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

Question

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (3x - 1)/(x(2x² + 1)²)

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 six T minus seven. That's divided by the denominator which is T multiplied by the quantity of three T squared plus two squared. Our answer choices are answer choice. A negative seven divided by four T plus 21 T divided by four multiplied by the quantity of three T squared plus two plus the quantity of 21 T plus 12 divided by two multiplied by the quantity of three T squared plus two squared. Answer choice B seven divided by four T plus 21 T divided by four multiplied by the quantity of three T squared plus two plus the quantity of 21 T plus 12 divided by two, multiplied by the quantity of three T squared plus two squared. Answer choice C negative seven divided by four T plus 21 T divided by two multiplied by the quantity of three T squared plus two plus the quantity of 21 T plus 12 divided by four multiplied by the quantity of three T squared. Plus two squared and answer choice D negative seven divided by four T plus T divided by two, multiplied by the quantity of three T squared plus two plus the quantity of 21 T plus 12 divided by two multiplied by the quantity of three T squared plus two squared. All right. Now, step one in decomposing into partial fractions is we want to see if we have a proper fraction. In the first place, we need the numerator to be less than the denominator for the degree. Excuse me, the numerator's degree less than the denominator's degree. The numerator's degree here is one, the denominator's degree. When we have a T, it's being multiplied by a three T squared, which is going to get multiplied by another three T squared. And so the degree of the denominator is five and one is definitely less than five. We're good. And the second thing we ask ourselves when beginning to decompose into partial fractions is whether our denominator is fully factored. And in this case, it is so we can proceed. Now, what we do is a, ask ourselves a few questions about the factors in the denominator that contain variables. In this case, the T first we ask ourselves if these are distinct or repeating factors and T there's only one factor of just T, that one is distinct but three T squared plus two. There are two of those because that factor is squared. So that one is repeating a little bit of both. Now, the next thing we ask ourselves is if the factors in the denominator are linear or quadratic key that is to the first power that one's linear three T squared plus two, that one's squared, that one's quadratic. So we have a linear distinct factor and we have a quadratic repeating factor. OK. So now we take that information, we set this fraction equal to, we are going to create one fraction for our distinct linear factor in the denominator that T and that gets to be the denominator here for this fraction. And then that's plus. Now this repeating factor gets two because it's to the second power because that factor is to the second power. So we have two more fractions. Now for our denominators for these two, we're going to take our factor three T squared plus two to the first degree for one of them. And then for the other, we're going to have the quantity of three T squared plus two squared for our denominator. And now we go back to that linear or quadratic question and that helps us establish our numerators for the T, the denominator that is linear, that is going to get a constant term for its numerator. So we don't know the constant term. We're gonna put a capital A for the quadratic denominators, they are going to get linear numerators. So for three T squared plus two, our numerator is going to be a BT because T is our variable here plus C. And for our three T squared plus two squared denominator, that one also gets a linear numerator which will be DT plus E. And there, we've set it up now that we have set it up. It's just a ton of algebra to solve for ABC D and E. So the first step in solving for ABC D and E is we are going to multiply this whole thing by the denominator on the left hand side. So we're multiplying everything by T multiplied by the quantity of three T squared plus two squared. This will allow us to cancel out all of our denominators. So multiplying that by the fraction on the left, our denominator cancels that out and we're left with just six T minus seven on the left hand side of the equal sign. Now taking our A divided by T and multiplying it by the T multiplied by three T squared plus two squared, the TS cancel out. So we have the A being multiplied by the quantity of three T squared plus two squared and bring down our plus. Now we have the quantity of that numerator BT plus C and the denominator of BT plus C cancels out one of the three T squared plus two factors, but it leaves one and it also gets multiplied by the T still. So that's being multiplied by T and by three T squared. Plus two just to the first degree. Now bring down our plus and then our numerator of, we'll put it in parentheses. DT plus E and then the three T squared plus two squared in its denominator cancels out the three T squared plus two squared and it just gets multiplied by the T. Now, it's a whole lot of foiling and distributing so we can get started on that, bringing down our six T minus seven equals. Now, we can bring down the A and we can start off by foiling because we can take the quantity of three T squared plus two multiplied by itself three T squared plus two. I'm going to do the foiling and distributing fairly quickly. If you need any reviews on foiling or anything, please definitely check out previous lesson videos. So A is multiplied by the quantity of nine T races to the fourth power and then it's going to be plus 12 T squared plus four. And now we can take the tea and distribute it to the BT plus C. That's going to be BT squared plus CT. And