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^2 + 2x + 1)

Accepted Answer

Hello. Today we're going to be simplifying the following rational expression by using partial fraction decomposition. So what we are given is T squared over the quantity T squared plus six T plus nine. Now, one thing to note is that the exponents in both the numerator and the denominator are of equal value. So that means that we're going to use long division to first simplify the rational expression. Now, in a previous video, we have discussed on how to do long division of two polynomials. So if you need help with that process, I highly recommend that you go back and check out those videos. But if you simplify this using long division of polynomials, you should get the quantity of one minus the remainder which is six T plus nine over the quantity T squared plus six T plus nine. For now, we're gonna go ahead and leave this quantity of one alone. And we want to go ahead and simplify six T plus nine over T squared plus six T plus nine using partial fraction decomposition. So in order to use partial fraction decomposition, we want to make sure that our denominator is in a fully factored form. Currently, our denominator can be simplified or factored using the trial and error method. If you were to simplify this trinomial, you should get the denominator of 60 plus nine over the quantity T plus three times T plus three. And because you have two instances of T plus three, you can go ahead and combine this together and rewrite the denominator as T plus three squared. So now that we have our denominator in a fully factored form, we can go ahead and write the partial fractions. Now, one thing to note is that T plus three squared is a repeating linear combination or linear quantity. So in order to write the partial fractions, we're going to write the partial fraction of the first quantity of T plus three. Now, since T plus three is linear, the first partial fraction is going to be a over T plus three. Now we want to write a partial fraction for the repeating quantity and that is going to give us B over T plus three squared. So one thing to note is that whenever you have repeating quantities in the denominator, you want to go ahead and represent the partial fraction and get a partial fraction for each exponent represented. So in this case here, because we had an exponent of two of this linear quantity, we got two partial fractions. So now that we have the partial fractions, let's go ahead and simplify this equation in order to solve for A and B. In order to do that, we're going to multiply everything by a lowest common denominator. And in this case here, the lowest common denominator between the three quantities is going to be T plus three squared. So we're going to multiply each quantity by the L CD. In order to simplify the equation, that is going to give us 60 plus nine over T plus three squared, multiplied by T plus three squared over one. And that is going to equal to A over T plus three times T plus three squared over one plus B over T plus three squared multiplied by T plus three squared over one. Now, what we're going to do is we're going to use cross reduction to eliminate the denominators. In the first term, we have T plus three squared in both the numerator and denominator. So we can go ahead and eliminate those quantities. Now, in the second term, we have one instance of T plus three in the denominator and two instances of T plus three in the numerator, we can go ahead and eliminate one instance from both the numerator and denominator. And in the last term, we can just eliminate T plus three squared from both numerator and denominator. That is going to leave us with the equation six, T plus nine is equal to A times the quantity T plus three plus B times one, which is just B. Now, what we're going to do is we're going to simplify the right hand side of the equation. In order to simplify, we're going to distribute A into the quantity T plus three that is going to give us six T plus nine is equal to A times T plus three A plus B. Now, what we're going to do is we're going to group up our constants together and we can use this equation to make a system of equations that will help us solve for A and B. So this is where the tricky part of partial fraction decomposition comes in. We need to go ahead and let the coefficients of the T terms in both the left and right hand side equal to each other. So on the right hand side, T has a coefficient of A and on the left hand side T has a coefficient of six. So that is going to give us an equation of A is equal to six. And that is also going to be one of the coefficients we need for A and B. Now, what we wanna do is we want to let the constant on the right hand side equal the constant on the left hand side. And that is going to give us the second equation which is going to be three A plus B equal to nine. Now keep in mind that we already have a quantity for A A is equal to six. So if we plug this into the second equation, we get three times six plus B is equal to nine, six times three is going to give us 18. And if we subtract 18 to both sides, we get the value of B is equal to negative nine, which is the second quantity that we need. So now that we have the coefficients of A and B, we can go ahead and plug this in to the partial fractions we came up with. Now, keep in mind that the partial fractions we made was A over T plus three plus B over T plus three quantity squared, plugging in A and B is going to give us N six because A is equal to six over T plus three minus nine over the quantity T plus three squared. So this is going to be the partial fractions for the remainder from the beginning. But keep in mind that in the beginning of the equation, we had this quantity of one minus the partial fraction decomposition we just came up with. So now we need to go ahead and put these partial fractions into the original, into the original quotient. That is going to give us one minus the partial fractions, which is six over T plus three minus nine over T plus three squared. If we distribute the negative one into the partial fractions, that is going to give us one minus six over T plus three plus nine over the quantity T plus three squared. And this is going to be the fully simplified form of the original rational expression. And with that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Factoring Polynomials

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