Hi there!
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⇒ [tex]- \frac{1}{2(x + 3)} + \frac{1}{2(x - 3)}[/tex]
Step by Step :
∴ Factor [tex]x^2 - 9[/tex] : [tex]( x+ 3 ) ( x - 3)[/tex]
[tex]= \frac{3}{(x +3)(x - 3)}[/tex]
∴ Create the partial fraction template using the denominator [tex]( x + ) ( x - 3)[/tex]
[tex]\frac{3}{( x + 3) ( x -3)} = \frac{a^0}{x + 3} + \frac{a^1}{x - 3}[/tex]
∴ Multiply the equation by the denominator.
[tex]\frac{3(x+3)(x-3)}{( x+3)(x-3)} = \frac{a^0(x +3)(x-3)}{x + 3} + \frac{a^1 (x+3)(x-3)}{x - 3}[/tex]
∴ Simplify.
[tex]3 = a^0 ( x - 3) + a^1 ( x+3)[/tex]
∴ Solve the unknown parameters by plugging the real roots of the denominator : - 3,3
⇅
∴ Solve the denominator root -3 : [tex]a^0 = -\frac{1}{2}[/tex]
∴ For the denominator root 3 : [tex]a^1 = \frac{1}{2}[/tex]
[tex]a^0 = -\frac{1}{2} , a^1 \frac{1}{2}[/tex]
∴Plug the solutions to the partial fraction parameters to obtain the final result.
[tex]\frac{(-\frac{1}{2}) }{x + 3} + \frac{\frac{1}{2} }{ x- 3}[/tex]
∴ Simplify
[tex]- \frac{1}{2(x + 3)} + \frac{1}{2(x - 3)}[/tex]
Hope this helped you!
