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Current time:0:00Total duration:10:35

Video transcript

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 it I've constructed a little problem here it's 6x squared let me make sure my pin is right 6x squared minus 19 X 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 yeah this is a little bit different because what do I do here I have this first-degree factor but it shows up twice and 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 will just add together because they have the same denominator and you could just do them as one variable you wouldn't have to have two separate variables here so this wouldn't make sense it's 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 now 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 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 you would get a partial or you would get a rational function that would end up having you know 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 in the numerator when you add just these two parts let me let me clarify what I'm saying now this is just for intuition if you were to add just these two parts here you would get you know 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 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 isolate actually let me erase some of this stuff that I've written let me erase all of that alright 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 the common denominator here is X minus 1 times X minus 2 squared all right 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 we have to multiply a by a times X minus 2 squared right if we cancel 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 and one of these X minus 2's 4 down here then you're just left with B over X minus 2 plus C times X minus 1 right 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 have to write it smaller 6x squared minus 19 X 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 numerator is equal to each other and try to solve for a B and C let me write that so we have 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 19 X plus 15 there you go and now we have to solve for them and we can do it just the way we've done the previous problems we can pick choices for X that make things cancel out nicely so if we wanted to solve for a and we want to make the B and the C cancel out if 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 becomes 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 - 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 we have this would be 0 so we could use that to solve for C so if we say that X is equal to 2 X is equal to 2 then we get this term would be 0 because that's 0 this whole expression zero because X minus 2 would be zero 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 3819 times 2 plus 15 let's see 24 plus 15 is 39 minus 38 that is equal to 1 we're in the homestretch and now for B there's no obvious way to make the other two cancel out without making the be canceled 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 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 it'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 so C times minus 1 is equal to if well if x is 0 then this is 0 this is 0 is equal to 15 now we just solve for B we get 8 let's see this is plus 2 B all right minus 1 times minus 2 is plus 2 minus oh and we shouldn't have written the C here we know what C is C is 1 right so this is just a 1 so 1 times minus 1 is minus 1 is equal to 15 and let's see we have 2 B is equal to let's 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 learnt we've solved for which is 2 so that 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 you know bla bla bla bla bla bla bla some polynomial up there and it was all over X - I don't know some number X minus a to the I don't know to the 10th power or let me get you know to the to the way yes sure to the 10th power if we wanted to partially 4 if we want 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 4 plus B times X minus a squared + and you go buhbuh buhbuh you'd have ten terms plus you know I don't know what the tenth letter of the alphabet is maybe it's H or I or something I don't know let me 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 3 variables on a on a on an actual exam it's probably something like i showed you in the lot previous example or 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