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

Question

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (x³ + 4)/(9x³ - 4x)

Accepted Answer

Hello. Today we're gonna be using partial fraction decomposition to simplify the following rational expression. So what we are given is T cubed plus 7/4 T cubed minus T. Now one thing to note is that the leading exponent in the numerator has the same value as leading exponent in the denominator. What this means is that we need to simplify the expression by using long division. Now, in a previous video, we have discussed how to long divide two polynomials together. So if you need help with the long division, I recommend checking out the previous videos. But if you long divide the current expression, what you can rewrite it as is 1/4 plus 1/ times the quantity T plus over four cubed minus T. Now, what we wanna do is we want to use partial fraction decomposition to simplify this part of the expression. And whenever we use partial fraction decomposition, we always want to make sure that our denominator is in its simplest form. Currently, both of the terms contain at least a multiple of T. So we can factor out A T from the denominator and we can rewrite that portion of the expression as T plus 28 over T multiplied by the quantity four T squared minus one. Now, four T squared minus one can be further factored by using factoring of the difference of squares. And we can factor the denominator further to rewrite the expression as T plus 28 over the quantity T times two T plus one multiplied by two T minus one. Now that we have the simplified form of the denominator, we can go ahead and create our partial fractions. So let's create a partial fraction for the first multiple of T T is a non repeating linear term, which means that the first partial fraction can be written as a over T next two T minus one. I'm sorry, two T plus one is a non repeating linear quantity. So the second partial fraction can be written as B over the quantity two T plus one and the final, the final factor for T which is two T minus one is a non repeating linear quantity. So the final partial fraction can be written as C over two T minus one. Now, what we wanna do is you want to simplify this statement to eliminate all the denominators. That way we can solve for our A B and C coefficients. In order to do that, we want to multiply that statement by the lowest common denominator. Now, the lowest common denominator is going to be the product of all the different denominators. And in this case here, the lowest common denominator is going to be T multiplied by the quantity two T plus one multiplied by two T minus one. So let's go ahead and multiply each of these terms by the lowest common denominator that is going to give us T plus over T multiplied by two T plus one, multiplied by two T minus one multiplied by the L CD, which is going to be T times two T plus one times two T minus 1/1. And that is equal to A over T multiplied by the L CD, which is T times two T plus one times two T minus one over one plus B over two T plus one multiplied by the L CD, then plus C over two T minus one multiplied by the L CD. So now what we wanna do is we don't want to use cross reduction to eliminate the denominators. On the left hand side, we have two, I mean, we have T times two T plus one times two T minus one in both the numerator and denominator. So we can eliminate those quantities. In the first partial fraction. We have T in both the numerator and denominator. So we can eliminate those quantities. In the second partial fraction. We can eliminate two T plus one from the numerator and denominator. And in the third partial fraction, we can eliminate two T minus one from the numerator and denominator that is going to leave us with a statement T plus 28 is equal to a multiply by two T plus one times two T minus one plus B multiplied by T times two T minus one plus C multiplied by T multiplied by two T plus one. Now, there's a bit of simplification that we can do to this statement first, if we multiply two T plus one times two T minus one, we can go ahead and rewrite this as a multiplied by the quantity four T squared minus one. If we multiply B N T together, we can simplify this to give us BT multiplied by two T minus one and we can multiply C N T together to give us C T multiplied by two T minus one. Now, what we wanna do is we want to distribute our, a term into four T squared minus one. Our BT term to two T minus one and our C T term to two T plus one. This is going to simplify the statement to give us T plus 28 is equal to four A T squared minus A plus two BT squared minus BT plus two C T squared plus C T. Now, what I recommend doing is grouping up all of our T square terms together, all of our T terms together and all of our constants together on the right hand side. And what we get on the right hand side is four A T squared plus two B T squared plus two C T squared plus negative BT plus C T and then plus negative A. And what we can do is we can factor out the T squared terms from the first group and the T term from the second group that is going to simplify the equation to give us T plus 28 is equal to T squared multiplied by the quantity four A