anonymous
  • anonymous
Find the partial fraction decomposition of the rational function. 9x^2 − 3x − 9 x^4 + 3x3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
are you looking for an answer or the process?
anonymous
  • anonymous
both
anonymous
  • anonymous
well first your need to set up the equivalence \[\frac{9x^2-3x-9}{x^3(x+3)}=\]\[\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}\]

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anonymous
  • anonymous
ok
anonymous
  • anonymous
Thats whats confusing me
anonymous
  • anonymous
what is confusing me is that this interface is buggy
anonymous
  • anonymous
now it works...
anonymous
  • anonymous
yea
anonymous
  • anonymous
then rewrite that as \[9x^2-3x-0=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]
anonymous
  • anonymous
that should be a 9 where that 0 is
anonymous
  • anonymous
k
anonymous
  • anonymous
then rewrite that as \[9x^2-3x-9=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]
anonymous
  • anonymous
so are we good to this point?
anonymous
  • anonymous
yes
anonymous
  • anonymous
now we can choose x=0 and determine that C=-3
anonymous
  • anonymous
ok
anonymous
  • anonymous
and next let x=-3 to determine that D=-3
anonymous
  • anonymous
are you good with what I am doing?
anonymous
  • anonymous
yea i get it
anonymous
  • anonymous
now however we are going to need to choose two different values of x and get a system of two equations in two variable tosolve for A and B
anonymous
  • anonymous
alright
anonymous
  • anonymous
so I chose x=1 and got 3=A+B
anonymous
  • anonymous
Ok
anonymous
  • anonymous
next I chose x=-1 and got 3=A-B
anonymous
  • anonymous
adding those equations together I get 6=2A which says A=3
anonymous
  • anonymous
then using A=3 in either of the other two either of the previous two equations I determine that B=0
anonymous
  • anonymous
Now finally substituting those values for A,B,C,D we get\[\frac{9x^2-3x-3}{x^3(x+3)}=\]\[\frac{3}{x}-\frac{3}{x^3}-\frac{3}{x+3}\]
anonymous
  • anonymous
ok i see
anonymous
  • anonymous
Thats alot!
anonymous
  • anonymous
your welcome, glad you followed it. Long process...
anonymous
  • anonymous
oh yea

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