Precalculus, Chapter 7, 7.4, Section 7.4, Problem 34

(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) + B(x-3)(x+2) + C(x-3)(x-2)
Then, plug-in the roots of each factor.
For the factor (x-2), its root is x=2.
2+6 = A(2-2)(2+2)+B(2-3)(2+2)+C(2-3)(2-2)
8=A(0)(4)+B(-1)(4)+C(-1)(0)
8=-4B
8/(-4)=(-4B)/(-4)
-2=B
For the factor (x + 2), its root is x=-2.
-2+6= A(-2-2)(-2+2) + B(-2-3)(-2+2)+C(-2-3)(-2-2)
4=A(-4)(0)+B(-5)(0)+C(-5)(-4)
4=20C
4/20=(20C)/20
1/5=C
And for the factor (x-3), its root is x=3.
3+6=A(3-2)(3+2) + B(3-3)(3+2) + C(3-3)(3-2)
9=A(1)(5) + B(0)(5) + C(0)(1)
9=5A
9/5=(5A)/5
9/5=A
So the partial fraction decomposition of the given rational expression is:
(9/5)/(x-3) + (-2)/(x-2)+ (1/5)/(x+2)
And this simplifies to:
= 9/(5(x-3)) - 2/(x-2) +1/(5(x+2))

To check, express the three fractions with same denominators.
9/(5(x-3)) - 2/(x-2) +1/(5(x+2))
= 9/(5(x-3))*((x-2)(x+2))/((x-2)(x+2)) - 2/(x-2)*(5(x-3)(x+2))/(5(x-3)(x+2)) + 1/(5(x+2))*((x-3)(x-2))/((x-3)(x-2))
= (9(x-2)(x+2))/(5(x-3)(x-2)(x+2))- (10(x-3)(x+2))/(5(x-3)(x-2)(x+2))+((x-3)(x-2))/(5(x-3)(x-2)(x+2))
=(9(x^2-4))/(5(x-3)(x-2)(x+2)) - (10(x^2-x-6))/(5(x-3)(x-2)(x+2))+(x^2-5x+6)/(5(x-3)(x-2)(x+2))
Now that they have same denominators, proceed to add/subtract them.
= (9(x^2-4) - 10(x^2-x-6) + x^2-5x+6)/(5(x-3)(x-2)(x+2))
= (9x^2-36-10x^2+10x+60+x^2-5x+6)/(5(x-3)(x-2)(x+2))
=(5x+30)/(5(x-3)(x-2)(x+2))
= (5(x+6))/(5(x-3)(x-2)(x+2))
=(x+6)/((x-3)(x-2)(x+2))

Therefore, (x+6)/((x-3)(x-2)(x+2))=9/(5(x-3)) - 2/(x-2) +1/(5(x+2)) .