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

Question

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

Accepted Answer

Hello, today we're gonna be using partial fraction decomposition to simplify the following rational expression. So what we are given is eight P plus four over the quantities P plus two times five P minus six. So before we start our partial fraction decomposition, we always want to make sure that our denominator is in a fully factored form. Currently, our denominator is fully factored. So we can go ahead and immediately set up our partial fractions. So we want to set up a partial fraction for every factor of P in the denominator. The first factor of P is going to be P plus two. This is a non repeating linear quantity and that is going to give us the partial fraction of A over P plus two. Now, we want to set up a partial fraction for the second factor of P which is going to be five P minus six. This is also a non repeating linear quantity and this is going to give us the partial fraction of B over five P minus six. So now what we want to go ahead and do is you want to solve for the coefficients of A and B in order to do that, we can go ahead and simplify this overall equation. We set up by multiplying the entire thing by a lowest common denominator in order to eliminate the denominators. Now, the lowest common denominator in this case is going to be P plus two times five P minus six. So we're going to multiply each term of the equation by the L CD. This is going to give us eight P plus four over P plus two times five, P minus six times the L CD, which is P plus two times five, P minus 6/1. And that is going to equal to A over P plus two times the L CD, which is P plus two times five P minus 6/ plus B over five, P minus six times the L CD, which is P plus two times five P minus six all over one. Now, we can go ahead and use cross reduction to eliminate the denominators. In the first term, we have P plus two times five P minus six in both the numerator and denominator. So we can just go ahead and eliminate those quantities. In the second term, we have P plus two in both numerator and denominator. So we can just go ahead and eliminate that. And in the third term, we have five P minus six in both numerator and denominator. So that's going to get eliminated. So what we are left with is the equation A P plus four is equal to A times the quantity five P minus six plus B times the quantity P plus two. So now what we want to go ahead and do is we want to simplify the right hand side of this equation. In order to do that, we're going to distribute A into five P minus six and distribute B into P plus two. This is going to give us the equation A five A P minus six A plus B P plus two B. Now, at this point, I would recommend reordering your terms. So that way you have all the P terms together and all the constant terms together reordering the right hand side is going to give us five A P plus B P minus six A plus two B. Now, what we can go ahead and do is we can factor out A P from the first two terms and group up our constants together doing this allows us to rewrite the equation as eight P plus four is equal to P times five A plus B plus the constant, which is negative six A plus two B. So this is where the tricky part of partial fraction decomposition comes in. We want to go ahead and use this equation to set up a system of equations in order to solve for A and B. Well, in order to do that, we want the coefficients of P on the right hand side which in this case is plus B to equal to the coefficients of P on the left hand side, which in this case is going to be eight. So we can go ahead and set up the first equation as five A plus B is equal to eight. And we're gonna do the same thing with the constant, the constants on the right hand side are going to equal to the constants on the left hand side. And we're going to get the second equation which is negative six A plus two B is equal to four. Now, we can go ahead and solve this as a system of equations. Let's go ahead and start by eliminating the B term. Now in the second equation, we have positive two B and the first equation we have positive B. If we multiply the first equation by negative two, we can rewrite the first equation as negative 10 A minus two B is equal to negative 16. And we're going to be adding the second equation which is negative six A plus two B is equal to four, negative two B plus two B is going to cancel each other out negative 10 A minus six A is going to give us negative 16 A and negative 16 plus four is going to give us negative 12. If we divide both sides of this equation by negative 16, we get A is equal to positive 12/16 which is going to simplify to give us 3/4. And this is going to be the first value for A. So now that we have a value for A, we can plug this into either of the equations in order to get a value for B. If we plug an A to the first equation, we get five times 3/4 plus B is equal to eight, five times 3/4 is going to give us 15/4 plus B is equal to eight. And if we subtract 15/4 to both sides of the equation, that is going to give us the value of B is equal to 17/4, which is going to be the second coefficient for our partial fractions. Now that we have the values of A and B, we can go ahead and plug this into our partial fractions. Recall that the partial fractions we came up with is A over P plus two plus B over five P minus six. If we plug in A and B, we get 3/4 over P plus two plus 17/4, over five P minus six. So what we currently have is a sum of complex fractions. A shortcut into simplifying the pro the complex fractions is to take the denominator in the overall numerator and move it to the overall denominator. If we do this to both of the complex fractions, we can rewrite the partial fractions as 3/ times P plus two plus 17/4 times five P minus six. And this is going to be the fully simplified form of the rational, of the original 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. (4x + 2)/((x + 2)(2x - 1))

Key Concepts

Partial Fraction Decomposition

Factoring Denominators

Setting Up and Solving Equations for Coefficients

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