Current time:0:00Total duration:13:02
0 energy points
Sal performs partial fraction expansion upon (10x²+12x+20)/(x³-8). Created by Sal Khan.
Video transcript
Let's see if we can tackle a more complicated partial fraction decomposition problem. I have 10x squared plus 12x plus 20, all of that over x to the third minus 8. The first thing to do with any of these rational expressions that you want to decompose is to just make sure that the numerator is of a lower degree than the denominator, and if it's not, then you just do the algebraic long division like we did in the first video. But here, you can do from [UNINTELLIGIBLE] the highest degree term here is a second-degree term, here it's a third-degree term, so we're cool. This is a lower degree than that one. If it was the same or higher, we would do a little long division. The next thing to do, if we're going to decompose this into its components, we have to figure out the factors of the denominator right here, so that we can use those factors as the denominators in each of the components, and a third-degree polynomial is much, much, much harder to factor than a second-degree polynomial, normally. But in this case, there's something that should hopefully pop out at you-- if it doesn't immediately, hopefully what I'm about to say will make it pop out at you in the future-- is you should always think about what number, when you substitute into a polynomial, will make it equal to 0. And in this case, what to the third power minus 8 equals 0? And hopefully 2 pops out at you. And this is something you can only do really through inspection or through experience. And you'll immediately see 2 to the third minus 8 is 0, so 2 is a 0 of this, or 2 makes this expression 0, and that tells us that x minus 2 is a factor. So we can rewrite this right here as 10x squared plus 12x plus 20 over x minus 2 times something something something something. We don't know what that something is yet. And I just want to hit the point home of why this is true, or what the intuition vibe has true. If 2 makes this 0, 2 should also make the factored expression 0. And we know that 2 would make this factored expression 0, because when you put a 2 right here, this factor become 0, so it'll make the whole thing 0. And so that's why, that's the intuition where, if you substitute a number here and it makes this 0, you do x minus that number here, and we know that that will be a factor of the thing. Well, anyway, the next step if we really want to decompose this rational expression is to figure out what this part of it is, and the way to do that is with algebraic long division. We essentially just divide x minus 2 into x to the third minus 8 to get this, so let's do that. So you get x minus 2 goes into x to the third-- and actually, what I'm going to do is, I'm going to write-- I leave space for the second-degree term, which is 0, the first-degree term, and then minus 8 is the constant term, so minus 8-- I like to put all my degree terms in their appropriate columns. You don't have to when you do algebraic long division. So x minus 2 goes into x to the third, or x goes into x to the third, how many times? You just have to look at the highest degree terms. Well it goes into it x squared times-- I'll put it in the x squared column-- x squared times x minus 2, x squared times x is x to the third, x squared times minus 2 is minus 2x squared. And now we want to subtract that from that, so I could just-- or I could add the negative, I always find that easier-- so these cancel out, so I'm left with 2x squared minus 8. How many times does x minus 2 go into 2x squared minus 8? You just look at the highest degree terms, x goes into 2x squared 2x times, so plus 2x, 2x times x is 2x squared, 2x times minus 2 is minus 4x. Subtract that from that, so we could-- or we could just add the negative-- and then we have 4x minus 8, x goes into 4x four times, 4 times x minus 2 is 4x minus 8, and there's no remainder, and this is further confirmation that x minus 2 definitely was a factor of x to the third minus 8 and the other factor is this right here. x squared plus 2x plus 4 times x minus 2 is equal to x to the third minus 8. So now we can write our original problem. Actually, let me write-- well, let me write it down here. What was it, it was 10-- let me switch to another color-- it was 10x squared-- I have a bad memory-- plus 12x plus 20, over x minus 2 times-- what was that-- x squared plus 2x plus 4. So all the work we did so far is just to factor out that x to the third minus 8, but now we can actually do some partial fraction expansion, or partial fraction decomposition. So this is going to be equal to-- and this is the interesting point-- this is where we diverge or advance a little bit from what we did in the first video. This is going to be equal to a constant over this denominator, over x minus 2, plus-- and this is a little different than what you were going to see, what you saw in the last video-- plus an x-term, some coefficient times an x-term, plus c, so bx plus c, over this. x squared plus 2x plus 4. And you might say, Sal, how did you know to do