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

Question

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

Accepted Answer

Hello everybody. I today, today we're going to be looking at this question that states for the following rational expression, apply partial fraction decomposition where our expression is one T or one divided by. So our expression will be a fraction and it will contain one in the new narrator and that one will be divided by T and the denominator will be t multiplying open parentheses, four T plus one close parentheses and open another parentheses multiply that will be multiplying by two T squared plus one closed parentheses. So that will be our rational expression. Now, our asteroids consist of all will consist of three different fractions, either performing an addition or subtraction um amongst themselves. Now, when we want to approach this question, we have to focus on, they want us to apply partial fraction decomposition. So what is, what is that? What is the composition? Basically, we're gonna take this one fraction that they gave us this rational expression. And that is that consists of one fraction and we're going to decompose it. And what I mean by that is we're going to try to separate it into multiple fractions. So we're going from one into multiple. And that's why our choice consists of three different fractions because we're decomposing our one fraction. Now to do that, we're going to set our rational expression equal to three different fractions as well. And we're going to be solving for the numerator of each fraction. So the first fraction we'll have is so we'll have our expression equal to a on the numerator divided by T and the numerator, the num the denominator of each fraction will be one of the terms in the denominator of our expression. So in this case, we have a divided by T and we'll have that plus B divided by the next term, which will be 40 plus one, which is the, the first parentheses. And then we'll have that plus C T plus D in the numerator divided by two T squared plus one. So why is the last term C T plus D enumerator? Well, it's because our denominator is in a quadratic form. So the first two terms, we have a team which has no other power except one. And we have a four plus one which again has no other power except one. And then in the last term, we have two T squared plus one. So when you have a square or something like that in a quadratic form, the numerator has to be changed to account for that. So in this case, we have C T plus D, if it was an X, it would be C X plus D and so on. Now, I'm going to scroll down on my screen a bit since I know this question is going to take a bit of work just so I'm able to utilize as much room as possible. Now, the first thing I'm gonna do is multiply each side of my equation by the numerator in my expression. So I'll have on the left side, I'll have an open bracket and inside of that open bracket, I'll have a, the T that we had in, in the, I'll have our new denominator basically. So I'll have the T multiplying the four T plus one, multiplying the two T squared plus one. And that entire thing will mean multiplying our expression, which was one divided by T multiplying a, the four T plus one, multiplying the two T squared plus one. And I'll do the same thing to the other side of our equation. So I have a divided by T plus B divided by four T plus one plus C T plus D divided by two T squared plus one. And that entire right hand side of our equation will be multiplied by T multiplying four T plus one, multiplying two T squared plus one. So we do that because now, for example, in the left hand side of the equation, if we multiply the left hand side of the equation by the denominator, the denominator will cancel. And we'll have, we'll be left over with just the one in the numerator. So we have that one equals. And then if we go turn by turn on the right hand side, first, we'll have, we'll be able to cancel the denominator of our A. So the T will cancel with the first team in our multiplication. And we'll have that A, we have one equal, a multiplying four T plus one, multiplying two T squared plus one. And that is because the T cancels, but the four plus one and the two T squared plus one don't cancel. So then we'll have that plus our numerator in the second term, which was A B and in this case, that B was divided by 40 plus one. So in this case, the four plus one will cancel and we'll have B multiplying the T which didn't cancel and multiplying the two T squared plus one, which also didn't cancel. Now, for the last term, we had C T plus D in the numerator and that will not be multiplying. So the denominator, the two T squared plus one will cancel since we're multiplying and will not be left multiplying the T and multiplying four T plus one. So now we've canceled everything that we need to and we have no fractions left over. So now there's different routes that you can take here. You can go ahead and foil right away. So you can distribute every parentheses and every multiplication or you can try to do some strategic plugging in and see where it takes from there. So in my case, I'm going to first do