Let's see if we can learn athing or two about partial fraction expansion, orsometimes it's called partial fraction decomposition. The whole idea is to takerational functions-- and a rational function is just afunction or expression where it's one expression divided byanother-- and to essentially expand them or decomposethem into simpler parts. And the first thing you've gotto do, before you can even start the actual partialfraction expansion process, is to make sure that thenumerator has a lower degree than the denominator. In the situation, the problem,that I've drawn right here, I've written right here,that's not the case. The numerator has the samedegree as the denominator. So the first step we want to doto simplify this and to get it to the point where thenumerator has a lower degree than the denominator isto do a little bit of algebraic division. And I've done a video on this,but it never hurts to get a review here, so to do that, wedivide the denominator into the numerator to figure out theremainder, so we divide x squared minus 3x minus 40 intox squared minus 2x minus 37. So how many times? You look at the highest degreeterm, so x squared goes into x squared one time, one timesthis whole thing is x squared minus 3x minus 40, and now youwant to subtract this from that to get the remainder. And see, if I'm subtracting, soI'm going to subtract, and then minus minus is a plus, a plus,and then you can add them. These cancel out. Minus 2x plus 3x, that's x. Minus 37 plus 40,that's plus 3. So this expression up here canbe rewritten as-- let me scroll down a little bit-- as 1 plus xplus 3 over x squared minus 3x minus 40. This might seem like some typeof magic thing I just did, but this is no different than whatyou did in the fourth or fifth grade, where you learned howto convert improper regular fractions into mixed numbers. Let me just do a littleside example here. If I had 13 over 2, and I wantto turn it into a mixed number, what you do-- you can probablydo this in your head now-- but what you did is, you divide thedenominator into the numerator, just like we did over here. 2 goes into 13. We see 2 goes into 13 sixtimes, 6 times 2 is 12, you subtract that from that,you get a remainder of 1. 2 doesn't go into 1, sothat's just the remainder. So if you wanted to rewritethis, it would be the number of times the denominator goes intothe numerator, that's 6, plus the remainder overthe denominator. Plus 6-- plus 1 over 2. And when you did it inelementary school, you would just write 6 1/2, but 6 1/2 isthe same thing as 6 plus 1/2. That's exactly the samething we did here. The denominator went to thenumerator one time, and then there was a remainder of x plus3 left over, so it's 1 plus x plus 3 over this expression. Now we see that that numeratorin this rational expression does have a lower degreethan the denominator. The highest degree here is 1,the highest degree here is 2, so we're ready to commence ourpartial fraction decomposition. And all that is, is taking thisexpression up here and turning it into two simpler expressionswhere the denominators are the factors of this lower term. So given that, let'sfactor this lower term. So let's see. What two numbers add up tominus 3, and when you multiply them, you get minus 40? So let's see. They have to be differentsigns, because when you multiply them you get anegative, so it has to be minus 8 and plus 5. So we can rewrite this up hereas-- I'll switch colors-- 1 plus x plus 3 over xplus 5 times x minus 8. 5 times 8 is minus 40-- 5 timesnegative 8 is minus 40, plus 5 minus 8 is minus 3,so we're all set. Now I'll just focus onthis part right now. We can just rememberthat that 1 is sitting out there out front. This is the expression wewant to decompose or expand. And we're going to expand itinto two simpler expressions where each of these are thedenominator-- and I will make the claim, and if the numberswork out then the claim is true-- I'll make the claim thatI can expand this, or decompose this, into two fractions wherethe first fraction is just some number a over the first factor,over x plus 5, plus some number b over the secondfactor, over x minus 8. That's my claim, and if I cansolve for a and b in a way that it actually does add up tothis, then I'm done and I will have fully decomposedthis fraction. I guess is the way-- Idon't know if that's the correct terminology. So let's try to do that. So if I were to add thesetwo terms, what do I get? When you add anything, you findthe common denominator, and the common denominator, the easiestcommon denominator, is