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Exercises 5–8, use inverse matrices to find the partial fraction decomposition.$$ rac{x-3}{x(x+3)}= rac{A}{x+3}+ rac{B}{x}$$...

Question

Exercises 5–8, use inverse matrices to find the partial fraction decomposition.$$ rac{x-3}{x(x+3)}= rac{A}{x+3}+ rac{B}{x}$$

Exercises 5–8, use inverse matrices to find the partial fraction decomposition. $$\frac{x-3}{x(x+3)}=\frac{A}{x+3}+\frac{B}{x}$$



Answers

Exercises 5–8, use inverse matrices to find the partial fraction decomposition.
$$\frac{x-3}{x(x+3)}=\frac{A}{x+3}+\frac{B}{x}$$

So in this problem for being asked to find the partial fraction decomposition of are given fraction. Well, in order to do this, we first must factor are denominator. Well, when we go, the factor are denominator. We have a GC off of an ax. So my new Marie there is going to stay to save X minus three over. I'm gonna faster on the next. And then I'm left with X squared plus three. Well, that can't be factored any fervor. So now I can get I can start my partial fractions decomposition. So our first factor in that denominator is just this linear term of X. So my first fraction will be a over Rex. Now, for my second fraction, I have my eyes next factor, which is X squared plus three, which is an irreducible quadratic factor. So to put this in the partial fraction decomposition in the numerator, I'm gonna have to have bx plus c all over X squared plus three. And again, keep in mind. This is all equal to X minus three all over x times X squared plus three. So what I'm gonna do next as I'm going to multiply both sides of our equation by our common denominator. So I'm gonna multiply by X times X squared plus three to both sides. And the reason why I'm doing this is because that way, on the right hand side of the equation, my denominator is going to cancel each other up. Now, let's look on the left hand side. Well, we have the first fraction a over X times are common denominator. While already has the X term is just missing the X squared plus three term. So we're gonna need to multiply a five x squared plus three. Now, for our second term, we have BX plus C all over X squared plus three. Well, it already has the X squared plus three factor, but it's missing the extra. So I'm gonna need to multiply my new Marie there BX plus C by X, and it's equal to just X minus three. So on the left hand side, I'm going to go ahead and distribute Well, that's going to give me a X squared plus three a. And then I'm gonna have plus beat X squared plus c x. And it's all equal to X minus three. So I'm going to next Get my life terms together. So I'm gonna start with my ex Square terms. Why have a X squared plus BX word? So again, a X squared plus BX where then I'm gonna get my ex terms together, which in this case, is just plus C X And then my constant term is just plus three. So if I look at those x square terms, I have a X squared plus bx word, so that should give us the coefficient of our X squared term in our solution. But there is no X word which tells us that a plus B must equal to zero. Now, if I look at that ex term C X, that should equal to the coefficient of the exterminator. Answer will. The coefficient of that is just one. So now we know that C is equal to one now for a cost. Interns, I've got the fringes, Dale. Our constant term is just three A well, that should equal to the constant term in our answer, which is negative. Three. So we know that three a must equal to negative three, which means that a must equal to negative one. So now we have see, we have a and now I can get B by substituting negative one into my equation. A plus B a zero. So I'll have negative one. Plus B is equal to zero, which means that B is equal to one. And now that I have A, B and C, I could substitute these values back into my prop partial fractions. So our final answer is going to be negative. One over X plus one X plus one all over X squared, plus three now in that new Marie that you don't have to put the one in front of the ex, but it's perfectly fine if you d'oh!

We begin by, I'll turn the denominator so they haven't spliced five x plus three. And this is equal to what the composition process began. So the alone a host and then there No and B X plus three many more supply both sides by the mist comin denominator. Okay. And so that will give us for X. Plus three is equal to eight times its plus three Cosby Times X plus, Dr. All right, we cannot let ex cool too. Negative. Three. No, that that would make this expression go to 00 And so, um, we have we can sell for B. Basically. So we have four times minus three plus three eyes equal to be times two donated. Three plus five is too. Okay, so this will enquire that if you do the arithmetic, you'll get b A sickle to negative 92 Good. So, similarly, we can that x equal to 95. And this would allow us this would make this expression go to zero, be zero. And Israel losses so far, eh? So we have four times negative. Five plus three is equal to, um a I'm staying too. Okay, so this would imply that a is equal to 17 over two. When you did a written, Who's to say This is big? You have everything we need, the unknowns. And so I would decomposition would be 17 divided by two over X. That's five minus nine over too over. Uh, thanks last year, that's all.

So we have this one once again, it's proper because the generator, the eggs in the numerator has a the biggest power of accident. Numerator is one the biggest out of eggs that nominators too. So all we gotta do is factor the denominator. And so you have this one, right? Two numbers. When you add, you get in. Uh, six. When you add you getting negative, five multiply, you get in. Uh, positive six. So is negative. Six. So your two numbers are explained. It three explains to because negative three times negative to his positive. Six negatives. Three minus two is negative. Five. Then we're just gonna replace the numerator by two constants yet to determine. All right. And then we're gonna solve the new Raiders. So we have just the new readers to evaluate, right? This is a thanks ways to you. Plus me. Thanks. Minus three. Okay, so this is most playing, Wilson. This is multiplying that, and then we're gonna eliminate, so we're gonna let XB two. So that leads me to eliminate a right. So there's gonna be three times to minus one a two minus you than plus B. Uh, do you minus three that this is zero. And this is to you three times to use. Six minus one is five. Right? And then this is negative, B because two minus three is negative one. So be is negative. Five. All right, How do you find a We have to choose eggs such that we're gonna eliminate. Be. So when you choose eggs here to be positive three, you can eliminate the right. So this is gonna be three times three minus one, because a three minus two plus B three minus three, three minutes, 30 So we're getting a right, because this is three times three is nine minus one is pain, and this is just a because three minus two is one. Right? So a is eight. And these negative five. So, what we have here, I'm just getting rid of this and sit in my ace eight. Right? And these negative five

Okay. In order to determine the decomposition of the partial fraction, we know that there we know they're two constants and be multiply both sides by X minus eight times Expose three we obtain X minus eight times X plus three Gives us just be over X plus three. Giving us the equations reacts 1 79 There's a terms ex was be terms are X minus eight. Plugging ex sequence ate into the equation We eventually obtained be equals eight, which gives us negative five over X months eight plus eight over X Plus three, which we send right onto eight over X plus three minus five over axe mons Hate.


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