(5x²+17x+15)/((x+2)²(x-3))≡
A/(x+2)+B/(x+2)²+C/(x-3).
Therefore:
A(x+2)(x-3)+B(x-3)+C(x+2)²=
Ax²-Ax-6A+Bx-3B+Cx²+4Cx+4C=
(A+C)x²+(-A+B+4C)x+(-6A-3B+4C)≡
5x²+ 17x+ 15.
So, A+C=5, C=5-A; -A+B+4C=17; -6A-3B+4C=15.
4C=17+A-B=15+6A+3B, 2=5A+4B, B=(2-5A)/4.
We now have B and C in terms of A, so substitute:
-A+(2-5A)/4+20-4A=17,
-4A+2-5A+80-16A=68.
14=25A, A=14/25; C=5-A=111/25; B=(2-5A)/4=-1/5.
We can write these as decimals: A=0.56, B=-0.2, C=4.44.
Partial fractions:
0.56/(x+2)-0.2/(x+2)²+4.44/(x-3).