Mathematics
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OpenStudy (anonymous):
Find the partial fraction decomposition of the rational function.
9x^2 − 3x − 9
x^4 + 3x3
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OpenStudy (anonymous):
are you looking for an answer or the process?
OpenStudy (anonymous):
both
OpenStudy (anonymous):
well first your need to set up the equivalence \[\frac{9x^2-3x-9}{x^3(x+3)}=\]\[\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}\]
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
Thats whats confusing me
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OpenStudy (anonymous):
what is confusing me is that this interface is buggy
OpenStudy (anonymous):
now it works...
OpenStudy (anonymous):
yea
OpenStudy (anonymous):
then rewrite that as
\[9x^2-3x-0=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]
OpenStudy (anonymous):
that should be a 9 where that 0 is
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OpenStudy (anonymous):
k
OpenStudy (anonymous):
then rewrite that as
\[9x^2-3x-9=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]
OpenStudy (anonymous):
so are we good to this point?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
now we can choose x=0 and determine that C=-3
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
and next let x=-3 to determine that D=-3
OpenStudy (anonymous):
are you good with what I am doing?
OpenStudy (anonymous):
yea i get it
OpenStudy (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
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OpenStudy (anonymous):
alright
OpenStudy (anonymous):
so I chose x=1 and got 3=A+B
OpenStudy (anonymous):
Ok
OpenStudy (anonymous):
next I chose x=-1 and got 3=A-B
OpenStudy (anonymous):
adding those equations together I get 6=2A which says A=3
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OpenStudy (anonymous):
then using A=3 in either of the other two either of the previous two equations I determine that B=0
OpenStudy (anonymous):
Now finally substituting those values for A,B,C,D we get\[\frac{9x^2-3x-3}{x^3(x+3)}=\]\[\frac{3}{x}-\frac{3}{x^3}-\frac{3}{x+3}\]
OpenStudy (anonymous):
ok i see
OpenStudy (anonymous):
Thats alot!
OpenStudy (anonymous):
your welcome, glad you followed it. Long process...
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OpenStudy (anonymous):
oh yea