that? Well, what you do is you look at the degree of the denominator. This is a first-degree denominator, and you say, OK. The degree of the numerator in this part of the fraction, I guess you could call it, is going to be 1 less than that. So this first degree-- so this is going to be a 0-degree or a constant term-- Here the degree is 2. The degree is 2, so the degree of its numerators is going to be 1, and since its degree is 1 it could still have a constant term, which is a 0-degree term, so you get bx plus c. And you know, maybe it does end up just being a constant, in which case we can solve for b and it'll just be 0. And actually, there's another thing you might be asking, is hey, why don't we factor this further, and you can try, but if you look at it by inspection, you could kind of in your head do the quadratic formula and you'll see that you only get imaginary roots here. So this actually isn't more-- this isn't factorable in the reals anymore. There no more real 0's to this. So we have factored this as much as we can. So the key now to complete our partial fraction expansion is to just solve for the a, b's, and c's, and we'll do it exactly the same way we did it in the last video. So what we want to do is essentially add these two things. So if we add-- let me write this on the left hand side-- so 10x squared plus 12x plus 20, over x minus 2 times x squared plus 2x plus 4, is equal to-- if we add these two things, we want to get a common denominator, which is x minus 2 times x squared plus 2x plus 4, just multiply the two things, and that'll be, when we add the common denominator, then it's a times this, x squared plus 2x plus 4, plus bx plus c, times this, times x minus 2. Well the denominators are the same, so the numerators have to equal each other, so we have-- I'll rewrite it-- 10x squared plus 12x plus 20 is equal to a times x squared plus 2x plus 4 plus bx plus c, all of that times x minus 2. This might look like a hairy problem, and if you do it kind of the slow way that I do in the last video, it would take you forever, it would be three equations and three unknowns, very very hairy, but we can do as a system pretty much the same way we did it the first one. Let's pick x values that simplify this and allow us to solve for one of these a, b's, or c's at a time, and the immediate thing is what can I-- what x value can I substitute that will cancel out at least two of the variables, two of the a, b, or c,? And if I put x is equal to 2, then this thing becomes 0, which will make this whole expression 0, and I'll just be left with an a and the x's, which I'll know is 2, because I'm picking x to be equal to 2, so let's do that. So if x is equal to 2, then what do I get? I get 10x squared, so I get 10 times 4, that's 2 squared, plus 12 times 2, which is 24, plus 20, is equal to a times 2 squared is 4, plus 4, plus 4. And then all of this just becomes 0, because I picked x is equal to 2. So this is what, 10 times 4 is 40, plus 44, that's 84, is equal to 12a, divide both sides by 12, we get a is equal to 7. We're making progress. So now let's see. We know what a is, can we pick any other x values that will make the b disappear? Or the c. Well, we can make the b disappear if we make x is equal to 0, so let's try that out. So if we say-- let me pick a new, vibrant color, a yellow-- if we make x equal to 0, then the left hand side of this, this is 0,0, we just left with-- we're just left with 20, is equal to a, which is just 7, times 0 plus 0 plus 4, 7 times 4, plus-- bx plus c, the bx just disappears because b times 0 is 0, plus c times 0 minus 2, c times minus 2-- scroll down a little bit-- so we get 20 is equal to 28 minus 2c, subtract 28 from both sides, minus 8 is equal to minus 2c, c is equal to 4. We're almost done. Now let me rewrite our equation up here, with the a's, and let me see if I can do this. All right. So now I have 10x squared plus 12x plus 20 is equal to 7, 7 times x squared plus 2x plus 4 plus bx plus c is 4, times x minus 2. Almost ran out of space. So now we have x's and b's, so to solve for b, we can really just substitute any value of x that makes our math reasonably easy that doesn't make the b disappear. So we can't pick x is equal to 0, but a good number is 1, so if x is equal to 1, we have 10 plus 12 plus 20 is equal to 7 times 1, plus 2 plus 4, that's 7, plus b times 1 is just b, plus 4, times 1 minus 2, is minus 1. So this is 22, 42 is equal to 49 minus b minus 4-- see, you can, I don't want to mess up the math-- so we can say 42 is equal to 45 minus b, subtract 45 from both sides, you get minus 3 is equal to minus b, and we get b is equal to 3, divide both sides by minus 1. And so on our original problem up here, we now know that b is equal to 3, c was equal to 4, and a is equal to 7. So the partial fraction decomposition of this, we're now done, is 7 over x minus 2 plus 3x plus 4, over x squared plus 2x plus 4. Well, that was a pretty tiring problem, and you can see, the partial fraction decomposition becomes a lot more complicated when you have a higher degree denominators, but hopefully you found that a little bit useful.