0
$\begingroup$

For the equation,

x-1/(x+1)(x-2)^2 = A/(x+1) + B/(x-2) + C/(x-2)^2

            x-1 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)

Using x = -1 gives A = -2/9

Using x = 2 gives C = 1/3

I am however unable to find the value of B

$\endgroup$
  • $\begingroup$ Consider $x=0$ and use the values of A and C. $\endgroup$ – mfl Sep 28 '16 at 12:58
  • $\begingroup$ Or use $x = 1$. Alternatively, since it shall hold for all $x$, compare the coefficients of $x^2$ on both sides. $\endgroup$ – Daniel Fischer Sep 28 '16 at 12:59
  • $\begingroup$ Or insert any value different from $-1$ and $2$ into $x$ $\endgroup$ – Peter Sep 28 '16 at 13:02
  • $\begingroup$ you should look at a simple proof, the general algorithm is then obvious $\endgroup$ – reuns Sep 28 '16 at 13:55
1
$\begingroup$

$$\frac { x-1 }{ \left( x+1 \right) { \left( x+2 \right) }^{ 2 } } =\frac { A }{ x+1 } +\frac { B }{ x+2 } +\frac { C }{ { \left( x+2 \right) }^{ 2 } } \\ x-1=A{ \left( x+2 \right) }^{ 2 }+B\left( x+1 \right) \left( x+2 \right) +C\left( x+1 \right) \\ x-1=\left( A+B \right) { x }^{ 2 }+\left( 4A+3B+C \right) x+4A+2B+C\\ \\ \\ \quad \begin{cases} A+B=0 \\ 4A+3B+C=1 \\ 4A+2B+C=-1 \end{cases}\Rightarrow \begin{cases} A=-B \\ -4B+3B+C=1 \\ -4B+2B+C=-1 \end{cases}\Rightarrow \begin{cases} -B+C=1 \\ -2B+C=-1 \end{cases}\Rightarrow \begin{cases} B=2 \\ A=-2 \\ C=3 \end{cases}\\ \\ \\ $$

$\endgroup$
1
$\begingroup$

As an alternative to solving a linear system, you may compute a couple of limits, since by assuming $$f(x)=\frac{x-1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2} \tag{1} $$ we have: $$ A=\lim_{x\to -1}(x+1)\,f(x) = \lim_{x\to -1}\frac{x-1}{(x+2)^2}=-2\tag{2} $$ $$ C=\lim_{x\to -2}(x+2)^2\,f(x) = \lim_{x\to -2}\frac{x-1}{x+1}=3\tag{3} $$ and $$ f(x)-\frac{A}{x+1}-\frac{C}{(x+2)^2} = \frac{2}{x+2} \tag{4} $$ (i.e. $B=2$) follows by direct computation.

$\endgroup$

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.