I let s = -1:
1 + 2(-1) = -3*(-1-2)(-1)^2 + 5/4*(-1-1)(-1)^2 + C*(-1-1)(-1-2)(-1)^2 + 1/2*(-1-1)(-1-2)
-1 = 9 -5/2 + 3 + 6C
-21/2 = 6C
-7/4 = C
Hmm. You should have this equation:
$\displaystyle 1+2s=A(s-2)s^{2}+B(s-1)s^{2}+C(s-1)(s-2)s+D(s-1)(s-2).$
You can see that there is only an $\displaystyle s$ multiplying C, not a $\displaystyle s^{2}.$
Carry this correction all the way through to see how that changes things.
I got s^2 at the end of C using $\displaystyle \dfrac{1+2s}{(s-1)(s-2)(s)^2}=\dfrac{A}{s-1}\,\dfrac{(s-2)(s)^2}{(s-2)(s)^2}+\dfrac{B}{s-2}\,\dfrac{(s-1)(s)^2}{x(s-1)(s)^2}
+\dfrac{C}{s}\,\dfrac{(s-1)(s-2)(s)^2}{(s-1)(s-2)(s)^2}+\dfrac{D}{s^2}\,\dfrac{(s-1)(s-2)}{(s-1)(s-2)}.$
Given that $\displaystyle \dfrac{3s}{(s-2)(s+1)(s^2+4)}=\dfrac{A}{s-2}\,\dfrac{(s+1)(s^2+4)}{(s+1)(s^2+4)}+\dfrac{B}{s +1}\,\dfrac{(s-2)(2^2+4)}{(s-2)(2^2+4)}
+\dfrac{Cs+D}{s^2+4}\,\dfrac{(Cs+D)(s-2)(s+1)}{(Cs+D)(s-2)(s+1)}$
Can you confirm that C = -17/10 and D = -3/10
the answers say C = -9/20 D = -3/20
You should have
$\displaystyle \dfrac{3s}{(s-2)(s+1)(s^2+4)}=\dfrac{A}{s-2}\,\dfrac{(s+1)(s^2+4)}{(s+1)(s^2+4)}+\dfrac{B}{s +1}\,\dfrac{(s-2)(s^2+4)}{(s-2)(s^2+4)}
+\dfrac{Cs+D}{s^2+4}\,\dfrac{(s-2)(s+1)}{(s-2)(s+1)};$
You'll see that there should be an s in a coupled of places, not a 2. Also, no need to multiply by Cs+D top and bottom, since that's in the numerator, and plays no role in determining the common denominator.
What did you get for A and B?