In Exercises 16–24, write the partial fraction decomposition of e...

Question

In Exercises 16–24, write the partial fraction decomposition of each rational expression. (4x^2 - 3x - 4)/x(x + 2)(x - 1)

Accepted Answer

Welcome back. I'm so glad you're here. Were given a rational expression, let's rewrite that. We have X squared minus two X plus one over these three factors X times X plus one times x minus one. And before we start thinking about how we're going to decompose that, let's take a look at the numerator and see if we could factor that at all. So we take a look and we could we could do X times X and we could do minus one minus one. Will give us that negative two, X and the plus one. So we can rewrite that With the new numerator as X -1 times X -1. And that means that we can cancel out one of the factors in the denominator. Great. That's a little bit easier. So let's rewrite that. We've now got x minus one in the numerator and X times X plus one in the denominator. Great. Now, our next step, we're going to look at the denominator and see what form we're going to use for partial fraction decomposition. And it's got to factors here, they are both linear. They are both non repeating their prime. We can't factor those out anymore. So we're going to set this equal to a over the first factor which is X plus B over the second factor, which is X plus one. And now it's just a matter of solving for A and B. We're going to take all three of these terms. And we're going to multiply them by the lowest common denominator which is going to be the denominator of the term on the left hand side. So we're going to multiply everything times X times X plus one. And rewriting that now over on the left hand side, that denominator is going to cancel out what's being multiplied by the numerator. So we'll just have x minus one on the left equals for the A. Well the X in the denominator is going to cancel out the X being multiplied by the numerator. So it's only the A. Is only going to be multiplied by the X plus one plus our B term. That X plus one in the denominator of R. B is going to cancel out the X plus one being multiplied by the B. So it's only being multiplied by the X. And now let's distribute that a term. So x minus one equals a, X plus A. And bring down our plus B. X. Let's rewrite this with our X terms together. So x minus one equals a, X plus b, X plus A. And now we can factor an X out of our two X terms. So X times A plus B. And then just bring everything else down plus a. And that's all equal to X -1. Great. This is the format we are looking forward to say. We know that the coefficients of the terms on both sides of the equal sign are going to be equal to each other. That means the coefficient of this X term here on the left is equal to the coefficient of this X term over here. The coefficient of that, even though it comes afterwards is A plus B. So we can say A plus B. And I'm writing the right hand side on the left hand side. Now A plus B equals and the coefficient of this x term on the left is one. So I'm going to write one, appear on the right hand side. Now let's look at our constant terms. Well on the left hand side are constant, is negative one. So I'll write that on the right hand side up here negative one. And the coefficient constant here is A. So A equals negative one. Well, this is great. We have solved for a We know that A equals negative one and we can plug that in for our first equation here, negative one plus B equals one will add wonderful sides. And so B equals two. Great B equals two A equals negative one. And now we can just plug those in on top. So they will have a negative one over X and b will have a two over X plus one. And we'll rewrite that one more time down here. Negative one over X plus two over X plus one. That's as simplified as that one is going to get. We look at our answer choices and that matches with answer choice. D Well done. We'll catch on the next one

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