decomposition of fractions

[cjswonderfulgirl]

i can't even find B in this problem....i think i'm going about it right, though
x^4-5x^3-11x^2+26x-5/(x+2)^2 (x-7)
so in order to make it so the numerator has a smaller(or the same) power as the denominator, i used long division, which produced
x-2 + 19x^2+102x+51/(x+2)^2 (x-7)=A/x-7 +B/x+2 + C/(x+2)^2
from which i got
19x^2+102x+51/(x+2)^2 (x-7)=A(x+2)^2 +B(x-7)(x+2) +C(x-7)
then i got
19x^2+102x+51/(x+2)^2 (x-7)=(A+B)x^2 +(4A-5B+C)x +(4A-14B)
and then
19=A+B
102=4A-5B+C
51+4A-14B-7C
but then somehow i got 157=-81 B which doesn't divide...where did i go wrong?

 

[galactus]

Long division results in

\(\displaystyle \frac{7x^{2}+6x-61}{(x-7)(x+2)^{2}}+x-2\)

Now, set it up

\(\displaystyle \frac{A}{x-7}+\frac{B}{(x+2)^{2}}+\frac{C}{x+2}=7x^{2}+6x-61\)

Just don't forget about the x-2 when you're done.

 

[cjswonderfulgirl]

so then is my answer going to be 4/x+7 +3/x+2 +5/(x+2)^2?

 

[galactus]

Yes, it is. Except that's x-7 and

\(\displaystyle \frac{4}{x-7}+\frac{5}{(x+2)^{2}}+\frac{3}{x+2}+x-2\)