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Intro to partial fraction expansion
Sal explains what partial fraction expansion is by rewriting (x²-2x-37)/(x²-3x-40) as the sum of 1 and two rational expressions with linear denominators. Created by Sal Khan.
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- I'm confused how he can just substitute in 8 and -5 for x. I get why but not how that works. Doesnt that mean that the expanded fraction only applies to when x is equal to 5 or -8? But thats indeterminate.... I don't know(23 votes)
- Unlike a system of equations where there are only unknown constants as variables, this kind of problem has unknown constants (what you're solving for) and a independent variable. Because of this, the statement must be true for all x. Thus, if it works for one specific x, it works for all x. We choose those two particular x values because they simplify the math a lot.(35 votes)
- How do we know that a decomposed part of the fraction (the A or B to solve for) is going to be a constant term? I understand the general method of decomposition, but why can we assume that?(13 votes)
- THe degree of the denominator always has to be bigger than the degree of the numerator. Therefore, if there is a LINEAR term (ax+b) in the denominator, then the numerator has to be a constant (A, B, C, whatever you wanna call it).
If there is a QUADRATIC term in the denominator (ax^2 +bx+c), then the numerator can be either LINEAR OR CONSTANT, since a quadratic equation has degree 2. When you do this in a decomposition, then you put Bx+C in the denominator (or whatever variables you wanna use). If the numerator ends up being a constant, then B=0.
By the way, something like (x+a)^2 counts as LINEAR.
Hope that helped :)(12 votes)
- The question is i have, why does the degree on the numerator needs to be lower than the degree of the denominator? can someone give an explanation? Thanks(11 votes)
- The degree of the numerator doesn't necessarily have to be lower than the degree of the denominator. But if we make it so, the work would be much more easier. Lets say for example: If the degree is more in the numerator , then instead of only adding constants , variables should also be added like A+Bx in the numerator. Too many unknowns=too much confusion(1 vote)
- How do you know which factor to put under the letter A. What would've happened if you put A/(x-8)+B/(x+5) instead?(8 votes)
- Does that mean A/(x-8)+B/(x+5) is equivalent to A/(x+5)+B/(x-8)?(3 votes)
- What is the purpose of partial fraction expansion? In what areas of math does this apply and facilitate?(5 votes)
- This is good for calculus, and differential equations.(8 votes)
- Why is it allowed to set x to different arbitrary values?(6 votes)
- x is a variable; you can set it to be any arbitrary value you feel like. It's important to note that he would have arrived at the same values for A and B had he chosen some other, less convenient values for x as well.(5 votes)
- What happens if the denominator has a factor which is also present in the numerator?Should the factors be cancelled?(3 votes)
- Hello there, if you mean by factors then you can cancel. However, do not mix factors with terms such as
(5x)/(x) (you can cancel) but (5+x)/x. in this you cannot cancel the x(3 votes)
- The main problem where I see a real leap to something new is5:10in the video. (x+5)(x-8) becomes "magically" (x+5) +(x+8). This is very confusing and there is no previous video that I can find where this is ever done.(3 votes)
- I take it you meant that (x+5)(x-8) "becomes" (x+5) + (x-8) ?
That's not what is going on. Those are the denominators he has decided to use, that is what this video is about, and he has to go through the steps outlined in the rest of the video to make that happen for his chosen denominators.(3 votes)
- At5:10, how would you know each value accounts for a variable? When would you use Bx+C?(3 votes)
- Do u know when what grade we mostly learned this? Maybe 5th grade or 6th?(1 vote)
- I am not from the US, but judging from the arrangement of the material in Sal's "missions", I would guess later than 6th grade. The work in the 6th grade mission is much easier and much earlier than this material.(3 votes)
Let's see if we can learn a thing or two about partial fraction expansion, or sometimes it's called partial fraction decomposition. The whole idea is to take rational functions-- and a rational function is just a function or expression where it's one expression divided by another-- and to essentially expand them or decompose them into simpler parts. And the first thing you've got to do, before you can even start the actual partial fraction expansion process, is to make sure that the numerator 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 same degree as the denominator. So the first step we want to do to simplify this and to get it to the point where the numerator has a lower degree than the denominator is to 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, we divide the denominator into the numerator to figure out the remainder, so we divide x squared minus 3x minus 40 into x squared minus 2x minus 37. So how many times? You look at the highest degree term, so x squared goes into x squared one time, one times this whole thing is x squared minus 3x minus 40, and now you want to subtract this from that to get the remainder. And see, if I'm subtracting, so I'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 can be rewritten as-- let me scroll down a little bit-- as 1 plus x plus 3 over x squared minus 3x minus 40. This might seem like some type of magic thing I just did, but this is no different than what you did in the fourth or fifth grade, where you learned how to convert improper regular fractions into mixed numbers. Let me just do a little side example here. If I had 13 over 2, and I want to turn it into a mixed number, what you do-- you can probably do this in your head now-- but what you did is, you divide the denominator into the numerator, just like we did over here. 2 goes into 13. We see 2 goes into 13 six times, 6 times 2 is 12, you subtract that from that, you get a remainder of 1. 