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. (2x^2 -18x -12)/x³- 4x

Accepted Answer

Welcome back. I am so glad you're here. We are given the expression to be squared minus B minus six, divided by b cubed minus B. We are asked to express this in its partial fraction decomposition. Alright, We'll need to recall quite a bit from previous lessons on partial fraction decomposition. And the first thing we'll do is take a look at the denominator. We notice the denominator is not factored yet. So that's our first step. I'll copy over the numerator and then we can factor a B out from both terms in our denominator. That will leave us with B times B squared minus we can recognize B squared minus 36 as a difference of squares. So recalling previous lessons on a difference of squares, we know that we can set two sets of parentheses for our factors. Then we're going to take the square root of both of those terms. The square root of b squared is B. So B goes in the first position in both sets of parentheses and take the square of 36. That is 66 goes in the second position in both sets of parentheses. And then by definition one is plus one is minus all right. That is fully factored and I apologize if my bees look like sixes, I will try to say it every time. So now the next things we ask ourselves about the factors and our denominator is first. Are these linear or are they quadratic? And each of those B's in all three factors, they have a degree of one. So all three of those are linear. Next we ask ourselves if these are distinct or repeating factors and those are all different from each other. They are all distinct. And the last question we ask ourselves is, how many are there? Well there are three of them. So now we're going to set this equal to three fractions all added together. And then working on our denominators first, our denominators are going to be the factors of our previous fractions. So the first fractions denominator is that factor B. The second fractions denominator is B plus six. And the third fractions denominator is B minus six. And then our numerator because all three factors and our denominators are linear. Our numerator czar all going to be constant terms. So the first fractions numerator will be a, the second fractions numerator will be a capital B. And the third fractions numerator will be a capital C. Alright, we have set that up and now what we need to do is solve for a B and C. And then we're going to come back up here and plug in our values for a B and C. And that will be our answer for partial fraction decomposition. So in order to solve for a B and C, we're going to need to multiply everything by the common denominator. The common denominator is what we pull from the left hand equation. So that's going to be B times B plus six times B minus six. That has to get multiplied by all four of those enumerators. So I will start to write that we have to be squared minus b minus six times B times B plus six times b minus six. All over B times B plus six times b minus six equals A times B times B plus six times B minus six over. Just little B plus Big B times Little B times B plus six times B minus six. All over, B plus six plus C times little B times B plus six times B minus six. And that is over B minus six. And now the glorious part canceling out factors that occur in both the numerator and the denominator. The first fraction on the left hand side, there's a B B plus six and a B minus six. The first fraction on the right hand side. Just the little B's cancel out the middle fraction on the right hand side. Just the B plus six Factors cancel out. And the third fraction on the right hand side. The B -6 factors cancel out now. Rewriting this with what we've got left, starting with two B squared minus B minus six equals a big capital A times parentheses B plus six parentheses, B minus six plus Big B times Little B times B minus six plus capital C times little B times B plus six. Right now let's do distribution where we can we're going to multiply these factors together Where we have parentheses to get rid of our parentheses. So I'll copy down the two B squared minus B minus six equals. I'm just going to keep the A. For now and we know from what we did with difference of squares, that B plus six times B minus six equals that difference of squares, B squared minus 36 that we had up in our original equation. And then plus Big B times. Now distributing the little B to the B minus six will be sorry B times B is B squared and B times six is minus six B plus and again distributing that little bee keeping the big C. There. So little B times Little B is B squared plus B times six is six B. Okay, now we can distribute our big letters. So I'll copy down the left hand side again and we've got a times B squared. So that's a B squared minus A. Times 36 is so minus 36. A. Now plus Big B times little B squared is B B squared minus Big B. Times little six B