anonymous
  • anonymous
Partial Fraction decomposition: (x^2)/(x^4-2x^2-8)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
can you factor the denominator?
anonymous
  • anonymous
Yes i factored denominator: (x^2+2)(x-2)(x+2)
anonymous
  • anonymous
ok then we have \[\frac{x^2}{(x^2+2)(x-2)(x+2)}\] \[=\frac{Ax+B}{x^2+2}+\frac{C}{x-2}+\frac{D}{x-2}\] as a fist step

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anonymous
  • anonymous
actually i think in this case \(A=0\) but it doesn't need to be
anonymous
  • anonymous
would (A/x-2) + (B/x+2) + ((Cx+D)/(x^2+2)) be correct as well?
anonymous
  • anonymous
now add \[Ax+B(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2\]
anonymous
  • anonymous
yes of course
anonymous
  • anonymous
it 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
  • anonymous
Yes that's completely fine. thank you
anonymous
  • anonymous
so 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
  • anonymous
the 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
  • anonymous
that is really the basic gimmick is it clear what i did? you get \[D=\frac{1}{6}\] pretty much in your head
anonymous
  • anonymous
one minute i am reading your work
anonymous
  • anonymous
ok take your time i am going to get a snack if it is not clear, ask
anonymous
  • anonymous
I understand what you did. I worked the problem and got D=1/6
anonymous
  • anonymous
ok good
anonymous
  • anonymous
now we can use the same gimmick only replace \(x\) yb \(-2\)
anonymous
  • anonymous
Ok, so that cancels out the first and last term. I can now solve for C. C=-1/6?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and now I plugin C and D into the first equation to solve for A & B?
anonymous
  • anonymous
you can do that, but first lets think
anonymous
  • anonymous
Ok..
anonymous
  • anonymous
the first part is \[(Ax+B)(x+2)(x-2)\]
anonymous
  • anonymous
if you multiply that out mentally you get an \(Ax^3\) term there is not \(x^3\) so \(A=0\)
anonymous
  • anonymous
Sorry I do not understand why Ax^3 = 0
anonymous
  • anonymous
because the numerator is \(x^2\)
anonymous
  • anonymous
no \(x^3\) in it
anonymous
  • anonymous
that means \(A=0\)
anonymous
  • anonymous
So the exponent in the denominator cannot be greater than x^2?
anonymous
  • anonymous
look at the original question
anonymous
  • anonymous
Ok
anonymous
  • anonymous
or even just this part \[(Ax+B)(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2\]
anonymous
  • anonymous
on the right hand side of the equal sign, there is only a little lonely \(x^2\) no \(x^3\) at all
anonymous
  • anonymous
on the left side there will be, when you multiply out, one \(x^3\) term it will be \(Ax^3\)
anonymous
  • anonymous
So A must equal 0 in order for the equation to be true
anonymous
  • anonymous
ooh no scratch that, i am totally wrong, sorry
anonymous
  • anonymous
No worries.
anonymous
  • anonymous
there are other \(x^3\) terms there i forgot about them
anonymous
  • anonymous
so yeah, substitute back what you know to solve for \(A\) and\(B\)
anonymous
  • anonymous
you will find that \(A=0\) in any case
anonymous
  • anonymous
\[x^2 = Ax+B(x+2)(x-2) - (1/6)(x^2+2)(x-2)+(1/6)(x^2+2)(x+2)\]
anonymous
  • anonymous
correct?
anonymous
  • anonymous
yeah, now lets find \(B\)
anonymous
  • anonymous
without doing a raft of solving a system of equations, which is the other long and tedious method of doing this
anonymous
  • anonymous
\[(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2\]
anonymous
  • anonymous
what will the constant be? the term without any \(x\) or any \(x^2\)?
anonymous
  • anonymous
if it is not clear what i mean, let me know
anonymous
  • anonymous
Ok i will work on the problem and let you know.
anonymous
  • anonymous
Ok, I am confused. I am stuck on the part where you said "what will the constant be? the term without any x"
anonymous
  • anonymous
from here \[-\frac{1}{6}(x^2+2)(x-2)\] when you mutiply out, you will get a number without an \(x\) namely \[-\frac{1}{6}\times 2\times (-2)=\frac{2}{3}\]
anonymous
  • anonymous
from here \[\frac{1}{6}(x^2+2)(x+2)\] you will also get \(\frac{2}{3}\)
anonymous
  • anonymous
Ok. That makes sense.
anonymous
  • anonymous
from here \[(Ax+B)(x+2)(x-2)\] then number will be \(4B\)
anonymous
  • anonymous
]scratch that it will be \[-4B\]
anonymous
  • anonymous
Ok makes sense.
anonymous
  • anonymous
on the right hand side there is no constant at all,
anonymous
  • anonymous
that means \[\frac{2}{3}+\frac{2}{3}-4B=0\]or \[4B=\frac{4}{3}\] making \[B=\frac{1}{3}\]
anonymous
  • anonymous
Ok. Why does the right side equal 0 though?
anonymous
  • anonymous
ok 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
  • anonymous
the constant is 1
anonymous
  • anonymous
no
anonymous
  • anonymous
Sorry I mean coefficient. yes there is no constant
anonymous
  • anonymous
right
anonymous
  • anonymous
you got \[(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2\] the contant on the left is \[\frac{4}{3}-4B\] and on the right it is \(0\)
anonymous
  • anonymous
Makes sense now.
anonymous
  • anonymous
that makes \(B=\frac{1}{3}\)
anonymous
  • anonymous
ok good
anonymous
  • anonymous
You made the constants equal to each other.
anonymous
  • anonymous
So I get the answer: [(1/3)/(x^2+2)] + [(-1/6)/(x+2)] + [(1/6)/(x-2)] ?
anonymous
  • anonymous
yes 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
  • anonymous
that is one way to write it, yes
anonymous
  • anonymous
Another way to write it would be [1/3(x^2+2)] - [1/6(x+2)] + [1/6(x-2)] ?
anonymous
  • anonymous
So would my answer that I put be correct as well?
anonymous
  • anonymous
normally you would write \[\frac{1}{3(x^2+2)}-\frac{1}{6(x+2)}+\frac{1}{6(x-2)}\]
anonymous
  • anonymous
it is the same only in this topic do you see compound fractions allowed as an answer, but yes, both are the same
anonymous
  • anonymous
Ok that makes sense. Thank you so much for you time I appreciate it. Good luck with your studies
anonymous
  • anonymous
yw

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