Algebra (all content)
Partial fraction expansion: repeated factors
Sal performs partial fraction expansion upon (6x²-19x+15)/(x-1)(x-2)². Created by Sal Khan.
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- When he started the question and at the end, he said it was hairier. What's that supposed to mean?(1 vote)
- I'm surprised to see that the slang "hairy" has the same meaning as its equivalent in brazilian portuguese, where here "hair" is translated as "cabelo" and "hairy/hairier" could be translated as "cabeludo".
Even more surprising to me, was that i used this terms to describe this whole partial fraction expansion methodology, as such strenuous exercise of algebra.(4 votes)
- at2:10I don't understand how is that if you ad B/(x-2) + C/(x-2)^2 you would get an expression something X + something over (x-2)^2(19 votes)
- B/(x-2) + C/(x-2)^2
Multiply B/(x-2) by (x-2)/(x-2) to get an equivalent fraction with the denominator (x-2)^2
B(x-2)/(x-2)^2 + C/(x-2)^2
Now we can add the numerators directly
( B(x-2) + C ) / (x-2)^2
Now expand the numerator
( Bx - 2B + C ) / (x-2)^2
There's your "something X + something over (x-2)^2 "(23 votes)
- I apologize in advance if this question is too generalized. I have been methodically advancing. Some topics are more difficult than others. I am bewildered by this topic. Completely. I passed over it in algebra but I am determined to master this topic. I've watched the videos (all of the ones leading up to this topic as well) multiple times. I think my problem is that I can't abstract this. The answer is more complex than the question. I don't want to move on. Any suggestions?(18 votes)
- When you add two fractions with different denominators, you find the common denominator, do the calculations, and then add them. You end up with a single fraction.
Partial fractions are the reverse of this process. You start with a single fraction and split it into one or more fractions with different denominators.
The reason you would bother doing this is that some computations (especially once you get into calculus) can be greatly simplified by breaking up a fraction in this manner.(21 votes)
- when must I use the form (Bx + c ) /... or (Bx^2 +C) / ... ????
and how many forms are there? What's the rule of that?(6 votes)
- (Bx + C) is for when you have an irreducible quadratic term (ax^2 + bx + c) in the denominator (possibly in the form: (x^2 - c)). In this problem the term is (x - a)^2, a subtle difference. In this case you need a fraction for each degree of the term.
So you get: B1 / (x - a) + B2 / (x - a)^2
If you have a quadratic that you can't reduce that is also repeated: (ax^2 + bx + c)^n
You will need to combine the irreducible quadratic approach with the repeated linear approach:
(B1 * x + C1)/(ax^2 + bx + c) + (B2 * x + C2)/(ax^2 + bx + c)^2 . . . (Bn * x + Cn)/(ax^2 + bx + c)^n
I hope that's clear.(12 votes)
- In the first Partial Fraction vid, you said that the degree in the nominator has to be less than the degree in the denominator (if that isn't the case, you'd have to do long division). Here clearly the degree in the nominator is greater, but it all worked out anyway. Why is that ?(3 votes)
- If you expand the denominator and do all the multiplication, you will see that you get a x^3, so the denominator has order 3 while the numerator has order 2.(14 votes)
- Why is it C/(x-2)^2 and not (Cx+D)/(x-2)^2. In the previous video, didn't he say that the numerator had to be one degree less than the denominator. In the first scenario it is two degrees less(4 votes)
- If the denominator is in the form ax^2+bx+c, then you are absolutely right and the numerator should be Cx+D. However, in this case, C / (x-2)^2 goes with the B / (x-2) just before it. When the two partial fractions are added, you get (B(x-2) + C)/(x-2)^2. As you can see, the numerator of this combined term is one degree less than the denominator. Sal explains how this works at2:24.(5 votes)
- in what circumstance would this be applicable to a real world problem?(3 votes)
- So this might help answer your question. Recently, I've been going through all the calculus modules on Khan Academy and I just reached a part of Differential Equations where I needed this skill (which is why I'm here). So far in my life, I have not needed to know how to do this because you're right - all on its own, this skill has very few real life uses. However, once you start doing some of the more advanced calculus problems, this is an essential technique for solving real world calculus problems.(7 votes)
The partial fraction videos were very helpful but i need to know how to solve for a_1,a_2 when I have (7s+12)/(s+2)^2.
thank You.(3 votes)
- The partial fraction decomposition for a repeated linear factor is of the form:
(7s+12)/(s+2)^2 = A/(s+2) + B/(s+2)^2. Multiply through by (s+2)^2 and you get:
7s + 12 = A*(s+2) + B
There are various techniques from here... You can substitute in values of s to find A, B, or both. In this case s=-2 gives you -14 + 12 = A*0 + B or B = -2. But s=-1 gives you -7 + 12 = A + B or A + B = 5.
