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

Question

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (-x⁴ - 8x² + 3x - 10)/((x + 2)(x² + 4)²)

Accepted Answer

Welcome back. I am so glad you're here. We are asked to decompose the given rational function in partial fractions. Our rational function is in the numerator. We have negative seven X to the fourth minus 28 X squared plus five, X plus seven. That's all divided by, we have the quantity of X plus seven multiplied by the quantity of X squared plus two squared. Our answer choices are answer choice A seven divided by the quantity of X plus seven plus five divided by the quantity of X squared plus two squared. Answer choice B negative seven divided by the quantity of X plus seven plus five divided by the quantity of X squared plus two squared. Answer to ac negative seven divided by the quantity of X plus seven minus five divided by the quantity of X squared plus two squared. And answer choice D seven divided by the quantity of X plus seven minus five divided by the quantity of X squared plus two squared. All right. Step one. When we're decomposing into partial fractions, we want to make sure that we have a proper fraction here. So we need the degree of the denominator to be larger than the degree of the numerator. The degree of the numerator is four. What's the degree of our denominator? Well, we've got one X here and then we have an X squared that will be squared, so that'll be X to the fifth and four is smaller than five. So this is a proper fraction and we may proceed. Yay. OK. Next, what do we do? We set our given fraction equal to and now we take a look at our factors in the denominator. So first, starting with X plus seven, we ask ourselves is that linear or quadratic? It is linear. It's to the first degree. So we are going to have that be the denominator of. So we put X plus seven in the denominator a fraction with a constant term in the numerator because the denominator is linear, we need a constant. So we'll use capital A for the numerator. OK? We've got that one. Now let's look at our second factor. We ask ourselves is this linear or quadratic? Well X squared that is quadratic. So we will be dealing with quadratic. Next, we ask ourselves is this distinct or repeating and this one is repeating because this factor is squared. So how do we deal with that? We are going to put two more fractions because that one is repeating the denominator of the first. While the second fraction that we have here, the first of our two blink ones is going to be the X squared plus two, the factor just to the first degree and the denominator of the second one is going to be that quantity of X squared plus two. But this time squared, those are our denominators. And now, because it's quadratic, we are going to have a linear numerator. So the numerator for the first one, the X squared plus two is going to be B X plus C. And the numerator for the second one, the X squared plus two squared is going to be D X plus E. And that's how we set this up. The rest is a ton of algebra. But if you've set it up like this, you've set it up properly. Great. OK. Now, the next thing we do is we're just, we're solving for ABC D and E and it's going to take absolutely forever, but we've got this. So we are going to get rid of our fractions first of all. And we do that by multiplying by the factors that are included for all the denominators. So essentially by the denominator on the left hand side. So multiplying everything by the quantity of X plus seven, multiplied by the quantity of X squared plus two squared. And so now rewriting this on the left, we will just have the numerator, the denominator gets completely canceled out. So we'll have negative seven X to the fourth minus X squared plus five X plus seven. That's equal to. Now, on the right, for our first fraction A divided by X plus seven, the X plus sevens cancel out. So we still have the A and now that's multiplied by the quantity of what's left is X squared plus two squared. And now we have plus our next numerator B X plus C. We can put that in parentheses. The denominator X squared plus two cancels out one of the X squared plus two factors, but it's still going to get multiplied by X plus seven and one of the X squared plus twos just to the first degree. Now, we can bring down our final numerator, the plus D X plus E and that is going to fit. Nope, that we can put on another line plus D X plus E put that in parentheses and the denominator is going to cancel out both of the X squared plus two factors and it's just multiplied by the X plus seven factor. OK. So we got rid of the fractions great. And it was just so much foiling and distribution note that the left hand side does get brought down, but I'm not going to bring it down every time. We just don't have space. So this is equal to the right hand side. Let's do some of the foiling here. We've got this X squared plus two factor that is squared. So we can bring down the A and then I am going to very quickly do all of the foiling and distributing. So if we've got X squared plus two multiplied by X squared plus two, that is going to be first X to the fourth plus outsides plus insides is four, X squared plus, last is four. Now, we can foil the X plus seven multiplied by X squared plus two. So I think I need to bring this down slightly. Ok. Now, plus I'll bring down this B X plus C factor. And now foiling the XX plus seven multiplied by X squared plus two will be X cubed plus two, X plus seven, X squared plus 14. And now foiling are D X plus E multiplied by X plus seven will have plus D X squared plus seven D X plus E X plus seven E. All right. And now we've still got more distribution to do, we can distribute this A to the X to the fourth plus four X squared plus four. So we will have A X to the fourth plus four A X squared plus four A. Now, we can distribute the B X to the X cubed plus two, X plus seven X squared plus 14. So we'll