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.

  • 1
    $\begingroup$ How to format your questions. $\endgroup$ – user137731 May 27 '15 at 23:05
  • 2
    $\begingroup$ On the left side of the equation $x^2-2x+3=...$ , the denominator is missing. $\endgroup$ – Peter May 27 '15 at 23:07

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.

  • $\begingroup$ 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

If you substitute $x=2i$, this gives


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 $x^3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.