plus two B plus two C plus T multiplied by negative B plus C plus negative A. Now, here's the tricky thing about solving for A B and C. We're gonna go ahead and set the coefficients of T squared on both the left and right hand side equal to each other. We're gonna set the coefficients of the T terms equal to each other and we're gonna set the constants equal to each other. And the reason why we do this is because we want to create a system of equations to help us solve for A B and C will notice that we don't have a T squared term on the left hand side of the equation. What this means is that the T squared term on the left hand side has a coefficient of zero. And the first equation that we can come up with is for A plus two B plus two C is equal to zero. Now let's go ahead and set the coefficients of the T terms together on the left hand side T has a coefficient of one. And the second equation we can make is negative B plus C is equal to one. Finally, we're going to set the constants equal to each other. And in this case, that's going to be negative A equal to 28. Now, in the third equation, we can solve for a by dividing both sides of the equation by negative one. And that is going to give us A is equal to negative which is going to be the first coefficient. Now that we have a value for a, we can plug it in to the first equation. And that will give us four multiplied by negative 28 plus two B plus two C equal to zero four, multiplied by negative 28 is going to give us negative 112. And if we add both sides of the equation by 112 we can create a new equation which is two B plus two C is equal to 112. We can further simplify this by dividing each of the terms by two. And we can reduce this equation to B plus C is equal 2 56. This is going to be our new equation one. And notice that our second equation is in the form of a polynomial comprised of B N C terms. So we can go ahead and sol for B N C by adding our new equation one with the equation two. And that is going to give us B plus C is equal to 56 plus negative B plus C is equal to one. If we add these terms together, the B terms are going to cancel each other out C plus C is going to give us two C and 56 plus one is going to give us 57. And if we divide both sides of the equation by two, we get C is equal to the value of 57/2. This is going to be the second coefficient we need for the partial fractions. Now that we have a value for C, we can plug this into equation two to sol for B. So plugging in C two equation two is going to give us negative B plus 57/2 equal to one. If we subtract both sides of the equation by negative 57/2, that is going to give us the value negative B is equal to negative 55/2. And if we divide both sides of the equation by negative one, we get B is equal to 55/2. So what we just saw for is the coefficient of A B and C. And now we can go ahead and plug this in to the original partial fractions that we came up with earlier. Now let's make a note we know that A is equal to negative we know that B is equal to 55/2. And we know that C is equal to 57/2. And the original partial fractions we came up with is A over T plus B over two T plus one plus C over two T minus one. If we plug these values into the partial fractions, we get negative 28 over tea plus 55/2, over two T plus one plus S 57/2, over two T minus one. Now, for the terms that have the fractions in the numerator, the easiest way or trick to simplifying, this is to take the denominator in the numerator and push it down to the overall denominator and we can rewrite the partial fractions as negative 28 plus 55/2 multiplied by the quantity two T plus one plus over the quantity two multiplied by two T minus one. So this is going to be the simplified form of our partial fractions. But keep in mind that earlier, we were simplifying T plus 28/4 T cubed minus T. So we need to take our current partial fractions and plug it in for this value. This is going to give us plus 1/4 multiplied by partial fractions which is negative 28 over T plus 55/2, multiplied by two T plus one plus 57 over two multiplied by two T minus one. Now, what we're going to do is we're going to distribute 1/4 to each of the quantities and that is going to give us 1/ plus, I'm sorry, not plus minus 28/ plus 55/8, multiplied by two T plus one plus 57/8, multiplied by two T minus one. And for the second partial fraction 28 4 can be simplified to give us seven over T. So this is going to be this fully simplified form of the original expression. We have 1/4 minus seven over T plus 55/8 multiplied by two T plus one plus 57/8 multiplied by two T minus one. And with that being said, the answer to this problem is going to be D. 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. (x³ + 4)/(9x³ - 4x)

Key Concepts

Partial Fraction Decomposition

Factoring Polynomials

Setting Up and Solving Systems of Equations

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Find the partial fraction decomposition for each rational expression. See Examples 1–4. (x3 + 4)/(9x3 - 4x)

Key Concepts

Partial Fraction Decomposition

Factoring Polynomials

Setting Up and Solving Systems of Equations

Watch next

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (x³ + 4)/(9x³ - 4x)