`(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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