Find the partial fraction decomposition of the rational function. 9x^2 − 3x − 9 x^4 + 3x3

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Find the partial fraction decomposition of the rational function. 9x^2 − 3x − 9 x^4 + 3x3

Mathematics
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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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Other answers:

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

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