now we can foil our BT squared plus CT by three T squared plus two. So bring down our plus and foiling that we've got three BT to the fourth power multiplied by or excuse me plus three CT cubed plus our two BT squared plus two CT. And now we can distribute this T to our DT plus E. So we have plus DT squared plus ET. Now, the last distribution that we need to do is our A to the nine T to the fourth plus T 12 T squared plus four. So we can bring down six T minus seven equals nine A T to the fourth plus A T squared plus four A. Now bring down the rest plus three BT to the fourth plus three CT cubed plus two BT squared plus two CT plus DT squared plus ET and now we want to put these in descending order for their degrees. So we're going to want to look for the T to the fourth, then our T cubes, et cetera. So let's start with six T minus seven equals, we have nine A T to the fourth and then we have plus three BT to the fourth. And that's it for T to the fourth. For T cubed, we only have plus three CT cubed. And for T squared, we have plus 12 A T squared plus two BT squared plus DT squared. Now T to the first power we have plus two CT plus ET. And for the constant term, it's just plus four A, all right. Now that these are in order we want to factor out their coefficients for each of our TS to the various degrees. So I'll bring down six T minus seven again. And for our T to the fourth powers, we factor out open parentheses, nine A plus three B closed parentheses multiplied by T to the fourth power. Now, for our T cubes, we only have one T cubed. So we can put plus three CT cubed. For T squared, we have plus open parentheses. Factoring out 12 A plus two B plus D multiplied by T squared. Factoring out the coefficients for our tita, the first power we have plus open parentheses, two C plus E closed parentheses multiplied by T. And then our constant term just plus four A. Now we are going to set the coefficients from each side of the equal sign equal to each other for like terms. And then we will create a system of equations and solve for ABC D and E. So for the first row of our system of equations, we'll match up the coefficients of T to the fourth taking from the right hand side. First, we have nine A plus three B and that's equal to from the left hand side, the coefficient of T to the fourth. Well, there is no T to the fourth. So the coefficient is zero for row two, we're going to take the coefficients of T cubed from the right hand side. It's three C that's equal to, on the left hand side, there's no T cubed. So again, equal to zero row three, the coefficients of T squared from the right, we have 12 A plus two B and then plus D and on the left, no T squared equal to zero row four, the coefficients of T. From the right, we have two C plus E and from the left, we do have a coefficient of T to the first power that's six. So it's equal to six row five are coefficient or are constant terms on the right, we just have four A, so that's equal to, on the left that's negative seven. And we have set up our system of equations. And this one's, this one's nice because we can just start solving right away. For some of these row five, we're given that four A equals negative seven, dividing both sides by four, we get a equals negative seven fourths who we got a solved. Now, we can use that and plug it into equation one. We've got nine multiplied by not A but now negative seven fourths plus three B equals zero. For the sake of time, I am going to just do all the, I, I've done the computation ahead of time. You're welcome to pause it and compute these and get these solved. But for the sake of time, if you go through and do all of that math for B, we get an answer of fourth. And now we've got A and B and we can use that to solve for D in our third row from our system of equations. We take 12 multiplied by negative seven fourth plus two multiplied by B which is 4th. And then that's going to be plus D is equal to zero. Solving for D, we get 21 halves. Now that we've got a B and D, uh, we could also solve for C pretty quickly looking at row for row two, we've got three C equals zero. We'll divide both sides by three C equals zero. Great. Now, we can, because we know C, we can solve for E on row four, we've got two multiplied not by C but by zero plus E equals six. And we get E equals six. So now that we have solved for all of those, it's just a matter of plugging ABC D and E in to our partial fraction decomposition that we set up. And if we take A, which is negative seven fourths and plug that in for a divided by T, we get negative seven divided by four T, then plus for BT plus C. Well, our B is 21 4th plus C is zero. So it's just 21 4th in the numerator which simplifies 2, 21 divided by or multiplied by that denominator which is three T squared plus two. And then our DT plus E. So we have plus RDT is 21 halves, T plus E is six. And now we've got a mix in our numerator. And so if we take 21 halves, T plus six, what we wanna do is factor out the half. And so that would give us one half multiplied by the quantity of 21 T plus 12. And then in our numerator, we'll have that 21 T plus divided by, in the denominator, the two from the one half multiplied by the quantity of three T squared plus two squared. And that is it, that is our partial fraction decomposition. Well done. We look at our answer choices and this matches with answer choice. A all right. Well done. We'll catch you on the next one.

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (3x - 1)/(x(2x² + 1)²)

Key Concepts

Partial Fraction Decomposition

Factorization of Polynomials

Handling Repeated Quadratic Factors

Watch next

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (3x - 1)/(x(2x2 + 1)2)

Key Concepts

Partial Fraction Decomposition

Factorization of Polynomials

Handling Repeated Quadratic Factors

Watch next

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (3x - 1)/(x(2x² + 1)²)