Alternatively you can expand the right hand side to get the polynomial Ax + (2A + B). Matching this against the terms on the left, we learn that A = 7 and 2A+B=12. Easily solved from there.(3 votes)
- If N(x)= x^2(x-1)(x^2+x+1) we get = A/x + [B/(x^2)] + C/(x-1) + (Dx+E)/(x^2+x+1).
My question is about the second term in .
Since x^2 is 2-grade shouldnt the B instead be BX+constant?(2 votes)
- I believe you only need the BX+constant when a factor in the denominator has a degree higher than one. In this case, (x-2)^2 is a repeated linear factor. The factor itself, x-2, is still just linear.(3 votes)
- In the previous video, he said that your numerator of each part of the decomposed fraction must be one degree less that its denominator. E.g (Bx+C)/(x^2+2x+4)
Why when wrote the decomposed fraction out did he just write the constant C over the (x-2)^2 denominator. Shouldn't it have been Cx+D?(2 votes)
- I read an answer to a question similar to yours over here. Basically you have to look at the factors of the denominator for the original fraction. So in the previous question, the one of the factor is x^2+2x+4, which can no longer be factored further. So the numerator for that partial fraction must have Bx^(2-1)=Bx.
In this video, the factor is (x-2)^2. The factor is still x-2 so even if it repeats as C/(x-2)^2, you do not need a x^2 term in the numerator.
Why this works or the general proof for this is beyond me and I hope someone else can explain this!(2 votes)
There's one more case of partial fraction expansion or decomposition problems that you might see, so I thought I would cover it. And that's the situation where you have a repeated factor in the denominator. So let's see, I've constructed a little problem here. It's 6x squared. Let me make sure my pen is right. 6x squared minus 19x plus 15. All of that over x minus 1 times-- and this is the thing that makes this one especially interesting-- x minus 2 squared. So you might say, gee, this is a little bit different. Because what do I do here? I have this first degree factor, but it shows up twice. It doesn't make sense, for example, it wouldn't make sense to do this. A over x minus 1 plus B over x minus 2 plus C over x minus 2. Because if you did this, the B and the C would just add together, because they have the same denominator. You could just view them as one variable. You wouldn't have to have two separate variables here. So this wouldn't make sense as a partial fraction expansion of this. You might want to maybe square this and view it as a second degree term, and do it like you did in the previous example. But then you wouldn't have fully decomposed this problem. And so the answer here-- and I'll just show you how to do it, I'll maybe leave it to you a little bit to think about why it works-- is to decompose it almost like this, but instead of having C over x minus 2, you're going to have C over x minus 2 squared. And I'll try to give you a little intuition about why that happens. So the decomposition here is going to be A over x minus 1 plus B over x minus 2 plus C over x minus 2 squared. And the intuition here is that if you were to-- let's ignore this term right here. But if you were to add these two terms right there, you would get a rational function that would end up having something times x plus something. And so it would be consistent with what we did in the second partial fraction video, where if you would have a second degree term on the bottom, and the numerator, when you add just these two parts-- let me clarify what I'm saying. And this is just for intuition. If you were to add just these two parts here, you would get something times x plus something, all of that over x minus 2 squared. And that's consistent with what we did in the previous video, where we said if we have a second degree term in the denominator-- which this really is if you were to expand this-- you should have a first degree term in the numerator. So I'll leave that there. That's a little bit of a nuance, and it's good to think about why this works. But with that said, let's just solve this problem. Let me erase some of this stuff that I've written. Let me erase all of that. All right. Ready to work on this problem. So given that this is the expansion, now we just have to solve for A, B, and C. So if we were to add these three fractions, the common denominator is x minus 1 times x minus 2 squared. This is a factor of that, so the least common multiple of this and this is just this. So it's just x minus 2 squared there, and that's why we just did this to begin with. And the numerator is going to be-- let me do it in a different color-- A. What do we have to multiply A by? A times x minus 2 squared, right? If we canceled these out, you're just left with A over x minus 1. Plus B times x minus 1 times x minus 2. Right? If we cancel out the x minus 1's and