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. 5x^2 -6x+7/(x − 1) (x² + 1)

Accepted Answer

Hello. Today we're going to be performing partial fraction decomposition on the given expression. So what we are given is nine A squared minus A plus 57 over A -3 times a squared plus nine. Now, before we perform partial fraction decomposition, we want to make sure that our denominator is fully factored A minus three cannot be simplified any further and a squared plus nine is not factory bubble. So the denominator is fully factored. Now, let's take a look at each term and see what kind of partial fraction they produce. Eight minus three is a linear quantity as the leading coefficient has the exponent of one. So it's going to produce the partial fraction of a over a -3 And a square plus nine is a quadratic quantity as the leading coefficient has an exponent of two. So this produces the partial fraction B. A plus C Over a squared plus nine. Now, what we want to go ahead and do is sulfur A, B and C. However, in order to do that, we need to get rid of all these denominators. In order to get rid of the denominators, we can go ahead and multiply by the lowest common denominator and the L. C. D. In this case is going to be a minus three times a squared plus nine. So let's go ahead and just write this out. We have 90 squared -10 a plus 57 over a - times a squared plus nine times the L C. D. Which is a minus three times a squared plus 9/1. And that is equal to our first partial fraction which is a over a -3 times a minus three times a squared plus 9/1 plus our second partial fraction which is B. A plus C. Over a squared plus nine. Multiplied by the L. C. D. Which is a minus three times a squared plus nine. All of that over one. Now multiplying the left hand side by the L. C. D. Is going to completely cancel the denominator on the right hand side, multiplying the first partial fraction by the L. C. D. Is going to cancel out A minus three. Believe us with a square plus nine in the numerator. And on the second partial fraction multiplying by the L. C. D. Is going to cancel out the eight square plus nine but it's going to leave us with the quantity a minus three in the numerator and this gives us nine A squared -10 A plus 57 is equal to a times a square plus nine. Mhm Plus B. A Plus C. Times A -3. And now what we want to go ahead and do is simplify the right hand side in order to do that. We can go ahead and distribute a into a square plus nine and foil B. A plus C. With a minus three and doing this is going to give us nine A squared minus 10 A plus 57 is equal to a times a squared plus nine A plus B, A squared -3 BA Plus CA -3 C. And now what we want to go ahead and do is group up all of our like terms together. So we're gonna group up all of our a squared terms together, all of our eight terms together and all of our constants together. And doing this is going to give us nine A squared minus 10 A plus 57 is equal to a times a squared plus B. A squared plus negative three B A plus C A plus nine, eight minus three C. And now what we want to go ahead and do is just factor out any common factors. In our first group we have a squared as a common factor and in our second group we have a as a common factor. So factoring those out is going to give us nine A squared minus 10 A plus 57 is equal to a squared times a plus B plus a, times negative three B plus C Plus our constants, which are nine a -3 c. And now using this we can go ahead and create a system of equations so that we can be able to solve for a B and C. So how do we do that though? We want to go ahead and set the coefficients of our a square terms equal to each other. So on the right hand side the coefficients of a square term is going to be a plus B. And on the left hand side the coefficient of a squared is going to be nine. So what we have is A plus B is equal to nine. For our second equation, we're saying the coefficients are eight terms equal to each other. So on the right hand side the coefficients are negative three B plus C. And on the left hand side we have negative 10 as the coefficient of A. So what we have is negative three, B plus C equal to negative 10. And for a third equation, we're just setting all the constants on the right hand side, equal to the constant on the left hand side and that's going to give us nine a minus three, C is equal to 57. And now you're gonna want to solve the system of equations in order to get our A B and c coefficients. And by this point we should be comfortable with solving a system of equations. Once you solve the system of equations you should get the coefficients A is equal to six. B is equal to three and C is equal to negative one. And now, what you want to go ahead and do is take these values and plug them into our original partial fraction decomposition. Now recall that the partial fraction decomposition was a over a minus three plus B. A plus C over a squared plus nine. So replacing a with six is going to give us six over a -3. Replacing B with three is going to give us three A plus C. And replacing C with -1 is going to give us three a -1. And this is going to be the complete partial fraction decomposition of our given expression. And that means that the answer to this problem is going to be a. 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. 5x^2 -6x+7/(x − 1) (x² + 1)

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