the plugging in to see what I can do. So first I'm going to let equal zero and I'll have now that one equals a multiplying four, multiplied by zero plus one closed parentheses and then open parentheses, two multiplied by zero squared plus one closed parentheses plus and now we can stop there. Why? Well, if we plug in zero for our team, our term alongside B had A T in a parentheses, multiplying everything and we have zero as A T A zero is multiplying that whole term. So the B is eliminated. And the same thing for the term with the C and the D. So we plug in zero for T which is multiplying everything including the C and the D. So C and D are eliminated. So now we're left solving for A, so I'll have that one is equal to four. So we have one equal to a multiplying. And then in the first parentheses, we had four multiplied by zero, which is zero plus one. So we have one in that parentheses, multiplying two multiplied by zero squared. Zero, squared is zero, multiplied by two, is zero plus one is one once again and we be left over with one equals A or I'll just write it in the different notation. I'll write it as a equals one. And that is our first answer. So now we can decide, do we want to do the strategic plugging in again or do we want to do the foiling? In my case, I'm going to do the, for the plugging in once again. So I'm going to now say let t equal negative one divided by four and I'll have that one is equal to b multiplied by negative one divided by four. And ok, so how did I assume or how did I come to the conclusion of doing negative one divided by four? Well, I look at my other terms and I saw that A, the terms associated with A and the terms associated with C and D both have a parentheses that has four T plus one E inside of it. Now, if I can set that parentheses equal to zero, that means I can eliminate and the variable A, the variable C and the variable D now have you set four plus one equals zero, then T will be negative one divided by four. So I'm plugging in negative one divided by 44 T so that I can eliminate the variables AC and D. So I have one equals B multiplying negative one divided by four, multiplying two, multiplying negative one divided by four squared plus one inside of all of that inside of a parentheses. So now I can do some simplification and I'll have that one equals B multiplying open parentheses negative one divided by four closed parentheses, multiplying an open parentheses, one divided by eight plus one. So negative one divided by four squared, the negative will go away. One squared is 14 squared is 16. So we'll have one divided by 16, multiplied by two will be two divided by 16, which simplifies to one divided by eight. So now we can continue to simplify and I'll have that one equals B multiplied by negative one divided by four multiplied by. So I had one divided by eight plus one. That one, I can turn it into eight, divided by eight to have a full fraction and one divided by eight plus eight, divided by eight is nine divided by eight. And then we can continue to simplify and we'll have that one equals B multiplying an open parentheses negative nine divided by 32. And all we did there was multiply the fractions that we had. So the negative one divided by four and the positive nine divided by eight. And this will give us that be, is equal to negative 32 divided by nine. And we obtain that by moving the fraction negative nine divided by 32 from the right hand side with the B to the left hand side with the one. And since on the right hand side, we're multiplying on the left hand side, we'll be dividing. And since we have one, it would just write the reciprocal which will be negative 32 divided by nine. And that'll be our answer or B So now once again, I'm going to ask myself, do I want to plug in strategically or should I form? And now we've gotten to a point where I believe foiling will be our best route. Um If we want to try to solve for D N C and D, the C N D are together, so it gets a bit complicated. And if we want to eliminate answer or, or the terms so associated with A and associated with B, there's nothing we can really plug in because two T squared plus one is if we said that equal to zero, that would be the parentheses that we can set equal to zero. So if we set that equal to zero, we would have that two T squared equals negative one or T squared equals negative one half. And you can't square anything to get a negative answer. So that won't be, we won't be able to do that. So instead we'll go ahead and foil or distribute. So we'll have one and we're distributing the very first equation that we had with the A and the B and the C and D terms multiplying everything else. So we'll have that one is equal to eight A T cubed plus for A T plus two A T squared plus A plus two T Q B plus BT plus four T QC plus T squared C plus four T squared D. And I'll write the final term under that plus D T. So now that is what happens when we foil our equation where we had one equals a multiplying the four T plus one closed parentheses, multiplying the two T squared plus one plus B, multiplying T multiplying the two T squared plus one plus open parentheses, C T plus D closed