1. ## Partial Fraction Decomposition

Write the partial fraction decomposition of each rational expression.

(1) 3x/(x + 2) (x - 4)

(2) (2x + 4)/(x^3 - 1)

2. $\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{2(x+2)+(x-4)}{(x+2)(x-4)}.$

3. ## ok but....

Originally Posted by Krizalid
$\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{2(x+2)+(x-4)}{(x+2)(x-4)}.$
I see that the answer was found but how did you do it?

4. It's just writting the numerator by using factors of the denominator.

Or make it as $\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{a}{x+2}+\frac{b}{x-4}$ and find $\displaystyle a,b.$ (I can't provide more help, probably someone else's gonna help a bit more.)

5. Originally Posted by Krizalid
It's just writting the numerator by using factors of the denominator.

Or make it as $\displaystyle \frac{3x}{(x+2)(x-4)}=\frac{a}{x+2}+\frac{b}{x-4}$ and find $\displaystyle a,b.$ (I can't provide more help, probably someone else's gonna help a bit more.)
This is exactly how to solve the problem. Then multiply both sides by $\displaystyle (x+2)(x-4)$, yielding:

$\displaystyle 3x = a(x - 4) + b(x + 2)$

$\displaystyle 3x = ax - 4a + bx + 2b$

$\displaystyle 3x = (a+b)x + (2b - 4a)$

from which we deduce

$\displaystyle a + b = 3$
$\displaystyle 2b - 4a = 0$

and of course the solution is a = 1, b = 2, as Krizalid indicated.

6. Originally Posted by icemanfan
This is exactly how to solve the problem. Then multiply both sides by $\displaystyle (x+2)(x-4)$, yielding:

$\displaystyle 3x = a(x - 4) + b(x + 2)$
Alternatively, you may just plug in x = 4 and then solve for b. Then plug in x = -2 and solve for a.