Partial Fraction Expansion 1 Introduction to partial fraction expansion
Partial Fraction Expansion 1
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- Let's see if we can learn a thing or two about partial
- fraction expansion, or sometimes it's called partial
- fraction decomposition.
- The whole idea is to take rational functions-- and a
- rational function is just a function or expression where
- it's one expression divided by another-- and to essentially
- expand them or decompose them into simpler parts.
- And the first thing you've got to do, before you can even
- start the actual partial fraction expansion process, is
- to make sure that the numerator has a lower degree
- than the denominator.
- In the situation, the problem, that I've drawn right here,
- I've written right here, that's not the case.
- The numerator has the same degree as the denominator.
- So the first step we want to do to simplify this and to get it
- to the point where the numerator has a lower degree
- than the denominator is to do a little bit of
- algebraic division.
- And I've done a video on this, but it never hurts to get a
- review here, so to do that, we divide the denominator into the
- numerator to figure out the remainder, so we divide x
- squared minus 3x minus 40 into x squared minus 2x minus 37.
- So how many times?
- You look at the highest degree term, so x squared goes into x
- squared one time, one times this whole thing is x squared
- minus 3x minus 40, and now you want to subtract this from
- that to get the remainder.
- And see, if I'm subtracting, so I'm going to subtract, and then
- minus minus is a plus, a plus, and then you can add them.
- These cancel out.
- Minus 2x plus 3x, that's x.
- Minus 37 plus 40, that's plus 3.
- So this expression up here can be rewritten as-- let me scroll
- down a little bit-- as 1 plus x plus 3 over x squared
- minus 3x minus 40.
- This might seem like some type of magic thing I just did, but
- this is no different than what you did in the fourth or fifth
- grade, where you learned how to convert improper regular
- fractions into mixed numbers.
- Let me just do a little side example here.
- If I had 13 over 2, and I want to turn it into a mixed number,
- what you do-- you can probably do this in your head now-- but
- what you did is, you divide the denominator into the numerator,
- just like we did over here.
- 2 goes into 13.
- We see 2 goes into 13 six times, 6 times 2 is 12, you
- subtract that from that, you get a remainder of 1.
- 2 doesn't go into 1, so that's just the remainder.
- So if you wanted to rewrite this, it would be the number of
- times the denominator goes into the numerator, that's 6, plus
- the remainder over the denominator.
- Plus 6-- plus 1 over 2.
- And when you did it in elementary school, you would
- just write 6 1/2, but 6 1/2 is the same thing as 6 plus 1/2.
- That's exactly the same thing we did here.
- The denominator went to the numerator one time, and then
- there was a remainder of x plus 3 left over, so it's 1 plus x
- plus 3 over this expression.
- Now we see that that numerator in this rational expression
- does have a lower degree than the denominator.
- The highest degree here is 1, the highest degree here is 2,
- so we're ready to commence our partial fraction decomposition.
- And all that is, is taking this expression up here and turning
- it into two simpler expressions where the denominators are the
- factors of this lower term.
- So given that, let's factor this lower term.
- So let's see.
- What two numbers add up to minus 3, and when you multiply
- them, you get minus 40?
- So let's see.
- They have to be different signs, because when you
- multiply them you get a negative, so it has to
- be minus 8 and plus 5.
- So we can rewrite this up here as-- I'll switch colors--
- 1 plus x plus 3 over x plus 5 times x minus 8.
- 5 times 8 is minus 40-- 5 times negative 8 is minus 40, plus 5
- minus 8 is minus 3, so we're all set.
- Now I'll just focus on this part right now.
- We can just remember that that 1 is sitting
- out there out front.
- This is the expression we want to decompose or expand.
- And we're going to expand it into two simpler expressions
- where each of these are the denominator-- and I will make
- the claim, and if the numbers work out then the claim is
- true-- I'll make the claim that I can expand this, or decompose
- this, into two fractions where the first fraction is just some
- number a over the first factor, over x plus 5, plus some
- number b over the second factor, over x minus 8.
- That's my claim, and if I can solve for a and b in a way that
- it actually does add up to this, then I'm done and I will
- have fully decomposed this fraction.
- I guess is the way-- I don't know if that's the
- correct terminology.
- So let's try to do that.
- So if I were to add these two terms, what do I get?
- When you add anything, you find the common denominator, and the
- common denominator, the easiest common denominator, is to
- multiply the two denominators, so let me write this here.