2 doesn't go into 1, so that's just the remainder. So if you wanted to rewrite this, it would be the number of times the denominator goes into the numerator, that's 6, plus the remainder over the denominator. Plus 6-- plus 1 over 2. And when you did it in elementary school, you would just write 6 1/2, but 6 1/2 is the same thing as 6 plus 1/2. That's exactly the same thing we did here. The denominator went to the numerator one time, and then there was a remainder of x plus 3 left over, so it's 1 plus x plus 3 over this expression. Now we see that that numerator in this rational expression does have a lower degree than the denominator. The highest degree here is 1, the highest degree here is 2, so we're ready to commence our partial fraction decomposition. And all that is, is taking this expression up here and turning it into two simpler expressions where the denominators are the factors of this lower term. So given that, let's factor this lower term. So let's see. What two numbers add up to minus 3, and when you multiply them, you get minus 40? So let's see. They have to be different signs, because when you multiply them you get a negative, so it has to be minus 8 and plus 5. So we can rewrite this up here as-- I'll switch colors-- 1 plus x plus 3 over x plus 5 times x minus 8. 5 times 8 is minus 40-- 5 times negative 8 is minus 40, plus 5 minus 8 is minus 3, so we're all set. Now I'll just focus on this part right now. We can just remember that that 1 is sitting out there out front. This is the expression we want to decompose or expand. And we're going to expand it into two simpler expressions where each of these are the denominator-- and I will make the claim, and if the numbers work out then the claim is true-- I'll make the claim that I can expand this, or decompose this, into two fractions where the first fraction is just some number a over the first factor, over x plus 5, plus some number b over the second factor, over x minus 8. That's my claim, and if I can solve for a and b in a way that it actually does add up to this, then I'm done and I will have fully decomposed this fraction. I guess is the way-- I don't know if that's the correct terminology. So let's try to do that. So if I were to add these two terms, what do I get? When you add anything, you find the common denominator, and the common denominator, the easiest common denominator, is to multiply the two denominators, so let me write this here. So a over x plus 5 plus b over x minus 8 is equal to-- well, let's get the common denominator-- 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 minus 8 over this whole thing. I mean, if I just wrote this right here, you would just cancel these two terms out and you would get a over x plus 5. And then you could add that to the common denominator, x plus 5 times x minus 8, and it would be b times x plus 5. Important to realize, that, look. This term is the exact same thing as this term if you just cancel the x minus 8 out, and this term is the exact same thing as this term if you just cancel the x plus 5 out. But now that we have an actual common denominator, we can add them together, so we get-- let me just write the left side here over-- a over x plus 5-- I'm sorry. I want to write this over here. I want to write x plus 3 over plus 5 times x minus 8 is equal to is equal to the sum of these two things on top. a times x minus 8 plus b times x plus 5, all of that over their common denominator, x plus 5 times x minus eight. So the denominators are the same, so we know that this, when you add this together, you have to get this. So if we want to solve for a and b, let's just set that equality. We can ignore the denominators. So we can say that x plus 3 is equal to a times x minus 8 plus b times x plus 5. Now, there's two ways to solve for a and b from this point going forward. One is the way that I was actually taught in the seventh or eighth grade, which tends to take a little longer, then there's a fast way to do it and it 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 make this 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'm just picking an arbitrary x to be able to solve for this-- then this would become minus 5 plus 3-- let me just write it out, minus 5 plus 3-- is equal to a times minus 5 minus 8-- let me just write it out, minus 5 minus 8-- plus b times minus 5 plus 5. And I picked the minus 5 to make this expression 0. So then you get-- pick a brighter 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 b is 0, and then you divide both sides by minus 13, you get-- negatives cancel out-- you get 2 over 13 is equal to a, and now we can do the same thing up here and get rid of the a terms by making x is equal to 8. If x is equal to 8, you get x plus 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 to 13b, divide both sides by 13, you get b is equal to 11 over 13. So we were able to solve for our a's and our b's. And so we can go back to our original equation and we could say, wow. This just has to be equal to 2 over 13, and this just has to be equal to 11 over 13. So our original, our very original thing we wrote up here, can be decomposed into 1, that's this 1 over here, plus this, which is 2 over 13-- I'll just write it like this for now-- 2 over 13, over x plus 5. You could bring the 13 down here if you want to write it so you don't have a fraction over a fraction. Plus 11 over 13 times-- over x minus 8. And once again, you could bring the 13 down so you don't have a fraction over a fraction. But we have just successfully decomposed this pretty-- I don't want to say that we necessarily simplified it, because you could say, oh, we only have one expression here, now I have three-- but I've reduced the degree of both the numerators and the denominators. 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 really useful technique later on when you get to calculus, and actually, differential equations, because a lot of times it's much easier-- and I'll throw out a word here that you don't understand-- to take the integral or the antiderivative of something like this, then something like this. And later, when you do inverse Laplace transforms and differential equations, it's much easier take an inverse Laplace transform of something like this than something like that. So anyway, hopefully I've given you another tool kit in your-- or another tool in your tool kit, and I'll probably do a couple more videos because we haven't exhausted all of the examples that we could we could show for partial fraction decomposition.