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^3+x^2+2/(x² + 2)²

Accepted Answer

Hello. Today we're going to be performing partial fraction decomposition on the given expression. So the expression given to us is y cubed plus two, Y squared plus Y plus 12 over the quantity Y squared plus six. That quantity squared. Now, before we perform our partial fraction decomposition, we always want to make sure that our denominator is fully factored. Why squared plus six is not a fact arable quantity, but it is a quadratic quantity because the leading coefficient is raised to the power of two and this quadratic quantity is raised to the power of two. So what we have in the denominator is a repeating quadratic quantity and this is going to produce two partial fractions. The first one being a Y plus B over Y squared plus six plus C, Y plus D over Y squared plus six squared. Now, what we want to go ahead and do is solve for a B C and D. But in order to do that, we need to find a way to get rid of all of these denominators first. Well, in order to do that, we can go ahead and multiply by the lowest common denominator and in this case here the lowest common denominator is going to be y squared plus six squared. So again, we're gonna multiply every value here in this equation by the L C. D. So let's just go ahead and write that out. We have y squared plus two Y. I'm sorry, Y cubed plus two Y squared plus Y plus over Y squared plus six squared. And we're multiplying that by the L C. D. Which is Y squared plus six squared. And this is equal to our first partial fraction which is A. Y plus B. Over Y squared plus six, multiplied by the L. C. D. Which is Y squared plus six squared over one plus our second partial fraction which is C. Y plus D. Over Y squared plus six squared, multiplied by R. L. C. D. Which is Y squared plus six squared over one. Now multiplying the left hand side by the L. C. D. Is going to completely cancel the denominator and on the right hand side, multiplying the first partial fraction by the L. C. D. Is going to cancel our denominator but leave one instance of Y squared plus six in the numerator. And multiplying the second partial fraction by L. C D. Is going to completely cancel the denominator. So what we are left with is Y cubed plus two, Y squared plus Y plus 12 equal to the quantity A Y plus B. Times Y squared plus six plus C. Y plus D. Now what we want to go ahead and do is simplify the right hand side. And in order to do that we want to go ahead and foil a Y plus B with y squared plus six. Doing so is going to leave us with Y cubed plus two Y squared plus Y plus 12 equal to a Y cubed Plus six A. Y plus B. Y squared plus six B plus C. Y plus D. Now this is starting to look like a jumble of terms here. But what we want to go ahead and do is group up all of our y cubed terms. Why square terms why terms and constant terms together. And so doing this is going to give us y cubed plus two, Y squared plus Y plus 12 equal to a Y cubed plus B, Y squared Plus six A. Y plus C. Y Plus our constants which is six B plus d. Now what we want to go ahead and do is factor out a common factor from any of our given groups in six A Y plus C. Y. We have a common factor of Y. So we can go ahead and factor out why. And doing this is going to give us y cubed plus two, Y squared plus Y plus 12 equal to a y cubed plus B, Y squared Plus Y, Times six A Plus C plus six B plus D. And now we're gonna use this equation to set up a system of equations so that we can solve for a B C and D. Well how do we do that exactly what we want to go ahead and take the coefficients of y cubed and equate them with each other. We want to take the coefficients of Y squared and set them equal to each other. We're going to take the coefficients of Y and set it equal to each other. And we're taking the constants and sending them equal to each other. So here on the right hand side, the coefficient of y cubed is a. And on the left hand side the coefficient of y cubed is one because one times y cubed gives us y cubed. So our first equation is going to be A is equal to one. And that solves one of the coefficients that we need now for a wide square terms. The coefficient of y squared on the right hand side is B. And on the left hand side the coefficient is two. So our second equation is going to be B is equal to two which solves another one of our coefficients that we need for the partial fractions. Now our third equation is going to be the coefficients of Y, which on the right hand side is six A plus C. And on the left hand side of the coefficient of y again is just one. So we end up with the third equation which is six A plus C is equal to one. And finally we want to go ahead and set our constants equal to each other. So the fourth equation is going to be six, B plus D is equal to 12. Now you want to go ahead and use these equations to solve for the remaining coefficients and by this point we should be comfortable with solving a system of equations. So if you solve the system of equations, you get the coefficients is equal to one. B is equal to two, C is equal to negative five and D is equal to zero. And now what you want to go ahead and do is replace all the coefficients from our original partial fraction with the coefficients that we just solved for. So again, our original coefficients are is A Y plus B over Y squared plus six plus C, Y plus D over Y squared plus six squared. We know that A is equal to one, so they can be replaced with one, which just gives us Y B is equal to two, so B is going to be too, so we have Y plus two, C is equal to negative five. So we have negative five Y and D is equal to zero. So we can go ahead and replace D with zero. Now for the right hand, partial fraction, since we have negative five Y plus zero, that's just going to give us negative five Y. And if we move the negative to the front of the partial fraction, we can rewrite the second partial fraction as negative five, Y over y squared plus six squared. And this is going to be the complete partial fraction decomposition for a given expression. And that means the answer to this problem is going to be a. So I hope this video helps you in understanding how to do 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^3+x^2+2/(x² + 2)²

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Polynomial Factorization

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