more decomposition into fractions

[cjswonderfulgirl]

here's another one i just don't see where i went wrong... help would be soooo appreciated
the original problem:
3x^2 +8x+4/(x^2 -4)(3x-2)
so i got as far as:
3x^2 +8x+4/(x^2 -4)(3x-2)=Ax+B/x^2 -4 + C/3x-2
3x^2 +8x+4=Ax+B(3x-2)+C(x^2 -4)
3x^2 +8x+4=(3A+C)x^2 +(-2A+3B)x+(-2B-4C)

then i found that
3=3A+C
8=-2A+3B
4=-2B-4C

from which (it would be too long to type out i believe) i found C=-3 A=2 and B=4

so thus 3x^2 +8x+4/(x^2 -4)(3x-2)=2x+4/(x^2 -4) - 3/3x-2

and i really can't see where i went wrong, because i'm pretty sure i have a, b, and c correct... can someone pleas help?

 

[galactus]

Notice the difference of two squares?. \(\displaystyle x^{2}-4=(x+2)(x-2)\)

and \(\displaystyle 3x^{2}+8x+4=(x+2)(3x+2)\)

\(\displaystyle \frac{(x+2)(3x+2)}{(x+2)(x-2)(3x-2)}\)

\(\displaystyle \frac{A}{x-2}+\frac{B}{3x-2}=(3x+2)\)

Now, try again.

 

[cjswonderfulgirl]

i got A=0 for that, which i doubt is possible...i got it by
3x^2 +8x+4=A(x-2)(3x-2)+B(x+2)(3x+2)+C(x-2)(x+2)
substitute x=-2
3(-2)^2 +8(-2)+4=A(-2-2)(3<-2>-2)+0+0
12-16+4=A(-4)(-8)
0=-32A
which doesn't work...

 

[galactus]

Did you see the simplication I made?. You do not need all that because the x+2 cancels out.

\(\displaystyle \frac{A}{x-2}+\frac{B}{3x-2}=3x+2\)

\(\displaystyle 3Ax-2A+Bx-2B=3x+2\)

Like coefficients:

\(\displaystyle 3A+B=3\)

\(\displaystyle -2A-2B=2\)

Now, solve.

 

[cjswonderfulgirl]

so then would my final answer be
2/x-2 -3/3x-2?

 

[cjswonderfulgirl]

can anyone confirm that? i enter my answers online and i only get two chances before i get a new problem...and this one took me a good 45 minutes to work out...