parentheses, multiplying T multiplying four T plus one. So once you distribute all of that and your foil and everything, you will get what we just obtained. And now my next step will be to rearrange this term into something that will help us a bit better, will help us visualize this a bit better. So I rearrange it to have all the T cubed terms first, then all the T squared terms. Second. Finally, the uh the terms of the T and finally any terms that don't have any T next to it. And at the same time that I do that, I will be plugging in our A and R B. So I have one equals eight T cubed. Here, I plugged in one for a, so that A can be eliminated and we'll have a T cubed plus four. And if I plug in one for a, once again, we'll just be left with or actually, I'll, I'm going to rearrange as I said. So we'll have a T cubed minus 64 divided by nine T cubed. And once again, that's just plugging in the B and we'll have that plus four T cubed C plus two T squared plus T squared, C plus four T squared D plus minus 32 divided by nine T plus D T plus one. So it all looks like a lot of it. It looks very convoluted. But all we did here was just rearrange, like I said, all the T Q terms, first T squared terms second, then the terms with only A T and finally, the term with no T attached to it. And at the same time as we rearranged, we also plugged in our known values for A and for B A being one and B being negative 32 divided by nine. So now why did I rearrange this? Well, we know that the terms on the right hand side of our equation must equal the terms on the left hand side of our equation because they're equal to each other. We have that one equals this big convoluted mess that we have. So we know that the T cubed terms, the T squared terms and the T terms will add up to zero because there are no T squared or T cubed and T S in the left hand side of our equation, it's only the one. So we know those terms will have to add up to zero. So we can use that to our advantage. In this case, I'm going to focus on the T cubed terms because we have A T cubed minus 64 divided by nine T cubed plus four T cube C. So our only variable here is C and we're, we already have A and B and we're, we're trying to solve for C and D now. So I'm going to focus on a term that only has C and in this case, that term will be the T squared. So we'll have that eight minus 64 divided by nine plus four C equals zero. And I obtained that by just taking the number in front of every T squared. So we had a T or a A T cubed, sorry. So we had a T cubed. So I take that eight, then we had minus 64 divided by nine Q. So I take that minus 64 divided by nine. And finally, we had four T cubed C. So I take both the four and the C because the C is also a number. So I have to put plus four C and all of that will be equal to zero. Since again, there are no T squared terms on the left hand side of our equation. So now we can solve, solve for C and we'll have that negative 64 divided by nine plus four C equals zero or equals eight because we move the eight over from the right hand side to the left hand side by subtracting. So it'll be negative eight. So we'll have negative 64 divided by nine plus four C equals negative eight. And then we'll have four C equals negative eight plus 64 divided by nine. And that will be four C equals negative eight divided by nine. And if we divide both sides by four C is equal to negative eight divided by 36. So that will be our answer four C or actually, if we want to simplify that a bit more, we can put that scene is equal to negative two divided by nine. And all we did there was divide the numerator and the denominator by four. So we'll have C equals negative two divided by nine. So now we have a B and C and finally, I'm going to, so redeem using terms that only have D in them. And in this case, it'll be the term associated with A T, not A T cubed or A T squared. Just a singular T which was 40 minus 32 divided by nine T plus D T. So again, we'll take the number associated with each T. So we have four minus 32 divided by nine plus D equals zero. Once again the zero, because we have no T terms in our left hand side. So we'll have that 32 or negative 32 divided by nine plus D equals negative four or D equals negative four plus 32 divided by nine. And if we simplify that we'll have that D equals negative four divided by nine. And now we have our A R B R C and R D. So with that information, what we can do is plug in all those terms to our fraction or our equation that we had where we had a divided by T plus B divided by four T plus one plus C T plus D divided by two T squared plus one. And once we plug that in, we'll see our three fractions match with answer choice D. So I hope that was helpful. And until next time.

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

Key Concepts

Partial Fraction Decomposition

Types of Factors in the Denominator

Solving for Unknown Coefficients

Watch next

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

Key Concepts

Partial Fraction Decomposition

Types of Factors in the Denominator

Solving for Unknown Coefficients

Watch next

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