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# Partial fraction expansion

## Video transcript

let's see if you can tackle a more complicated partial fraction decomposition problem I have 10x squared plus 12x plus 20 all of that over X to the 3rd minus 8 the first thing to do with any of these fried these rational expressions that you want to I guess 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 the algebraic long division like we did in the first video but here you can do from expection the highest degree term here is the 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 is into its components we have to figure out the factors of the denominator right here so that we can you know 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 this case there's something 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 you should always think about what number when you substitute into a into a into a polynomial will make it equal to zero and in this case you say well what to the third power minus 8 equals zero and hopefully two pops out at you and this is something you can only do really through inspection or through experience and you remedial C to the third minus 8 is 0 so 2 is a zero of this so 2 is zero or two makes this expression zero and that tells us that X minus 2 is a factor so we can rewrite this right here as 10 x 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 with why this is true or what the intuition by it is true if two makes is 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 this factor becomes zero so it'll make the whole thing zero and so that's why that's the intuition where if you substitute a number here it makes this zero 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 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 eight to get this so let's do that so you get X minus two 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 zero the first degree term and then minus eight is a constant term so minus eight I like to put all my all my degree terms and they're appropriate columns well you don't have to when you do algebraic long division so X minus two 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 what goes into it x squared times I'll put it in the x squared column x squared times X minus two x squared times X is X to the third x squared times minus 2 is minus two x 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 two x squared minus eight how many times is X minus two go into two x squared minus eight 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 can or we could just add the negative and then we have 4x minus 8x goes into 4 X 4 times 4 times X minus 2 is 4x minus 8 and there's no remainder and this is further confirmation x- to definitely was a factor of x 2/3 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 we can write our original problem like actually let me write well I did let me write it down let me write it down here what was it it was 10 let me switch to another color it was 10 x squared I have a bad memory plus 12 X plus 20 plus 12 X plus 20 over X minus 2 times what was that x squared plus 2 X plus 4 x squared plus 2 X plus 4 so all the work we did so far is just to 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 advanced 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're going to see when 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 okay the degree of the numerator and the in the this this part of the fraction I guess you could call it it's going to be 1 less than that so this first degree since it's going to be a zero degree or a constant term here the degree is 2 right the degree is 2 so the degree of its numerator is going to be 1 and since this degree is 1 it could still have a constant term which is a zero degree term so you get BX plus C and you know maybe maybe it does end up just being a constant in which case we can solve for B and it'll just be 0 actually there's another thing you might be asking is like 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'll only get imaginary roots here so this actually isn't more this isn't factorable in the reals anymore there are no more real zeros to this so we 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 in the last video so if we just 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 then that'll be if when we add the comment then it's a times this x squared plus 2x plus 4 plus bx plus c 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 is a might look like a hairy problem and if you do it kind of the slow way that I do in the last video would take you forever it would be three equations and three unknowns very very hairy but we can do is a system oh 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 or CS at a time and the immediate thing is well what can i what x value can I substitute that will cancel out this at least two of the variables two of them a B or C and if I put X is equal to two then this thing becomes zero which will make this whole expression zero 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 so let's do that so if X is equal to 2 X is equal to 2 then what do I get I get 10 x squared so I get 10 times 4 all right 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 right 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 12 a divide both sides by 12 we get a is equal to 7 we're making progress all right 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 if we yellow if we make X equal to 0 then the left-hand side of this this is 0 0 we just left with where's yourself 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 2 C C is equal to 4 we're almost done now let me rewrite our equation up here so we with the A's and the let me see if I can do this all right so now I have 10 x squared plus 12x plus 20 is equal to 7/7 times x squared plus 2x plus 4 plus B X plus C is 4 times X minus 2 almost ran out of space so now we have X's and B 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 zero 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 4z you can let's see 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 you get B is equal to 3 divide both sides by minus 1 and so our original problem up here we now know that B is equal to 3 C was equal to 4 C is 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 3 X 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 higher degree denominators but hopefully you found that a little bit useful