Find the partial fraction decomposition for each rational express...

Question

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (x^2)/(x^4 - 1)

Accepted Answer

Welcome back. I am so glad you're here. We are asked to decompose the given rational function into partial fractions. Our given rational function is in the numerator, we have three T squared that's divided by, in the denominator seven multiplied by the quantity of T race to the fourth power minus one. Our answer choices are answer choice A three divided by 28 multiplied by the quantity of T plus one plus three divided by 28 multiplied by the quantity of T minus one plus three divided by 14 multiplied by the quantity of T squared plus one answer. Choice B negative three divided by 28 multiplied by the quantity of T plus one plus three divided by 28 multiplied by the quantity of T minus one plus three divided by 14 multiplied by the quantity of T squared plus one answer. Choice C negative three divided by 28 multiplied by the quantity of T plus one plus three divided by 14 multiplied by the quantity of T minus one plus three divided by 14 multiplied by the quantity of T squared plus one. And answers are traced D negative three divided by 14 multiplied by the quantity of T plus one plus three divided by 28 multiplied by the quantity of T minus one plus three, divided by 14, multiplied by the quantity of T squared plus one. All right. The first thing we ask ourselves when decomposing into partial fractions is the degree of our numerator less than the degree of our denominator. Or in other words, do we have a proper fraction? You, the degree of the numerator is two. The degree of the denominator is four and two is less than four. So yes, this is a proper fraction and we may proceed. The next thing we ask ourselves is is our denominator fully factored. And in this case, it's not. So we recognize that T race to the fourth minus one is a difference of squares. I'm going to factor this very quickly if you need a review of factoring, especially factoring differences of squares, definitely check out previous lesson videos. So we have seven multiplied by the quantity of T squared minus one multiplied by the quantity of T squared plus one. And we recognize that T squared minus one is another difference of squares. So this factor is even further to be seven multiplied by the quantity of T plus one multiplied by the quantity of T minus one multiplied by the quantity of T squared plus one. So now it's fully factored. And what we do is for each of the factors that have variables in them. In the denominator, we're going to ask ourselves two questions. First, are these distinct or repeating factors? T plus one? There's only one of those that's distinct T minus one. That's the only T minus one factor. It's distinct T squared plus one, also distinct. So all three of these are distinct factors. The next question we ask ourselves is, are these linear or quadratic factors? T plus one? That's to the first degree that's linear T minus one. That T is to the first degree, it's linear T squared plus one T is to the second degree that one's quadratic. Now with that information, because these three factors with variables are all distinct. We are setting our original fraction equal to three new fractions added together. And the denominators of each of those three fractions are going to be our three factors with the variables. So the first fraction has the denominator of T plus one. The second fraction has the denominator of T minus one and the third fraction has the denominator T squared plus one. And now for our linear denominators, they are going to have constant terms as their numerators. So for T plus one, we don't know what that constant term is. We're going to set it as a capital A for the T minus one. That linear denominator that gets a numerator of a capital B. And for our quadratic denominator, that one is going to have a linear numerator instead of a constant term like the other two. So that's going to be C multiplied by TT is the variable here. And then plus D, and that's the whole thing set up. Now, it's just a ton of algebra to solve for ABC and D. So if you've got it set up like this, that's great. Now, solving for ABC and D, the first thing we wanna do is get rid of our fractions. So we're going to multiply everything by the factors for our denominators that occur all everywhere. And so that will be this seven multiplied by the quantity of T plus one multiplied by the quantity of T minus one multiplied by the quantity of T squared plus one. So that's our left hand side fully factored. Now, when we multiply that by everything for the fraction on the left, the denominator cancels out all of that and we're left with just three T squared. That's equal to our, a fraction getting multiplied by all of that. Well, the denominator T plus one cancels out the T plus one factor. So it gets multiplied by the seven. We have seven, a multiplied by the quantity of T minus one multiplied by the quantity of T squared plus one or the B, its denominator T minus one cancels out the T minus one. So we have plus that B gets multiplied by the seven, we have seven B multiplied by the quantity of T plus one multiplied by the quantity of T squared plus one. Now, for our CT plus D, we can put plus, then we can put that in parentheses. CT plus D because that quantity gets multiplied by the seven also by the T plus one. And the quantity of T minus one, the T squared plus one and the denominator cancels out that factor. And now it's just a whole lot of foiling and distributing. So let's get started on that. We can bring down our three T squared on the left hand side, that's equal to, let's bring down our seven A and now we can foil the T minus one multiplied by the T squared plus one. I'm going to foil very quickly if you need a review. Definitely check out previous lesson videos. So seven A is multiplied by the quantity of T cubed plus T minus T squared minus one. And now we can bring down our plus seven B and we can foil T plus one multiplied by the quantity of T squared plus one. So seven B is multiplied by the quantity of T cubed plus T plus T squared plus one. And now we've got a lot going on here. We can bring down plus we can distribute the seven to the CT plus D. So we have the quantity of seven ct plus seven D. And now that's going to be multiplied by the quantity of when we foil T plus one multiplied by T minus one. Which will be T squared minus one. We've still got more distributing and foiling to do. We'll bring down the left hand side three T squared equals. Now, let's distribute