2x^3-x^2-50x+25=x^2(2x-1)-25(2x-1)=(2x-1)(x^2-25)=(2x-1)(x-5)(x+5)=A/(2x-1)+B/(x-5)+C/(x+5).
Ax^2-25A+2Bx^2+9Bx-5B+2Cx^2-11Cx+5C=13x^2-17x+30.
Equating terms: -25A-5B+5C=30 so -5A-B+C=6; 9B-11C=-17; A+2B+2C=13, so 5A+10B+10C=65. Add this to the first equation: 9B+11C=71 add to 9B-11C=-17 gives 18B=54, so B=3, and 11C=71-27=44, so C=4. -5A=6-1=5 and A=-1. Therefore the partial fractions are: 3/(x-5)+4/(x+5)-1/(2x-1).