In Exercises 16–24, write the partial fraction decomposition of e...

Question

In Exercises 16–24, write the partial fraction decomposition of each rational expression. (4x^3 + 5x^2 + 7x - 1)/(x^2 + x + 1)^2

Accepted Answer

Welcome back. I'm so glad you're here. We have got this rational expression here. Two X cubed minus four X squared minus three X plus one divided by X squared plus two X plus three. And that whole denominator is squared. We want to decompose this. So let's see what's going on in the denominator. Can we factor this anymore? We've got to do X X plus three. Okay. No we can't because they're both positive and they're not going to equal positive to when they add together or multiply to get three there so we can't factor that anymore meaning it is prime. We also know that it is a quadratic because of that too. And we know that it is a repeated prime quadratic because it's squared. So when we have this form we're going to set this equal to a X plus B over our denominator to the first power. So X squared plus two X plus three plus C. X plus D. Over our denominator as it is X squared plus two X plus three. All squared. So that's the form we're going to take and now we're just solving for a B. C. And D. Easy. Right Well there's there's a lot of steps but that's what we're doing. We're solving for A B C. And D. So the first thing we're going to do to solve for those is we're going to multiply everything by the common denominator which in this case is going to be X squared plus two X plus three. All squared. And for that first one it's just gonna um cancel out everything that's being multiplied by the top. So we'll just have to X cubed minus four X squared minus three X plus one. And then on the right hand side that first term X plus B. Well since there's only one X squared plus two X plus three in the denominator it's still going to get multiplied by another one on top. So X plus B. Times X squared plus two X plus three but only one of them. And then plus R. C. X plus D. Well that X squared plus two X plus three. All squared in the denominator is going to cancel everything that's being multiplied on the numerator. So great. We got rid of our fractions. Now let's distribute over here we can start distributing that A. X. So we've got a X cubed and then times that middle term plus two X squared. And times that third term plus three A. X. Now let's distribute the B. B times the X squared. So plus be X squared times the middle term plus to be X. And times that third term plus three B. Bring down the plus C. X. And the plus D. Will bring down the left hand side two X cubed minus four X squared minus three X. Plus one. Now on the right hand side let's put them in descending order of our X terms. So the A X cubed is the only X cubed plus R X squared terms. We've got to A X squared plus B, x squared. Now our X terms, we've got three of them plus three, A X plus to be X plus C. X. And now our constant terms plus three B plus D. And now let's factor out our X terms where we can we can't with the first one, A x cubed is the only X cubed but we can factor out our X squared. And that leaves us with two A plus B. Inside the parentheses, we can factor out an X. And that leaves us with three A plus two, B plus C. And then this one is like factoring out of one, it's just the three B plus D. Here and that's all still equal to two X cubed minus four, X squared minus three X plus one. And now we know that coefficients on the left and the right hand side are going to be the same because we've got the same degrees here, we've got an X cubed and X cubed and X squared and x squared and X and X and a constant. So the coefficients of those terms. Since these are equal, they're going to be the same thing. So we can set our x cubed coefficients to be equal. So coefficient of this one is 22 equals coefficient of this one is A. And I'm going to rewrite that with a on the left and the constant term on the right. But now we've got our a term, we've solved a equals two. Now let's look at the coefficients of X squared. Well x squared that's going to be negative four. I'm going to write that on the right hand side negative four equals. And the coefficient of the X squared term over here is to a plus B. It's a little different because the coefficient is on the right but it is still the coefficient of X squared. We'll leave that there for a moment. Our coefficient of our x terms well -3 and it was the longer one got three A plus two. B plus. I'm just going to scoot this over a little plus C equals negative three. And this one up here I'm going to scoot over just a little bit. So they line up That one equals -4. And now our constant terms, We've got a one equals three, B plus D. Well there's another one even so three B plus D equals one. And now we've already got our a term solved. So this isn't too bad. We can start plugging that in and solving for R. B terms. So we know we've got two times two plus B equals negative four. And so that's four plus B equals negative four and that'll give us B equals -8. Got two of them. We've got an A term and a B term. So we can look at that purple equation and start filling that in. So three times a day or two Plus two times our b. term which is negative eight plus C equals negative three. Right so we got six minus 16 plus C equals negative three. In other words negative 10 plus C equals negative three. Or see equals seven. And now Make that Clear C. Equal seven. Now we just gotta sell for R. D. And we know our b. Term. We can fill that in up here three times negative eight plus D. Equals one. Okay in other words that's negative 24 plus D. Equals one, Add 20 affordable sides. And we've got d. 25. We have solved for our A. B. C. And D. And now now we just fill them in. So looking back up here A instead of a we've got two X plus B. B. Is negative eight so two X minus eight C. Is going to be seven. So seven X plus D. Is 25. So in other words we've got two x minus eight over our X squared plus two X plus three plus seven X plus 25 over our X squared plus two X plus three squared. We're going to look at our answer choices and this matches with answer choice. A Well done we'll catch you on the next one

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In Exercises 16–24, write the partial fraction decomposition of each rational expression. (4x^3 + 5x^2 + 7x - 1)/(x^2 + x + 1)^2