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decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$

the way my teacher wants us to solve is by substitution values for x,

I set it up like this:

(after setting the variables to the common denominator and getting rid of the denominator in the original equation)

$x^2-2x+3= \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}$

1)Let $x=1$, after plugging in for $x$ I got $2=5B$, or $B=\frac{2}{5}$

2)let the $x=0$, I get $3=-4A+4B+D$, and if I substitute B and simplify I get $D=4A-\frac{7}{5}$. Furthermore, no matter what value I plug in I still end up with two variables and cant seem to find a way to eliminate one. It seems like I have to set it up as the triple system of equations but I just dont know how to apply it here. Would really appreciate if you guys could give me a hint on how to go about it.

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    $\begingroup$ How to format your questions. $\endgroup$ – user137731 May 27 '15 at 23:05
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    $\begingroup$ On the left side of the equation $x^2-2x+3=...$ , the denominator is missing. $\endgroup$ – Peter May 27 '15 at 23:07
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Hint : Start with $$\frac{x^2-2x+3}{(x-1)^2(x^2+4)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{CX+D}{x^2+4}$$

Multiply with $(x-1)^2(x^2+4)$ and then insert the special values.

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  • $\begingroup$ I did set it up like that, but i keep getting 2 unknowns and have no clue how to go down to one. $\endgroup$ – Jx1 May 27 '15 at 23:23
  • $\begingroup$ The insert of special values does not always give ALL solutions. What are the equations you got ? Did you get $4$ equations ? $\endgroup$ – Peter May 27 '15 at 23:25
  • $\begingroup$ after making x=1, I've got B=2/5, then I did x=0, and got D=4A-(8/5)+3 and im stuck at this point since no matter what value for X i put. i still have D and C as unknowns $\endgroup$ – Jx1 May 27 '15 at 23:27
  • $\begingroup$ Try to solve the rest using the value for $B$. $\endgroup$ – Peter May 27 '15 at 23:28
  • $\begingroup$ I did, still end up with D and C as unknowns $\endgroup$ – Jx1 May 27 '15 at 23:30
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If you substitute $x=2i$, this gives

$\;\;\;-1-4i=((2Ci+D)(2i-1)^2=(-3-4i)(2Ci+D)=(-3D+8C)-(6C+4D)i$

Therefore $8C-3D=-1\;\;$ and $\;\;6C+4D=4,\;$ so

$\hspace{.6 in}16C-6D=-2\;\;$ and $\;\;9C+6D=6\implies 25C=4\implies C=\frac{4}{25}$.

Then $D=1-\frac{3}{2}C=1-\frac{6}{26}=\frac{19}{25},\;\;$ and $\;\;A=-\frac{4}{25}\;$ since $0=A+C$ from the coefficient of $x^3$.

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