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moses_fj

  • 3 years ago

solve Partial Fractions (x-5)/(x^2+2x-3

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  1. ChristineToTheRescue
    • 3 years ago
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    |dw:1363121278806:dw| Looks like this, right?

  2. ChristineToTheRescue
    • 3 years ago
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    So basically, We begin by factoring the denominator.

  3. zepdrix
    • 3 years ago
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    \[\large \frac{x-5}{x^2+2x-3} \qquad = \qquad \frac{x-5}{(x-1)(x+3)}\]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.

  4. zepdrix
    • 3 years ago
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    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,\[\large x-5=A(x+3)+B(x-1)\]

  5. zepdrix
    • 3 years ago
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    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

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