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anonymous
 one year ago
Partial Fraction decomposition:
(x^2)/(x^42x^28)
anonymous
 one year ago
Partial Fraction decomposition: (x^2)/(x^42x^28)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you factor the denominator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes i factored denominator: (x^2+2)(x2)(x+2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok then we have \[\frac{x^2}{(x^2+2)(x2)(x+2)}\] \[=\frac{Ax+B}{x^2+2}+\frac{C}{x2}+\frac{D}{x2}\] as a fist step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually i think in this case \(A=0\) but it doesn't need to be

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would (A/x2) + (B/x+2) + ((Cx+D)/(x^2+2)) be correct as well?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now add \[Ax+B(x+2)(x2)+C(x^2+2)(x2)+D(x^2+2)(x+2)=x^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it doesn't matter since addition is commutative but since i spent all that time writing it in latex, would you mind using it so i don't have to start over?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes that's completely fine. thank you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now there are a couple of ways to proceed the easiest way is probably to replace \(x\) by \(2\) and see what you get

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first two terms drop out, you are left with \[D(2^2+2)(2+2)=2^2\] and you can solve that for \(D\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is really the basic gimmick is it clear what i did? you get \[D=\frac{1}{6}\] pretty much in your head

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one minute i am reading your work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok take your time i am going to get a snack if it is not clear, ask

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand what you did. I worked the problem and got D=1/6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we can use the same gimmick only replace \(x\) yb \(2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so that cancels out the first and last term. I can now solve for C. C=1/6?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and now I plugin C and D into the first equation to solve for A & B?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can do that, but first lets think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first part is \[(Ax+B)(x+2)(x2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you multiply that out mentally you get an \(Ax^3\) term there is not \(x^3\) so \(A=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I do not understand why Ax^3 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because the numerator is \(x^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the exponent in the denominator cannot be greater than x^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0look at the original question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or even just this part \[(Ax+B)(x+2)(x2)+C(x^2+2)(x2)+D(x^2+2)(x+2)=x^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on the right hand side of the equal sign, there is only a little lonely \(x^2\) no \(x^3\) at all

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on the left side there will be, when you multiply out, one \(x^3\) term it will be \(Ax^3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So A must equal 0 in order for the equation to be true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh no scratch that, i am totally wrong, sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there are other \(x^3\) terms there i forgot about them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so yeah, substitute back what you know to solve for \(A\) and\(B\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you will find that \(A=0\) in any case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2 = Ax+B(x+2)(x2)  (1/6)(x^2+2)(x2)+(1/6)(x^2+2)(x+2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, now lets find \(B\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0without doing a raft of solving a system of equations, which is the other long and tedious method of doing this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(Ax+B)(x+2)(x2)\frac{1}{6}(x^2+2)(x2)+\frac{1}{6}(x^2+2)(x+2)=x^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what will the constant be? the term without any \(x\) or any \(x^2\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it is not clear what i mean, let me know

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok i will work on the problem and let you know.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I am confused. I am stuck on the part where you said "what will the constant be? the term without any x"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from here \[\frac{1}{6}(x^2+2)(x2)\] when you mutiply out, you will get a number without an \(x\) namely \[\frac{1}{6}\times 2\times (2)=\frac{2}{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from here \[\frac{1}{6}(x^2+2)(x+2)\] you will also get \(\frac{2}{3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. That makes sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from here \[(Ax+B)(x+2)(x2)\] then number will be \(4B\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0]scratch that it will be \[4B\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0on the right hand side there is no constant at all,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that means \[\frac{2}{3}+\frac{2}{3}4B=0\]or \[4B=\frac{4}{3}\] making \[B=\frac{1}{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. Why does the right side equal 0 though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i was mistaken before when i was trying to explain why \(A=0\) but now i am not the original numerator is just \(x^2\) it has no constant in it i

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I mean coefficient. yes there is no constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you got \[(Ax+B)(x+2)(x2)\frac{1}{6}(x^2+2)(x2)+\frac{1}{6}(x^2+2)(x+2)=x^2\] the contant on the left is \[\frac{4}{3}4B\] and on the right it is \(0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that makes \(B=\frac{1}{3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You made the constants equal to each other.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I get the answer: [(1/3)/(x^2+2)] + [(1/6)/(x+2)] + [(1/6)/(x2)] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes sometimes you can get away without doing that, but not in this case because of the \(x^2+2\) which is never zero so you can't use the previous trick of substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is one way to write it, yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another way to write it would be [1/3(x^2+2)]  [1/6(x+2)] + [1/6(x2)] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So would my answer that I put be correct as well?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0normally you would write \[\frac{1}{3(x^2+2)}\frac{1}{6(x+2)}+\frac{1}{6(x2)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is the same only in this topic do you see compound fractions allowed as an answer, but yes, both are the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok that makes sense. Thank you so much for you time I appreciate it. Good luck with your studies
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