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

Question

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

Accepted Answer

Hello, today we're gonna be simplifying the following rational expression by using partial fraction decomposition. So what we are given is three, I'm sorry, five T minus two over T plus one times three T squared plus three. Now, whenever we want to use partial fraction decomposition, we always want to make sure that our denominator is fully factored. In this case here, the denominator is given to us in an already factored form. So we can go ahead and immediately set up our partial fractions. So in the denominator, we are given T plus one times three T squared plus three. The first quantity T plus one is a non repeating linear quantity. And that is going to give us the partial fraction of A over T plus one. Now the second quantity which is three T squared plus three, since it has an exponent of two, this is going to be a quadratic quantity. And since it's a non repeating quadratic quantity, we're going to get a second partial fraction, which is of the form B X plus C over three T squared plus three. So now that we have the partial fraction set up, we need to go ahead and solve for the coefficients of A B and C. In order to do that, we can use this entire equation. But let's go ahead and simplify the equation by eliminating the denominators. So we're going to need to multiply everything by a lowest common denominator. And this lowest common denominator is going to be the product of all the factors of the denominators. So we have T plus one, we have three T squared plus three and we have T plus one times three T squared plus three. That means the lowest common denominator in this case is going to be T plus one times three T squared plus three. So we're going to multiply each term in the current equation by this lowest common denominator, this is going to give us five T minus two over T plus one times three T squared plus three, multiplied by T plus one times three T squared plus three over one. And that is going to equal to A over T plus one times T plus one times three T squared plus 3/1. And that is plus B X plus C over three T squared plus three times the lowest common denominator which is T plus one times three T squared plus 3/1. Now, we can go ahead and use cross reduction in order to simplify each of the terms. So in the first term, we have the product of the lowest common denominator in both the denominator and numerator. So we can go ahead and immediately cross out those quantities. In the second term, we have T plus one in both the numerator and denominator. So we can go ahead and cross out those quantities. And in the last term, we have three T squared plus three in both the numerator and denominator. So that is going to allow us to just cross out those quantities. So what we are left with is five T minus two, equal to a times the quantity three T squared plus three plus the quantity B X plus C times the quantity T plus one. So we have the simplified equation that is going to help us solve for the coefficients of A B and C. So let's go ahead and simplify the right hand side of the equation. In the first group, we have a times the quantity three T squared plus three to simplify this, we're just going to distribute a into this group. And that is going to give us three A T squared plus three A. And in the second term, we're going to have to foil these two quantities together. So we're going to take the first term which is B X and multiply to T and to one that is going to give us B X T plus B X and we're going to multiply C to T and to one that is going to give us C T plus one. Now I would recommend grouping up our terms with respect to the variable T. And that is going to allow us to reorder the terms as three A T squared plus BT squared plus BT plus C T plus three A plus C. Now, what we're going to do is we're going to factor out T squared from the first two terms. We're going to factor out T from the middle two terms and we're going to group up our constants together. Doing this allows us to rewrite the right hand side as T squared times the quantity three A plus B plus T times the quantity B plus C plus our constants, which are three A plus C. And keep in mind that this is all still equal to five T minus two. Now this is where the tricky part of the partial fraction decomposition comes in. We're gonna go ahead and use this form of the equation to set up a system of equations in order to solve for A B and C. So in order to do that, we're going to take the coefficient of T square on the right hand side, which in this case is three A plus B and set it equal to the coefficient of T squared on the left hand side. But keep in mind that we don't have A T squared on the left hand side, we can go ahead and add one by giving it a coefficient of zero. So that means that the first equation that we can set up is going to be three A plus B equal to zero. Then we're gonna repeat this process with the T quantities that is going to be the coefficient of T on the right hand side, which is B plus C equal to the coefficient of T on the left hand side, which is going to be five. So we get a second equation which is B plus C equal to five. And finally, we're going to take our constant on the right hand side and set it equal to the constant on the left hand side. That is going to give us a third equation which is going to give us three A plus C equal to negative two. Now, what we want to go ahead and do is you want to solve this using a system of equations. So let's go ahead and take equation one and line it up with the equation too. That is going to give us three A plus B is equal to zero. And equation two is going to give us B plus C is equal to five. We can go ahead and eliminate the B term by multiplying the second equation by negative one. If we do this, we get three A plus B is equal to zero and we get negative B minus C is equal to negative five. Adding these two equations together is going to give us three A B minus B is going to cancel each other out minus C is equal to zero minus five, which is negative five. So what we have here is an equation in the terms of A and C and I'm gonna go ahead and label this as the fourth equation. But if you take a look at the third equation, we also have an equation of A N C. So let's go ahead and add these two equations together. So if we take the third equation, which is three A plus C is equal to negative two, and we add our new fourth equation which is three A minus C is equal to negative five. As you can see, adding these two equations together is going to cancel out the C term. So what we are left with is three A plus three A which is going to give us six A is equal to negative two minus five, which is going to give us negative seven. In order to solve for a, we're gonna divide both sides of the equation by six. And that is going to give us A is equal to negative 7/6. So this is going to be the first equation or I'm sorry, the first coefficient that we need for the partial fractions. Now we need to solve for the other two coefficients. Since we have a value for A, we can go ahead and plug that in to equation three in order to get a value for C. So plugging in A to C is going to give us three times negative 7/6 plus C is equal to negative two, three times negative 7/6 is going to give us negative 21 over six. And what we need to do now is we need to add 21/ to both sides of the equation. And doing this is going to give us the value of C is equal to 3/2. That is going to be the value for C that we need. And now that we have a value for C, we can plug this into equation two in order to solve for B, plugging into equation two is going to give us B plus 3/ is equal to five. If we subtract 3/2 to both sides of the equation, we get the value that B is equal to positive 7/2. So now that we have the coefficients for A B and C, let's go ahead and plug that in to the original partial fraction decomposition. Now keep in mind that the partial fraction decomposition is A over T plus one plus B X plus C over three T squared plus three. If we plug this in what we are at, what we end up with is negative 7/6 over T plus one plus 7/ T plus 3/2 over the quantity three T squared. Plus three. Now, normally we don't want to leave fractions in the numerator. So we can go ahead and just move the denominators of each of the fractions to the denominator of the overall fraction. Now, on the right hand side, since we have 7/2 T plus 3/2 T, we can go ahead and combine this into a single fraction doing this allows us to rewrite this as 70 plus 3/2. And moving each of the denominators to the overall denominator allows us to rewrite this as negative 7/6 times the quantity T plus one plus seven T plus three over the quantity two times three T squared plus three. Now there's one final step to do in this equation, we can go ahead and factor out a three from this quantity and multiply it with two. And that's going to allow us to rewrite the second denominator as six times the quantity T squared plus one. And this is going to be this fully simplified form of the rational expression. And with that being said, the answer to this problem is going to be B 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. (3x - 2)/((x + 4)(3x^2 + 1))

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Polynomial Factorization

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