is minus six. B B. Little B big B and then plus C. Times B squared is C. B squared plus six B times c. six BC. All right. Now, what we've got to do is combined, we want to regroup these so that we've got our B squared together. So we've got an A B squared, A BB squared and a CB squared. We also want to get our little bees together so that's a minus six BB and A plus six B. C. And our constant term minus 36 A. So we're going to rewrite these and group things together so copy down the left hand side again to b squared minus b minus six. I promise. We will use that soon. And now with the B squared. So that will be an A B squared minus sorry? Plus big B. Little B squared plus C. B squared. Now our little bees will be minus six B. Little B. Big B plus six little B big C. In our constant terms minus 36 A. Okay, we've grouped things together and for our B squared we'll be able to factor out a B squared. And for our little bees we can factor out that little B. So now we'll rewrite that. Bringing down the left hand side to B squared minus B minus six equals factoring out our b squared, B squared times A plus B plus C. Plus factoring out our little B. That will be a minus little B times minus six. Big B plus six C. And then bring down the -36 a. Okay, here's where again we recall from previous lessons on partial fraction decomposition that the coefficient of our terms on either side of the equation are equal to each other. So this B squared. On the left hand side has a coefficient of two. And this b squared on the right hand side has a coefficient of a plus B plus C. Those are equal to each other. I'm gonna bring this down a little bit. So we've got a plus B plus C from the right hand sides, B squared coefficient equals two from the left hand sides, B squared coefficient. Now the coefficients for regular B, regular B on the right hand side has the coefficient of negative six, B plus six. C plus six. E. Is equal to the coefficient of B on the left hand side, which is negative one. So those are equal to each other And then our constant terms on the left hand side are constant term is -6. And on the right hand side the constant term is - a. And we've just come up with three equations. This system of equations can help us solve for A B and C. Then we can plug those back in and we have our partial fraction decomposition. Alright, we're almost there. Now, the easiest thing to solve for with this is for that, eh we'll divide both sides by a negative and that will give us a equals negative six divided by negative 36. That will simplify to a positive 1/6. So A equals 1/6. I'm going to write that over here so that when we can when we scroll up we can still see it and that. Well I mean that helps a bit but it's going to be a little bit a few more steps before we can solve for B and C. So that first equation here, A plus B plus C equals two. We are going to rewrite that filling in for A. So A is 1/6 so we can put 1/ plus B plus C equals two and then we will subtract 1/ from both sides And that will give us a new equation, one of b plus c equals 11/6. So that's our first equation and I'm going to copy over our second equation which was negative six B plus positive six C equals negative one. So this system of equations has two variables. Let's do some work on the first equation and then we can add the two together. Let's eliminate the B variable. So we can solve for C. And I'm going to multiply everything in the first equation by six. And so six times B is six, B. Six times C is plus six C equals and 11. 6 times six is 11. And we can add those two equations together. Oops that plus goes right here Alright adding these together the beast cancel out and we have 12 c equals 10. And to solve for C, Divide both sides by 12 and we get C equals 10. 12 is 5/6. So I'll copy that down up here so we can see it when we scroll up. C. Equals 5/6. Alright we've got two of them now we can plug C. N. To our equation. Let's plug it into this first equation here where we have B. Plus C. Equals 11 6th. So that'll be B plus Instead of C. It'll be 5/6 And that equals 11 6th will subtract 5 from both sides And 11/6 -5/ equals 6/6 or one. So b equals one. Okay now back up here where we can see our answer choices. Okay I'm going to go back to where we have our partial fraction decomposition set up here and I'm going to fill in A. B. And C. There. So A. Is 1/6. So it's 1/6 divided by B. Plus big B. Is just one so one divided by B. Plus six plus big C. Is 5/6. So 5 6 divided by B -6. And we look at our answer choices and they've got this answer, they've just switched the order around for our b minus six denominator and R. B. Plus six denominator. And this matches with answer choice D. Well done, you've done it we'll catch you on the next one

In Exercises 9–42, write the partial fraction decomposition of each rational expression. (2x^2 -18x -12)/x³- 4x

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