- So a over x plus 5 plus b over x minus 8 is equal to--
- well, let's get the common denominator-- it's equal to
- x plus 5 times x minus 8.
- And then the a term, we would-- a over x plus 5 is the same
- thing as a times x minus 8 over this whole thing.
- I mean, if I just wrote this right here, you would just
- cancel these two terms out and you would get a over x plus 5.
- And then you could add that to the common denominator, x plus
- 5 times x minus 8, and it would be b times x plus 5.
- Important to realize, that, look.
- This term is the exact same thing as this term if you just
- cancel the x minus 8 out, and this term is the exact same
- thing as this term if you just cancel the x plus 5 out.
- But now that we have an actual common denominator, we can add
- them together, so we get-- let me just write the left side
- here over-- a over x plus 5-- I'm sorry.
- I want to write this over here.
- I want to write x plus 3 over plus 5 times x minus 8 is equal
- to is equal to the sum of these two things on top.
- a times x minus 8 plus b times x plus 5, all of that over
- their common denominator, x plus 5 times x minus eight.
- So the denominators are the same, so we know that this,
- when you add this together, you have to get this.
- So if we want to solve for a and b, let's just
- set that equality.
- We can ignore the denominators.
- So we can say that x plus 3 is equal to a times x minus
- 8 plus b times x plus 5.
- Now, there's two ways to solve for a and b from
- this point going forward.
- One is the way that I was actually taught in the seventh
- or eighth grade, which tends to take a little longer, then
- there's a fast way to do it and it never hurts to do
- the fast way first.
- If you want to solve for a, let's pick an x that'll
- make this term disappear.
- So what x would make this term disappear?
- Well, if I say x is minus 5, then this becomes 0, and
- then the b disappears.
- So if we say x is minus 5-- I'm just picking an arbitrary x to
- be able to solve for this-- then this would become minus 5
- plus 3-- let me just write it out, minus 5 plus 3-- is equal
- to a times minus 5 minus 8-- let me just write it out,
- minus 5 minus 8-- plus b times minus 5 plus 5.
- And I picked the minus 5 to make this expression 0.
- So then you get-- pick a brighter color-- minus 5 plus
- 3 is minus 2, is equal to-- what is this?-- minus 13a
- plus-- this is 0, right?
- That's 0.
- Minus 5 plus 5 is 0, 0 times b is 0, and then you divide both
- sides by minus 13, you get-- negatives cancel out-- you get
- 2 over 13 is equal to a, and now we can do the same thing up
- here and get rid of the a terms by making x is equal to 8.
- If x is equal to 8, you get x plus 3 is equal to 11, is equal
- to a times 0 plus b times-- what's 5-- 8 plus 5
- is-- plus b times 13.
- Their b looks a bit like a 13.
- And then you get 11 is equal to 13b, divide both sides by 13,
- you get b is equal to 11 over 13.
- So we were able to solve for our a's and our b's.
- And so we can go back to our original equation
- and we could say, wow.
- This just has to be equal to 2 over 13, and this just has
- to be equal to 11 over 13.
- So our original, our very original thing we wrote up
- here, can be decomposed into 1, that's this 1 over here, plus
- this, which is 2 over 13-- I'll just write it like this for
- now-- 2 over 13, over x plus 5.
- You could bring the 13 down here if you want to write
- it so you don't have a fraction over a fraction.
- Plus 11 over 13 times-- over x minus 8.
- And once again, you could bring the 13 down so you don't have
- a fraction over a fraction.
- But we have just successfully decomposed this pretty-- I
- don't want to say that we necessarily simplified it,
- because you could say, oh, we only have one expression here,
- now I have three-- but I've reduced the degree of both the
- numerators and the denominators.
- And you might say, well, Sal, why would I ever
- have to do this?
- And you're right.
- In algebra you probably won't.
- But this is actually a really useful technique later on when
- you get to calculus, and actually, differential
- equations, because a lot of times it's much easier-- and
- I'll throw out a word here that you don't understand-- to take
- the integral or the antiderivative of
- something like this, then something like this.
- And later, when you do inverse Laplace transforms and
- differential equations, it's much easier take an inverse
- Laplace transform of something like this than
- something like that.
- So anyway, hopefully I've given you another tool kit in your--
- or another tool in your tool kit, and I'll probably do a
- couple more videos because we haven't exhausted all of the
- examples that we could we could show for partial
- fraction decomposition.
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