have plus B X to the fourth plus two B X squared plus seven B X cubed plus 14 B X. Now, we can distribute this C to the X cubed plus two X plus seven X squared plus 14. So we'll have plus C X cubed plus two C X plus seven C X squared plus 14 C and then bringing down the rest, we'll have plus D X squared plus seven D X plus E X plus seven E, OK. Next, we want to group the like terms together, put these in order by degree. So we'll start off looking for the X to the fourth. So we'll have an A X to the fourth and plus B X to the fourth. So those two have been found and now we'll look for the X cubed. So we'll have plus seven B X cubed plus C X cubed. And that's it for the X cube. Now, let's look for our X squared will have plus four A X squared plus two B X squared plus seven C X squared plus D X squared. There are all of our X squared. Now, our X to the first powers, we have plus 14 B X plus two C X plus seven D X plus E X and there are all of the X to the first powers. And then finally, our constant terms, we'll bring these down here plus four A plus 14 C plus seven E OK. We've got them grouped. Now, we need to establish the coefficients for each of our X terms. So the coefficients for our X to the fourth, we have the quantity of A plus B multiplied by X to the fourth. All right. Now, I think I need to bring this down again, OK. Now, for our X cubed, we will have the quantity of seven B plus C multiplied by X cubed. Now, for our X squared, the coefficient is plus the quantity of four A plus two B plus seven C plus D all multiplied by X squared. Now, for the X to the first powers we have plus the quantity of our coefficient is 14 B plus two C plus seven D plus E. that's all multiplied by X. And then we can group our constant terms together the four A plus 14 C plus seven E. So we don't forget those all go together. Now, we are going to match the coefficients from the right hand side of the equation, which is what we've been working with with the coefficients from the left hand side of the equation. Remember there's, there's a left side too. So we look at the coefficients of our X to the fourth. From the left hand side, it's negative seven. From the right hand side, it's A plus B. So we can set up, we're beginning a system of equations. We have A plus B and that's equal to I'm leaving a bunch of space that's probably not even enough, leaving some space that's equal to negative seven. Now, the second row of our system of equations, we've got the X cubed constant terms. And from the right hand side, we have seven B plus C. From the left hand side, there is no X cubed. So it's equal to zero. Now, four R X squared we have on the right, we'll have four A plus two B plus seven C plus D. That's all equal to from the left, our coefficient of X squared is negative 28. And now our fourth row, the coefficients of the XS we've got from the right 14 B plus two C plus seven D plus E and that's all equal to from the left five. And for our final one, the constant terms we've got from the right hand side for a little more room four A plus 14 C plus seven E, that's all equal to from the left, going back up, we've got a positive seven and there, we've got a system of equations and solving this system of equations with this many variables. The way I personally chose to solve it is by setting up a matrix and then doing matrix solutions. You can use Gaussian elimination, Gauss Jordan elimination to come up with your solutions for ABC D and E, I'm going to show you how to set up the matrix. And then if you don't recall how to solve it, please go check out previous lesson videos. This would be twice as long if I tried to show all of the steps for the matrix, but I'll show you how to start it. So we're going to take all of the coefficients in our constants from the system of equations. That we have. So our first row is going to be one, the coefficient for a one, the coefficient for B zero, the coefficient of C zero, the coefficient of D zero, the coefficient of E and the negative seven, the constant term, the second row again, taking coefficients in constant terms will have 071000. For the third row, we'll have 42710, negative 28. For the fourth row, we'll have 0 14, 2715. And for the fifth row, we'll have 40 14, 077. And so then yeah, definitely check that out. Once you go through the whole process of solving that you will come up with solutions of E which is equal to five, you get D equal to zero, you get C equal to zero B equal to zero and a equal to negative seven. Now, what do we do with that? This very important information we've gleaned here. Well, we are going to take that and plug in for ABC D and E back up here where we started 300 years ago for ABC D and E. So plugging in for those R A was equal to negative seven. So we have a negative seven divided by that X plus seven. Now, B and C, those are both zero. So the numerator of this second fraction is zero. So that's zero. And we don't have to put it for the third fraction D is also equal to zero. So we don't have to put anything there. We just need the E and the E is a positive five. So we have plus five divided by and then our denominator, the quantity of X squared plus two squared. And that's it. That is our partial fraction decomposition. That's a lot of hard work. Very nice job. 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. (-x⁴ - 8x² + 3x - 10)/((x + 2)(x² + 4)²)

Key Concepts

Partial Fraction Decomposition

Handling Repeated Irreducible Quadratic Factors

Equating Coefficients to Solve for Unknowns

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Find the partial fraction decomposition for each rational expression. See Examples 1–4. (-x4 - 8x2 + 3x - 10)/((x + 2)(x2 + 4)2)

Key Concepts

Partial Fraction Decomposition

Handling Repeated Irreducible Quadratic Factors

Equating Coefficients to Solve for Unknowns

Watch next

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (-x⁴ - 8x² + 3x - 10)/((x + 2)(x² + 4)²)