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. (6x-11)/(x − 1)²

Accepted Answer

Welcome back. I am so glad you're here. We are given the expression to d minus 13, divided by D minus nine squared. And we are asked to express this in its partial fraction decomposition. Alright, so recalling from previous lessons on partial fraction decomposition, what we're going to do is look at our denominator D minus nine squared. We're going to ask ourselves a few questions about it. First, is this linear or quadratic? And that D. That's to the first power. So this is linear. And our next question is, is our factor in the denominator, distinct or repeating that D -9. It's repeating. There are two of them. And the last question is, how many factors are there? There are two. Just the D minus nine and the D minus nine. So with that information we are now going to set this up as equal to because we have two factors will have two fractions and we'll work on the denominators first Because that D -9 is repeating the first fractions, denominator will be D -9 and the second fractions denominator will be D -9 squared. And then because both of these are linear factors, we are going to have constant terms as our numerator. So the constant term for our first fractions numerator is A and the second fractions numerator is the constant B. That's it. That's the setup we needed to do now we just need to solve for A and B. And rewrite this part of our equation with A and B filled in once we've solved for them. So two sulfur A and B. Let's get rid of this fraction. We're going to do that by multiplying everything by our common denominator are common. Denominator is the denominator from the left hand side. So that's D minus nine squared. And now we can rewrite this. We've got to D -13 times D -9 squared. All over D minus nine squared. On the left hand side equals a. Times D minus nine squared over. Just D minus nine Plus b. Times D -9 Squared, 19 Goodness T -9 Squared. And the b. is over D -9 squared. Okay now the fun part canceling things out the fraction on the left hand side of the equal sign in the numerator and the denominator both have d minus nine squared. So those cancel out for our first fraction on the right hand side A. Is being multiplied by two d minus nines. And the denominator only has one D minus nine. So one of those minus nines disappears and one stays for B. That has a D minus nine squared in both the numerator and the denominator. And so those factors cancel out. So now simplified, we have two D minus 13 equals A times D minus nine plus. Just be all right. Let's distribute that A. To what's in the parentheses there will bring down to D minus equals A D minus nine A plus B. Now the next step is to group our D terms and our constant terms on the right hand side. But that's already done for us. Our D term is just this AD and our constant terms are this minus nine A plus B. So they're already grouped together. And now we recall from those previous lessons on partial fraction decomposition that the coefficient of R. D terms on each side of the equal sign are equal to each other. So from the right hand side the coefficient of the D. Term is A. A. Is equal to the coefficient of the D. Term on the left hand side which is two, so A equals two. Now, for our constant terms, we know that our constant terms are equal to each other. The right hand side constant terms are minus nine A plus B. And that's equal to the constant terms on the left hand side, which is negative 13. And now we have two equations set up and we can use the system of equations to solve for A. And B. And this one is nice because A is already solved for us A equals two and we can use that A equals two to substitute in for the A. And our second equation. So this will be a negative nine times two instead of a plus B equals negative 13, negative nine times two is negative 18 plus B equals negative 13. And now we'll add 18 to both sides. So the left hand side of the equation is B equals and the right hand side negative 13 plus 18 is five. So we've solved for A and B. This is great. That's what our goal was. And now we're going to take that and fill the A. And the B. End to our original right hand side of our equation. And that will be our partial fraction decomposition. So instead of a. In the numerator, it's two divided by D -9 plus and instead of being the numerator, it's five divided by D - squared. And that is our final answer. We look at our answer choices in this matches with answer choice. A. Well done. We'll catch you on the next one.

In Exercises 9–42, write the partial fraction decomposition of each rational expression. (6x-11)/(x − 1)²

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Polynomial Long Division

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