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ChristineToTheRescueBest ResponseYou've already chosen the best response.0
dw:1363121278806:dw Looks like this, right?
 one year ago

ChristineToTheRescueBest ResponseYou've already chosen the best response.0
So basically, We begin by factoring the denominator.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \frac{x5}{x^2+2x3} \qquad = \qquad \frac{x5}{(x1)(x+3)}\]Understand how that factors? Ok the next step is to split up those factors in the denominator. It will look like this.\[\large \frac{x5}{(x1)(x+3)} \qquad = \qquad \frac{A}{x1}+\frac{B}{x+3}\]Where A and B are some unknown constants.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Multiply both sides by the denominator on the left.\[\large \cancel{(x1)}\cancel{(x+3)}\frac{x5}{\cancel{(x1)}\cancel{(x+3)}} \qquad = \\ \qquad \frac{A}{\cancel{x1}}\cancel{(x1)}(x+3)+\frac{B}{\cancel{x+3}}(x1)\cancel{(x+3)}\] Leaving us with,\[\large x5=A(x+3)+B(x1)\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Then solve for A and B by `equating like terms`. Then plug the A and B back into this,\[\large \frac{A}{x1}+\frac{B}{x+3}\]And that's it! :D
 one year ago
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