distribute C into y squared minus four Y plus four. That's going to give us four, Y squared is equal to a Y squared Plus two. AY -2. AY minus four A Plus B. Y Plus two, B plus C, Y squared minus four C. Y plus four C. Now notice that we have plus two A. Y and minus two A. Y. So those terms are going to cancel with each other. Now this looks like a mess of terms. However, let's go ahead and group up our y squared terms together are white terms together and are constants together and doing this is going to give us four Y squared is equal to a Y squared plus C, Y squared plus B, Y -4 C, Y plus our constants, which is negative for a plus to be plus four C. And now what we can go ahead and do is factor out a Y squared in our first group and a Y in our second group. And that's going to give us four, Y squared is equal to why squared times the quantity A plus C plus Y times the quantity B minus four C plus our constants, which is negative for a plus two, B plus four C. And now we can go ahead and use this equation to set up a system of equations. So that way we can sulfur A B and C. So what we want to go ahead and do is that the coefficients of our white square terms equal to each other, the coefficients of our white terms equal to each other and our constant equal to each other. But notice that on the left hand side we don't have a white term and we don't have a constant, but we can go ahead and add those in because those coefficients are just going to be zero. So we can change the left hand side to B four, y squared plus zero, Y plus zero. And now we can go ahead and create a system of equations. Our first equation is going to be the coefficients of y squared set equal to each other on the right hand side, the coefficient of y squared is a plus C. And on the left hand side it is four. So what we have is a plus C is equal to four. For a second equation, we're going to set the coefficients of Y is equal to each other, so be -4. C is going to equal to zero and finally we're going to set the group of constants equal to each other and that equation is going to be negative for a plus two, B plus four, C is equal to zero. And now you can go ahead and use this system of equations to solve for your missing coefficients. Now we should be comfortable with solving a system of equations at this point. And if you solve the system of the equations, you should get a sequel to three, B is equal to four and C is equal to one. And now you want to go ahead and plug that into your original partial fraction decomposition. Recall that the original partial fractions was a over y minus two plus B. Over y minus two squared plus C over y plus two, replacing a with three is going to give us three over y minus two, replacing B is going to give us four over y minus two squared and replacing C is going to give us one over y plus two and this is going to be the complete partial fraction decomposition for our given expression. And that means that the answer to this problem is going to be be so. I hope this video helps you in understanding how to perform partial fraction decomposition, and I'll go ahead and see you all in the next video.

In Exercises 9–42, write the partial fraction decomposition of each rational expression. x^2/(x − 1)² (x + 1)

Watch next