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## Algebra (all content)

### Course: Algebra (all content) > Unit 13

Lesson 13: Partial fraction expansion# Partial fraction expansion

Sal performs partial fraction expansion upon (10x²+12x+20)/(x³-8). Created by Sal Khan.

## Want to join the conversation?

- Can someone explain the Bx+C idea?(30 votes)
- When decomposing into partial fractions, the numerator of each fraction needs to be less than the degree of the denominator. Thus, if the denominator is linear, then the numerator can only be constant. If the denominator is quadratic, then the numerator might be linear, or it might be constant. To allow for either possibility, you should use the most general linear expression: Bx + C. This allows for both possibilities, including the case in which B = 0 and the numerator is constant.(56 votes)

- At8:50, Sal "gets rid of" the (Bx+C)(x-2) term by making x equal 2. But the original fraction is not defined when x is 2. Are we allowed to substitute values of x for which the original fraction is undefined?(26 votes)
- That's great that you were able to answer your own question!(1 vote)

- I don't understand how he gets the Bx+C at5:50.. even after listening to the explanation several times. Can anybody explain that to me please?(9 votes)
- Because you cannot simply separate (x-2) and the (x^2 + 2x + 4) into two fractions without having the extra Bx + C. This is because the two new fractions do not have the same denominator and must be multiplied in. Essentially what you're trying to do it separate a large fraction, factor out the denominator and break it into two or three smaller fractions. However, you must still be able to multiply them all back out to get the same fraction as in the beginning, hence the Bx+C/(x^2 + 2x + 4) is there to multiply with the A/(x-2) to get the original fraction. So you would cross multiply and end up with A(x^2 + 2x + 4) + (Bx+C)(x - 2) = 10x^2 + 12x + 20(3 votes)

- Can decomposition of cubic equations be like this:

(x^3±y^3) = (x±y)(x^2∓xy+y^2)(3 votes)- I remember coming across a mnemonic for the sequence of signs in the factoring of the
**sum of cubes**and the**difference of cubes**.**sum of cubes**:`a^3+b^3=(a+b)(a^2-ab+b^2)`

The sequence of the signs for the**sum**of cubes is:

(+) : + , - , +**difference of cubes**:`a^3-b^3=(a-b)(a^2+ab+b^2`

The sequence of the signs for the**difference**of cubes is:

(-) : - , + , +

If you notice,

the**first**sign in the formula is the**Same**as the sum or the difference,

the**second**sign in the formula is the**Opposite**of the sum or the difference,

and the**third**sign in the formula is**Always Positive**.

That spells out the acronym**SOAP**(Same, Opposite, Always Positive) as our mnemonic.

I hope this enables you to remember which is which. This should help**clean it up**for you.*I previously offered this in “Difference of Cubes,” but I thought it would be appropriate here as well. Sorry for the repetition.*(8 votes)

- 1:35How does "if 2 is a zero" that leads to x-2? I don't see how one automatically springs from the other.(6 votes)
- The way I understand it, it's best to see this as a consequence of the remainder theorem. If you've been following the algebra 2 playlist you should've come across that. Here's the relevant video: https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-remainder-theorem-to-determine-coefficient-example

Basically, here's the rule: something is a factor of a polynomial`p(x)`

if dividing the polynomial by that something gives you no remainder (this is true not just for polynomials but anything, really!). The remainder theorem states that`(x-a)`

is a factor of a polynomial`p(x)`

if and only if`p(a) = 0`

(0 being the remainder here). For our exercise, we know that, if we can figure out a value for`x`

for which`x^3 - 8 = 0`

, we'll have found our`a`

and our corresponding factor`x-a`

. You can pretty simply see that this value has to be 2. So, from the remainder theorem:`p(2) = 0 <=> x-2 is a factor of p(x)`

.

Furthermore, remember that`x^3 - 8`

is also a difference of cubes. In factoring we learnt a handy little shortcut formula for both the sums and differences of cubes. For the difference, we know that`a^3 - b^3 = (a - b)*(a^2 + ab + b^2)`

. So you could've come up with Sal's answer, which he got through long division, simply by plugging numbers into that formula (provided you could remember it! :) ).

If anyone has any corrections or additions I'd love to hear them!(4 votes)

- Why does the numerator have to be a
**lesser**degree than the denominator?(6 votes) - At6:35, he points out that he didn't factor the (x^2 + 2x + 4) any further because it ends up with imaginary numbers. But I can think of plenty of situations where one COULD factor something further. What happens then? The next video didn't seem to answer this question for me. In the next video, the rule demonstrated only seems to apply when there is more than one of the SAME thing (in that case, it's multiples of x-2). What about when there 3 or more different things? If, for example, I have x^3 - 4x in my denominator, could I make it into (x)(x-2)(x+2)? Would I still use A, B and C numerators?(3 votes)
- It would decompose into something of the form A/x + B/(x-2) + C/(x+2). The degree of the numerator has to be one less than the degree of the denominator.(3 votes)

- At 12;23, he says divide both sides by minus 1. Could someone please explain why it couldn't have remained -b=-3?(0 votes)
- because in the question you have B not -B If you were to but -b=-3 into the question you will have to change a few signs, so the best way is to put b=3 and directly substitute. hope that helped(3 votes)

- you couldn't pull a root out of x^2 -X could you?(2 votes)
- I could. I'd factor out an x. Try it and see what happens.(2 votes)

- So i followed this step by step for the decomp. of 6y/(y^3-y^2+y-1) and got (3/(y+1))-((3y+3)/y^2+1). However, my homework is marking me incorrect. any idea where I went wrong?(2 votes)
- Could you please show your working? From the looks of the solution you got, you have made a small error with finding the factors of the denominator.(2 votes)

## 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.