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. (3x +50)/(x -9)(x +2)

Accepted Answer

Welcome back, I am so glad you're here. We are given the expression five X plus 19 divided by x minus one times X plus three. And we are asked to express this in its partial fraction decomposition. Alright, When we are dealing with partial fraction decomposition, we recall from previous lessons that we're going to take a look at the denominator and we're going to ask ourselves a few questions about the factors of the denominator. So we've got x minus one and X plus three. The first question is, are those factors linear or quadratic? And those are both with a degree of one. So they are both linear factors. Next we ask ourselves are these factors distinct or repeating and they're different from each other. They are distinct. And the last question is, how many factors are there? And there are two of them. And because there are two of them, we are going to set this fraction equal to two new fractions added together. The denominators of these two new fractions are going to be the two distinct factors from our previous fractions denominator. So the first new fractions denominator will be x minus one and the second new fractions denominator will be X plus three. And because these are both linear factors, both of their numerator are going to be constant terms. So for X -1, the numerator is going to be the constant term capital a and for the second fraction, the numerator will be capital B. And now we just need to solve for A and B. And once we've solved for A and B. Then we'll replace those here and this will be our partial fraction decomposition. Wonderful. Now, in order to solve for A and B, we are going to first get rid of these fractions and in order to do that we're going to multiply all three fractions by the lowest common denominator. The lowest common denominator comes from the denominator on the left hand side of the equation, so the lowest common denominators, X minus one times X plus three. And multiplying those two factors by everything that will look like F or five X plus times X minus one times X plus three. All over x minus one, that's not a minus x minus one times X plus three equals a capital A Times X -1 times x plus three. All over x minus one plus a capital B Times X - times x plus three. All over X plus three. And now the best part in any fraction where there is a factor occurring in both the numerator and the denominator, that factor gets canceled out. So for the fraction on the left hand side of the equal sign, the x minus one is cancel out and the X plus three's cancel out for the first fraction on the right hand side of the equation, the x minus one's cancel out. And for the second fraction on the right hand side of the equation X plus three's cancel out. So now this reads five X plus 19 equals capital A times X plus three plus capital B times X minus one. Now we can distribute those capital letters on the right hand side of the equation. So distributing the A. That will be A X plus three A. Plus distributing the B, b x minus B. And now I'd like to re group the X terms together and our constant terms together. So bringing down the left hand side, five X plus 19 equals now bringing the X terms together A X plus B, X. And bringing the constants together plus three A minus B. And the last step here on this section of the work is I'd like to factor out an X from our X terms. So five X plus 19 equals X times A plus B plus three a minus B. Now we recall from previous lessons on partial fraction decomposition that when we have like terms on either side of the equation, their coefficients are going to be equal to each other. So on the right hand side of the equation, the coefficients of x are a plus B. An A plus B is equal to the coefficient of X. On the left hand side of the equation, A Plus B is equal to five. Likewise, our constant terms are going to be equal to each other. So on the right hand side of the equation, three a minus B Is equal to the constant term on the left hand side of the equation, so that is equal to 19. Now we have two equations. And from this system of equations we can solve for A and B. And then fill them back in for our partial fraction decomposition. Alright well we can go ahead and add these two equations together because then RB terms will cancel out and we can solve for A. A plus three. A. Is going to give us four. A. B terms cancel out. And that equals 19 plus five, so that equals 24 Dividing both sides by four. We get an answer of a equals six. Alright now we can go back to equation one and substitute A. Six in for that. A. So instead of A plus B it's six plus B equals five and we can subtract six from both sides of the equation and we get B equals five minus six is negative one. So we have solved for A and B. Now we can fill those in to our partial fraction decomposition up here and replacing A with six. We have six divided by x minus one plus replacing B with negative one. So negative one divided by x plus three. We look at our answer choices and this matches with answer choice B. Well done. We'll catch you on the next one

In Exercises 9–42, write the partial fraction decomposition of each rational expression. (3x +50)/(x -9)(x +2)

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Factoring Polynomials

Watch next