In Exercises 9–42, write the partial fraction decomposition of ea...

Question

In Exercises 9–42, write the partial fraction decomposition of each rational expression. (x^2-6x+3)/(x − 2)³

Accepted Answer

Hello. Today we're going to be performing partial fraction decomposition on the given expression. So what we are given is to d squared minus 23 D plus 58 Over the quantity D minus five cubed. Now before we start our partial fraction decomposition, we always want to check to make sure that our denominator is fully factored. So D minus five is already in its simplest form as it cannot be factored any further. So our denominator is fully factored. Now let's go ahead and check to see what kind of partial fractions the denominator produces D minus five is a linear quantity as its leading coefficient has an exponent of one. However, this quantity is raised to the third power or in other words it's being repeated three times. So this linear quantity is going to produce three partial fractions and those partial fractions are going to be a over d minus five plus B, over d minus five squared Plus C Over D -5 Cubed. And now what we want to go ahead and do is sell for a B and C. However, in order to do that we need to get rid of all of our denominators. In order to get rid of the denominators, we need to multiply by the lowest common denominator and the L C. D. In this case is going to be d minus five cubed. So let's just go ahead and write out what it would be like to multiply everything by the L C D. So we have two D squared minus 23 D plus 58 Over D -3 d -5. I'm sorry cubed times the L. C. D. Which is d minus five cubed over one is equal to our first partial fraction which is a over d minus five times the L. C. D. Which is d minus five cubed over one plus our second partial fraction which is B over D minus five squared times the L. C. D. Which is d minus five cubed over one plus C. Over d minus five cubed times the L. C. D. Which is d minus five cubed over one. Now multiplying the left hand side by the L. C. D. Is going to completely cancel out the denominator. Multiplying our first partial fraction by the L. C. D. Is going to cancel out the d minus five in the denominator But leave us with D -5 squared in the numerator. Multiplying the second partial fraction by the L. C. D. Is going to cancel a denominator. Believe us with one instance of d minus five in the numerator and multiplying the last, the last partial fraction by the L. C. D. Is going to completely cancel out the denominator. So what we have is two D squared minus D plus 58 is equal to a times d minus five squared Plus B. Times D -5 Plus C. And now what we want to go ahead and do is simplify the right hand side. In order to do that we're gonna foil D minus five squared. And we're going to distribute B into d minus five. And doing this is going to give us to d squared minus 23 D plus 58 is equal to A. Times D squared minus 10 D plus Plus B. Times D -5 B Plus C. And there's one more thing we need to do. We need to go ahead and distribute A into d squared minus 10 D. Plus 25. And this is going to give us two D. Squared minus 23 D. Plus 58 is equal to A. Times D squared minus 10 A. D. Plus 25 a Plus B. Times D -5 B Plus C. And what we want to go ahead and do is just group up all of our common terms together. So we're gonna group up all of our D squared terms together all of our D. Terms terms together and all the constants together. And doing this is going to give us to D squared minus 23 D. Plus 58 is equal to a. D. Squared plus negative A. D. Plus five plus b. d. Plus 25 A - B Plus C. And now what we want to go ahead and do is just factor out the common term in our middle group as the common term is just going to be D. And this is going to give us two D squared minus 23 D. plus 58 is equal to a D squared plus D, times negative 10 A plus B Plus 25 a -5 B Plus C. And now we can go ahead and use this to set up a system of equations in order to solve for a B and C. And in order to do that, we're going to take the coefficients of R D square term and set them equal to each other. We're gonna take the coefficients of R. D term and set them equal to each other. And we're gonna take all of our constants and set them equal to each other. So our first equation is going to be the the coefficients of D squared on the right hand side, the coefficient is a and on the left hand side the coefficient of the squared is two. So we get A is equal to two, which solves one of our coefficients. For a second equation, we're taking the coefficients of R D terms, which on the right hand side is negative 10 A plus B. And the coefficient of the on the left hand side, which is negative 23 so what we get is negative 10 A plus B Is equal to -23. And for a third equation, we're taking all the constants on the right hand side and setting it equal to the constant on the left hand side and that's gonna be 25 A minus five B plus C is equal to 58. And we can go ahead and use the system of equations to solve for the coefficients. Now, by this point we should be comfortable with solving a system of equations at once. We do solve a system of equations. We're going to get A is equal to two, B is equal to -3 and C is equal to -7. And now we want to go ahead and plug in those coefficients into the original partial fraction decomposition. Now recall that our original partial fractions was a over d minus five plus B Over D - Squared plus C over d minus five cubed, replacing A with two is going to give us two over d minus five, replacing B with negative three is going to give us negative three over d minus five squared but we can go ahead and move the negative sign to the front of the fraction. So we get minus three over d minus five squared and we can go ahead and replace C with negative seven and move the negative sign to the front of the fraction. So we get minus seven over d minus five cubed. And this is going to be the complete partial fraction decomposition forgiven expression and that means that the answer to this problem is going to be be so I hope this video helps you in performing partial fraction decomposition and I'll go ahead and see you all in the next video.

In Exercises 9–42, write the partial fraction decomposition of each rational expression. (x^2-6x+3)/(x − 2)³

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Polynomial Long Division

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