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moses_fj
solve Partial Fractions (x-5)/(x^2+2x-3
|dw:1363121278806:dw| Looks like this, right?
So basically, We begin by factoring the denominator.
\[\large \frac{x-5}{x^2+2x-3} \qquad = \qquad \frac{x-5}{(x-1)(x+3)}\]Understand how that factors? Ok the next step is to split up those factors in the denominator. It will look like this.\[\large \frac{x-5}{(x-1)(x+3)} \qquad = \qquad \frac{A}{x-1}+\frac{B}{x+3}\]Where A and B are some unknown constants.
Multiply both sides by the denominator on the left.\[\large \cancel{(x-1)}\cancel{(x+3)}\frac{x-5}{\cancel{(x-1)}\cancel{(x+3)}} \qquad = \\ \qquad \frac{A}{\cancel{x-1}}\cancel{(x-1)}(x+3)+\frac{B}{\cancel{x+3}}(x-1)\cancel{(x+3)}\] Leaving us with,\[\large x-5=A(x+3)+B(x-1)\]
Then solve for A and B by `equating like terms`. Then plug the A and B back into this,\[\large \frac{A}{x-1}+\frac{B}{x+3}\]And that's it! :D