seven A, two T cubed plus T minus T squared minus one. We'll have seven A T cubed plus seven A T minus seven A T squared minus seven A. Now distributing seven B to T cubed plus T plus T squared plus one. We have plus seven BT cubed plus seven BT plus seven BT squared plus seven B. And lastly, we foil seven ct plus seven D multiplied by the quantity of T squared minus one. So we'll have minus seven ct. Oops, I didn't do that in the right order 1st. 1st. So plus seven CT cubed minus seven CT plus seven DT squared minus seven D that almost fits. I'll bring minus seven right here above it. And I mean, for foiling, when you multiply them all together, the order doesn't matter, but I want you to be able to follow along. OK. Next, we've done all of the distributing and foiling. Now, we want to put these in descending order for the T variables. So we want the T cubed, followed by the T squared, et cetera. Bring down the left hand side, three T squared equals now let's find our T cubs. We have seven A T cubed plus seven BT Q plus seven CTQ. And now for our T squared, we have minus seven A T squared plus seven BT squared plus seven DT squared. Now, for our TS to the first power plus seven A T plus seven BT minus seven CT. Now our constant terms minus seven A plus seven B minus seven D, right? They're in order. Now, the next step is we are going to factor out our coefficients for our T variables. So we can bring down three T squared equal to, for our T cubed variables. We want to factor out the coefficient here. So we have open parentheses, seven A plus seven B plus seven C closed parentheses, all being multiplied by T cubed. Now we'll factor out the coefficients for our T squared. So plus open parentheses, negative seven A plus seven B plus seven D closed parentheses multiplied by T squared. Now factoring out the coefficient of our T to the first power. So plus open parentheses, seven A plus seven B minus seven C closed parentheses multiplied by T. Now we can group all of our coefficients together. Just put them in parentheses. So plus open parentheses, negative seven A plus seven B minus seven D closed parentheses. Now what we're doing is for our coefficients, we are matching the coefficients on the right to the coefficients on the left of the like terms and setting up a system of equations so that we can solve for ABC and D. So for row one, we'll have the coefficients of our T cubes, we'll take from the right hand side first. So we'll have seven A plus seven B plus seven C leaving, leaving a space for the no DS equal to. Now, the T cubes from the left hand side, there are none. So the coefficient of T cubed on the left is zero. Now row two will be our T squared. So from the right, we have a negative seven A plus seven B plus seven D equal to. And then on the left, the coefficient of T squared is three row three, the coefficients of our TS to the first power from the right, we have seven A plus seven B minus seven C that's equal to on the left, there are no T to the first power. So the coefficient is zero and row four, those are our constant terms from the right, we have a negative seven A plus seven B minus seven D equal to on the left, no constant terms. So that's zero. OK. Now we have a system of equations and we can solve for ABC and D this one's nice because addition is going to work very well. If we see that rows one and row three are nearly identical except for the CS which are going to cancel out. When we add each other, we can add those two together to eliminate that C variable variable. So seven A from row one plus seven A from row three is A seven B from row one plus seven, B from row three is plus 14 B seven C from row one minus seven, C from row three is zero and that equals zero plus zero, which is zero. Now, very similarly rows two and four are nearly identical except for the DS which are going to cancel each other out when added. So adding rows two plus four negative seven A plus negative seven A is negative 14 A seven B plus seven B is plus 14 B seven D plus negative seven D is zero that equals three plus zero, which is three. Now adding these two new ones, the one plus three plus two plus four. When we add those together, the A's cancel out and we have 14 B plus 14 B. Well, that's 28 B equal to zero plus three which is three dividing both sides by 28 we get B equals 3 20 eights. Now that we've solved for B, we can just use substitution to go back in and solve for AC and D. I'm going to do this part quickly. You're welcome to pause the video and work out all of the math. But for the sake of time, if we're plugging in B which is 28th into this equation where we have 14 A plus 14 B, that would be 14 A plus 14 multiplied by the quantity of 3 28th equals zero. Well, solving for A, we're going to get a negative 3 28. Now that we've solved for A, we can plug in for A and B in our first equation to solve for C. If you take in row multiplied by a negative 3 plus seven multiplied by a positive 3 that's going to be zero plus seven C equals zero is so for C and C equals zero, we've got C and now we can use row two to solve for D plug in for A and B which we know we plug in a negative seven multiplied by a negative 28 plus seven, multiplied by a positive 3 plus seven. D equals a positive three. You go through and you solve for D and D equals 3/ and there's ABC and D. Now, what do we do with those? Well, we'll bring those up to a different part of the screen where we started and remembering that our goal was to solve for ABC and D and now we're going to plug those in to our original partial fraction decomposition that we set up. So we've got a divided by T plus one. Well, A is negative 3 28. So that's going to be a negative three in the numerator divided by 28 multiplied by the quantity of T plus one in the denominator. And now plus our B divided by T minus one. B is positive 3 28. So the numerator is three divided by, and the denominator 28 multiplied by the quantity of T minus one. And now for our CT plus D, well, C is zero. So that CT is gone. And we just have D in the numerator, D is 3/14. So plus we'll just have three in the numerator divided by 14, multiplied by the quantity of T squared plus one in the denominator. And that's it. That's our partial fraction decomposition. Well done. This is tricky. We look at our answer choices and this matches with answer choice. B well done. We'll catch you on the next one.

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (x^2)/(x^4 - 1)

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Factoring Polynomials

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