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`(x + 6)/(x^3 - 3x^2 - 4x + 12)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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`(x+6)/(x^3-3x^2-4x+12)`

To decompose this to partial fractions, factor the denominator.

`x^3 - 3x^2-4x + 12`

`= (x^3-3x^2) + (-4x + 12)`

`= x^2(x-3) - 4(x -3)`

`=(x-3)(x^2-4)`

`=(x-3)(x-2)(x+2)`

Then, write a fraction for each factor. Since the numerators are still unknown, assign a variable for each numerator.

`A/(x-3)`   ,   `B/(x-2)`   and   `C/(x+2)`

Add these three fractions and set it equal to the given fraction.

`(x+6)/((x-3)(x-2)(x+2)) = A/(x-3)+B/(x-2)+C/(x+2)`

To solve for the values of A, B and C, eliminate the fractions in the equation. So, multiply both sides by the LCD.

`(x-3)(x-2)(x+2) *(x+6)/((x-3)(x-2)(x+2)) = (A/(x-3)+B/(x-2)+C/(x+2))*(x-3)(x-2)(x+2)`

`x+6=A(x-2)(x+2) +...

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