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. (7x^2 - 7x + 23)/(x - 3)(x^2 + 4)

Accepted Answer

Welcome back. I'm so glad you're here. We have got this rational expression. We've got X squared plus two X minus divided by X plus two times X squared plus one. And we are going to decompose that, let's take a look at the denominator and that tells us what um form we're going to use. The denominator has a linear term here, the X plus two. And it's being multiplied by a quadratic X squared plus one. And that is prime. We can't factor that down anymore. So when we have this form, we're going to set this equal to a over our linear term which is X plus two plus bx plus C over R. Prime quadratic which is X squared plus one. Now let's get rid of our fractions. You multiply everything by the common denominator which is going to be X plus two times X squared plus one. And on the left hand side that denominator will cancel out. The two factors were multiplying it by so we'll just have X squared plus two, X minus 10. On the left, on the right are first term A over X plus two. Well, the X plus two's will cancel out so that they will just get multiplied by X squared plus one in the third term, the bx plus C. While the denominator X squared plus one will cancel out with the X squared plus one. And that B X plus C will just get multiplied by the X plus two. Now on the right hand side we can distribute. So we got a X squared plus A plus. Now we'll foil this be X squared plus to be X plus c, X plus two C. And on the left hand side will bring down the x squared plus two x minus 10. On the right hand side. Let's write those in descending order of our x values. So we'll group them with a X squared plus B, x squared. And next we'll have plus two, B, X or X terms plus C X. And now our constant terms plus A plus to see and bring down our left hand side, X squared plus two X minus 10. And now on the right hand side let's factor out our X term. So X squared times A plus B plus x, times two, B plus C plus. And this is just a plus two C equal to our left hand side. And now we know that coefficients, we've got x squared terms, we've got X terms and we've got constant terms. The coefficients are going to be equal so we can set them equal to each other. So we know this coefficient here is one, I'm going to write that on the right hand side and the coefficient here is a plus B. I'm going to write that on the left hand side. So we know that A plus B equals one for our x squared terms. Looking at the coefficients over X terms over here, it's two And over here it's two B plus c. And we've got our constant terms are going to be equal to each other, so negative 10 equals A plus two C. Now we've got a system of equations and it might help if we just rewrite this with a little bit of space so that our terms line up, there's no C term in that first one. And over here we'll just scoot this a term over and now we can line them up. We're going to try to get um solve for R A's or B's and our seas and we can use addition for this. We can multiply this one by negative one and then add it to our first term. So A plus negative A will cancel out the B will just carry down because there's no B term on that bottom one plus. And then we've got a negative to see now And negative one times negative, 10 is going to be positive. 10 plus are one of their is going to be 11. So now we have two equations with B and C terms and we can um add those together. Let's cancel out our C term. So let's multiply this one times two. So two times to be, that's going to be four, B plus B is five B, two times C. S, two, C minus two C. That cancels out and two times two is four plus 11 is 15. So we've got 5B equals 15. Or in other words we know that B Equals three. We've got one of them. Now we can plug that back into our first equation. That be the easiest one. We've got A plus three equals one And we have got a equals -2. Oh we're almost there. Now let's solve for C. We can plug it into our third equation, we can say that negative two plus two. C equals negative 10, add tunable sides. So we've got two C equals negative eight. Or in other words C equals negative four. Great. We've got our A. B. And C terms and now it's just a matter of plugging those back in up here. So we're going to have a negative two over X plus two. So negative two over X plus two. The B term is three times X plus C. That's going to be a minus four. So negative two over X plus two plus three X minus four over X squared plus one. We look at our answer choices and we have to look very carefully because none of them start with a negative two. So let's start with that second term. Okay three X minus four over X squared plus one minus two over X plus two. That's going to match with answer choice B. Well done. We'll catch you on the next one

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