to multiply the two denominators,so let me write this here. So a over x plus 5 plus bover x minus 8 is equal to-- well, let's get the commondenominator-- it's equal to x plus 5 times x minus 8. And then the a term, we would--a over x plus 5 is the same thing as a times x minus8 over this whole thing. I mean, if I just wrote thisright here, you would just cancel these two terms out andyou would get a over x plus 5. And then you could add that tothe common denominator, x plus 5 times x minus 8, and itwould be b times x plus 5. Important to realize,that, look. This term is the exact samething as this term if you just cancel the x minus 8 out, andthis term is the exact same thing as this term if you justcancel the x plus 5 out. But now that we have an actualcommon denominator, we can add them together, so we get-- letme just write the left side here over-- a over xplus 5-- I'm sorry. I want to write this over here. I want to write x plus 3 overplus 5 times x minus 8 is equal to is equal to the sum ofthese two things on top. a times x minus 8 plus b timesx plus 5, all of that over their common denominator, xplus 5 times x minus eight. So the denominators are thesame, so we know that this, when you add this together,you have to get this. So if we want to solvefor a and b, let's just set that equality. We can ignore the denominators. So we can say that x plus 3is equal to a times x minus 8 plus b times x plus 5. Now, there's two ways tosolve for a and b from this point going forward. One is the way that I wasactually taught in the seventh or eighth grade, which tends totake a little longer, then there's a fast way to do it andit never hurts to do the fast way first. If you want to solve for a,let's pick an x that'll make this term disappear. So what x would makethis term disappear? Well, if I say x is minus 5,then this becomes 0, and then the b disappears. So if we say x is minus 5-- I'mjust picking an arbitrary x to be able to solve for this--then this would become minus 5 plus 3-- let me just write itout, minus 5 plus 3-- is equal to a times minus 5 minus 8--let me just write it out, minus 5 minus 8-- plusb times minus 5 plus 5. And I picked the minus 5 tomake this expression 0. So then you get-- pick abrighter color-- minus 5 plus 3 is minus 2, is equal to--what is this?-- minus 13a plus-- this is 0, right? That's 0. Minus 5 plus 5 is 0, 0 times bis 0, and then you divide both sides by minus 13, you get--negatives cancel out-- you get 2 over 13 is equal to a, andnow we can do the same thing up here and get rid of the a termsby making x is equal to 8. If x is equal to 8, you get xplus 3 is equal to 11, is equal to a times 0 plus b times--what's 5-- 8 plus 5 is-- plus b times 13. Their b looks a bit like a 13. And then you get 11 is equal to13b, divide both sides by 13, you get b is equalto 11 over 13. So we were able to solvefor our a's and our b's. And so we can go back toour original equation and we could say, wow. This just has to be equal to2 over 13, and this just has to be equal to 11 over 13. So our original, our veryoriginal thing we wrote up here, can be decomposed into 1,that's this 1 over here, plus this, which is 2 over 13-- I'lljust write it like this for now-- 2 over 13, over x plus 5. You could bring the 13 downhere if you want to write it so you don't have afraction over a fraction. Plus 11 over 13 times--over x minus 8. And once again, you could bringthe 13 down so you don't have a fraction over a fraction. But we have just successfullydecomposed this pretty-- I don't want to say that wenecessarily simplified it, because you could say, oh, weonly have one expression here, now I have three-- but I'vereduced the degree of both the numerators and thedenominators. And you might say, well,Sal, why would I ever have to do this? And you're right. In algebra you probably won't. But this is actually a reallyuseful technique later on when you get to calculus, andactually, differential equations, because a lot oftimes it's much easier-- and I'll throw out a word here thatyou don't understand-- to take the integral or theantiderivative of something like this, thensomething like this. And later, when you do inverseLaplace transforms and differential equations, it'smuch easier take an inverse Laplace transform ofsomething like this than something like that. So anyway, hopefully I've givenyou another tool kit in your-- or another tool in your toolkit, and I'll probably do a couple more videos because wehaven't exhausted all of the examples that we could wecould show for partial fraction decomposition.