one of these x minus 2's from down here, you're just left with B over x minus 2. Plus C times x minus 1. If that cancels with that, we're left with C over x minus 2 squared. And all of this is going to be equal to-- that up there in blue is going to be equal to-- I have to write it smaller-- 6x squared minus 19x plus 15, all of that over x minus 1 times x minus 2 squared. And like we've done in every problem so far, the denominators are the same, so we can just set the numerators equal to each other and try to solve for A, B, and C. Let me write that. And I'm really just rewriting it. A times x minus 2 squared plus B times x minus 1 times x minus 2 plus C times x minus 1 is going to be equal to 6x squared minus 19x plus 15. There you go. And now we have to solve for them. And we can do it just the way we done the previous problems. We can pick choices for x that make things cancel out nicely. So if we wanted to solve for A, we want to make the B and the C cancel out. We just have to pick x is equal to 1. Because if x is equal to 1, this becomes 0, so C disappears. This becomes 0, so this whole term disappears, and we're just left with this and this. So let's do that. So when x is equal to 1, we get-- let me pick another color-- A times, 1 minus 2 is minus 1, squared is just 1. So it's A times 1. So we can just leave the A there. This becomes 0, because this term right here will be 0 if x is 1. This become 0, because that term would be 0 if x is equal to 1. So A is equal to 6 times 1-- so 6-- minus 19 times 1-- minus 19-- plus 15. And what is that? Minus 19 plus 15, that's minus 4, plus 6, that is equal to 2. There you go. Now let's try to solve for-- let's see, what can we do? If we make x equal to 2, the A will disappear. This whole term will disappear, because this would be 0. So we could use that to solve for C. So if we say that x is equal to 2, then we get this term would be 0, because that's 0. This whole expression is 0, because x minus 2 would be 0. And we're just left with C times 2 minus 1, so that's just 1-- so C is equal to 6 times 4, right? 2 squared. So it's 24, minus 38-- 19 times 2-- plus 15. Let's see, 24 plus 15 is 39, minus 38, that is equal to 1. We're in the home stretch. And now for B, there's no obvious way to make the other two cancel out without making the B cancel out. But we've solved for everything else, so we can really just pick an arbitrary value for x that'll make things easy to compute. So if we just pick x is equal to 0, that's always something that can clear out a lot of the hairiness of an algebra problem. x is equal to 0, not 3. x is equal to 0. Let me do that in a different color. If x is equal to 0, then what do we have? We have A, but A is 2, right? 2 times 0 minus 2 squared. So that's minus 2 squared, so that's 2 times 4. Plus B-- that's what we're trying to solve for; we already solved for everything-- B times minus 1, right? 0 minus 1 is minus 1. Times minus 2, plus C times minus 1 is equal to-- well, if x is 0, then this is 0, this is 0, is equal to 15. Then we just solve for B. We get 8 plus 2B, right? Minus 1 times minus 2 is plus 2. Oh, and we shouldn't have written the C here, we know what C is. C is 1. So this is just a 1. So 1 times minus 1 is minus 1, is equal to 15. And let's see, we have 2B is equal to, lets see, 16 is equal to 8, B is equal to 4, dividing both sides by 2. So we're done. So the partial fraction decomposition of this right here is A, which we've solved for, which is 2. So it equals 2 over x minus 1 plus B, which is 4-- plus 4 over x minus 2, plus C, which is 1, over x minus 2 squared. And what we did in this with the repeated factor is true if we went to a higher degree term. So if we had blah blah blah, some polynomial there, and it was all over x minus-- I don't know, some number. x minus a to the 10th power. Well, to the-- well, yeah, sure. To the 10th power. If we wanted to decompose this partial fraction, or expand it, it would be A over x minus a-- it's a different a. I'm just showing this. Plus B times x minus a squared, plus, and you go blah blah blah. You'd have 10 terms, plus-- I don't know what the 10th letter of the alphabet is; maybe it's H or I or something. Maybe it's J. J over x minus a to the 10th. So you can use that property in general, though you'll very seldom see something like this, because computationally it would take you forever to do it. The hairiest problem that you might see with three variables on an actual exam is probably something like I showed you in the previous example, or what I just showed you right now. Anything hairier than that, you'd probably end up using a computer. But you should know how to do it, because if you have a computer and you're solving a real world problem that is a lot more complicated than this, you should know how to do it. Anyway, hope you found that a bit helpful.