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

Question

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

Accepted Answer

Welcome back. I am so glad you're here. We are asked to decompose the given rational function in partial fractions. Our given rational function is in the numerator. We have three T plus two. That's divided by our denominator, which is the quantity of T plus five cubed. And our answer choices are answer choice A three divided by the quantity of T plus five squared plus 13 divided by the quantity of T plus five cubed. Answer choice B three divided by the quantity of T plus five squared minus 13 divided by the quantity of T plus five cubed. Answer choice C 13 divided by the quantity of T plus five squared minus three divided by the quantity of T plus five cubed. And answer choice D negative 13 divided by the quantity of T plus five squared plus three divided by the quantity of T plus five cubed. And when we are working with partial fraction decomposition, the first thing we ask ourselves is do we have a proper fraction? In other words, is the degree of the numerator less than the degree of the denominator. Here, the degree of our numerator is one, the degree of our denominator that T will be multiplied by itself three times. So is one less than three. Yes, it is, we can proceed. The next thing we ask ourselves is, is our denominator fully factored. And in this case, it is. So next, we ask ourselves some questions about the factors in the denominator. The first question we ask ourselves is, is this distinct or repeating? We only have one factor in the denominator. It's the T plus five, but it is repeating, it repeats itself three times. So this one is a repeating factor. The next question we ask ourselves is, is this linear or quadratic T in the T plus five, that's to the first degree, this is linear. So now that we have established that we set our fraction equal to. And now because this linear factor repeats itself three times, we're going to have three fractions on the right hand side. So three fractions being added together. Now the denominators for our three fractions will come from that T plus five cubed. The first fractions denominator will be T plus five, that factor. But to the first degree, the second fractions denominator will be the quantity of T plus five. But this time it is squared. And for the third fraction, we have still that same factor, the quantity of T plus five. But this time it is cubed and since it only goes up to cubed, that's as high as it gets. And so we've got our denominators. Now, our numerators because all three of these are linear are going to be constant terms. We don't know what they are yet. That's what we're trying to find out. But we will use capital A for the first one, the T plus five, we'll use a capital B for the numerator of the second one, the T plus five squared. And we'll use a capital C for the numerator of the third fraction. And now the rest of this is all just algebra trying to solve for A B and C. So we are going to do that by multiplying everything here by the denominator on the left hand side of the equation. So that's going to be multiplying everything by the quantity of T plus five cubed, that's going to get rid of all of these denominators. So multiplying T plus five cubed by our fraction. On the left hand side, we are going to have just the three T plus two because the denominator cancels that out. And that is equal to, we have our capital A, the numerator being multiplied by the T plus five in its denominator cancels out one of the T plus five factors. So that A gets multiplied by the quantity of T plus five squared. There's two of them left and we bring down plus and then our B, the B's denominator of the T plus five squared cancels out two of the T plus five factors. And it's, there's only one left to multiply it by. So B multiplied by T plus five and then we bring down our plus C. Well, the denominator of CT plus five cubed, it's can it cancels out all of the T plus five cubs? And so we just have the C, now we need to do some foiling and some distributing, we can bring down our three T plus two, which is equal to and that A is being multiplied by T plus five, multiplied by the quantity of T plus five. And I am going to foil that very quickly if you need a refresher on foiling. Definitely check out previous me lesson videos, bring down the A multiplied by the quantity of T squared plus 10 T plus 25. And then that B gets distributed to the T plus five. So we'll have plus BT plus five B then bring down our plus C. Now we can distribute the A to the T squared plus 10 T plus 25. I'll bring down three T plus two that's equal to A T squared plus 10 A T plus 25 A, bring down the rest plus BT plus five B plus C. Now we want to put these in descending order of their variables. On the right hand side, it won't take much rearranging for this one. So we'll bring down three T plus two on the left equal to. Now we want our T squared that's just A T squared. That's first there. Now we want our T to the first powers. So we'll have the plus 10 A T and plus BT. Now we want all of our constant terms, which is the rest plus 25 A plus five B plus C. Now that these are all in order, we want to factor out our coefficients for all of our TS to their degrees. So we can bring down three T plus two equals. And for that T squared, it's only the A T squared for the T to the first power we'll have plus open parentheses, 10 A plus B closed parentheses multiplied by T. And then we can put our coefficient or our constant terms, excuse me into a parentheses plus open parentheses. 25 A plus five B plus C just so we keep those together. Now we are going to take the coefficients from each side of the equal sign and set them equal to their like terms. So for our T squared from the right, the coefficient is A, we set that equal to the coefficient of T squared on the left. Well, there is no T squared on the left. So that coefficient is zero. So A is equal to zero. Now are coefficients of T to the first power. We're going to take 10 A plus B from the right, that's equal to now the coefficient of T on the left, that's three. And now for our third row, we can label these rows 12 and three. In our system of equations, we take our constant terms. So our constant terms from the right are 25 A plus five B plus C that's equal to the constant terms on the left. That's just a positive two. And now we've got a system of equations we can solve for A B and C. Well, A is already given to us A equals zero. Wonderful. Now, B that one's not too tough because when we plug in zero for A, in the second row 10 multiplied by zero plus B equals three, we can see that B equals three. Now we know A and B we can plug that into row three. So 25 multiplied by not A but zero plus five multiplied not by B but three plus C equals two. So five multiplied by three is 15. We have 15 plus C equals two. We subtract 15 from both sides and we get C equals a negative 13 and we've got A B and C solved. So now we're just plugging those into our original setup for the partial fraction decomposition. So A is zero. Well, zero, divided by T plus five is zero. So we don't have anything for that first fraction. Now B is three. So we have three divided by the quantity of T plus five squared. Then plus and CC is negative 13. So actually that plus becomes minus 13. Divided by the denominator is the quantity of T plus five cubed. And that is it, that is our partial fraction decomposition. 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. (2x + 1)/(x + 2)^3

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Polynomial Long Division

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