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

Accepted Answer

Welcome back. I'm so glad you're here. We've got this rational expression which is X over X plus five times x minus four. And we are going to decompose it. So let's take a look at the denominator. We've got two factors here, X plus five and x minus four and they're both linear and and neither one of them repeats. So when that's the case, in order to write this in partial fraction decomposition form, we're going to set this equal to a constant term A Over our first linear factor X plus five Plus be a constant term over our second linear factor, which is X -4. And now we're going to go through a bunch of steps to solve for A&B. First step is that we're going to get rid of the fraction. We're going to multiply everything by the least common denominator, which is going to be the denominator on the left hand side. So we're going to multiply everything times X plus five times X minus four. And I'll rewrite that with that. So we're going to have X over X plus five times x minus four. And it's going to be multiplied in the numerator by X plus five, X minus four. And then on the right hand side it equals a over X plus five. And the A is also going to be multiplied by that whole thing X plus five times X minus four and the B. So plus B over X minus four. The B on top the numerator gets multiplied by the X plus five X minus four. And now the most beautiful part of algebra is when things cancel out. So the factor X plus five on top cancels out with this X plus five on bottom x minus four cancels out x minus four. X plus five for the A cancels out the X plus five. And with the B the x minus four cancels out the x minus four. So now we can rewrite this on the left hand side, it's just X On the right hand side. We have a times X - plus and then we've got B times X plus five and now we'll distribute. So X equals a X minus four, A plus B, X plus five B. Let's rewrite those with the X terms grouped together. So we've got X equals a X plus b, x minus four, A plus five B, R two constant terms with no excess. And now with this A X plus b, X, we can factor an X out in front. So we've got X times A plus B. And I'm going to write these two terms grouped together plus negative for a plus five B. Those are two constant terms and we know that coefficients we've got terms on both sides of the equal sign. The coefficients of them are going to be equivalent because they're on two sides of an equal sign. And so the coefficient of our X term over here, which is this invisible one is the same as the coefficient of our x term on the right hand side, which is A plus B even though it comes after, it's still the coefficient of X. We could write it before. So we can set those two things equal to each other. And I'm going to write the A plus B on the left hand side. A plus b equals r coefficient from the left hand side down here one. So we've got one equation that we can work with here. And now we've got this other coefficient negative for a plus five B. That's our constant term. And there's no constant term on the left hand side which means our constant term on the left hand side is zero. So we can set a zero equal to our right hand sides, constant coefficient which is negative for a plus five B. And now we've got two equations and we can set up a system of equations to solve for A and B. And one way to do that here is we can just add them together. We'll multiply this first equation. Bye. four. So then the a terms will cancel out. So four times a and then four times B and four times one. So at four A minus four B. When we add those two together and the I terms will cancel out four B plus five B. Well that gives us nine B and four times one is four plus zero will give us four. We can solve for B B By itself. We know that B equals 4/9. And now we can just plug that back into the first equation U substitution. So we're looking at A plus and fill in our B term A plus 4/9 equals one. We can get that a term by itself and we know that one minus 4/9. While A equals 5/9. We have our A. And R. B term solved. Now. We just need to plug them back in to our equation up top. So A was 5/9 and be as 4/9. And we can just rewrite this. So it's a little bit cleaner down here. We know that 5/9 over X plus five Plus 4/ over X -4 is a rational expression in its partial fraction decomposition form. We look at our answer choices and that's going to match with answer choice B. Well done. We'll catch you on the next one.

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