Partial Fraction Expansion 2 A more complex problem
Partial Fraction Expansion 2
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- 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.
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