# Partial fraction expansion: repeated factors

## 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, 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.