resolve into partial fraction:((2x)/((x^2-1)(x-1)))

I'm not sure what you mean here. \[\frac{2x}{(x^2-1)(x-1)}=\frac{2x}{(x-1)^2(x+1)}\]

the first one is the one that needs to be solved into partial fractions

\[\frac{2 x}{(x-1)^2 (1+x)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}\]

@sam does the answer end ther isnt there perhaps more its a 7 marks question

\[\text{That's just a step,}\] \[\begin{array}{l} \text{Multiply both sides by }(x-1)^2 (x+1)\text{ and simplify:} \\ 2 x=A \left(x^2-1\right)+C (x-1)^2+B (x+1) \\ \text{You need to expand it } \\\end{array}\] \begin{array}{l} 2 x=-A+B+C+\left(A+C\right) x^2+\left(B-2 C\right) x \\ \text{compare the coefficient of x^2, x , and constant} \\ \text{Coefficient of x^2: }0=-A+B+C \\ \text{Coefficient of x: } 2=B-2 C \\ \text{Constant: } 0=A+C \\ \end{array} \[\text{then solve by matrix or substitution}\]

@.Sam. is exactly right where are you stuck? solving the system?

yes solving the system

did you try making a matrix for this? that's how I think I would do this

|dw:1338161220050:dw|sorry, forgot some zeros :P

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