moses_fj Group Title solve Partial Fractions (x-5)/(x^2+2x-3 one year ago one year ago

1. ChristineToTheRescue Group Title

|dw:1363121278806:dw| Looks like this, right?

2. ChristineToTheRescue Group Title

So basically, We begin by factoring the denominator.

3. zepdrix Group Title

$\large \frac{x-5}{x^2+2x-3} \qquad = \qquad \frac{x-5}{(x-1)(x+3)}$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 Group Title

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 Group Title

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