help with partial fractions

\[\frac{9x^2+19x+56}{x(x-8)(x-7)}=\frac{A}{x}+\frac{B}{x-8}+\frac{C}{x-7}\] =\[\frac{A(x-8)(x-7)+Bx(x-7)+Cx(x-8)}{x(x-8)(x-7)}\] since the denominators are the same then the numerators are the same \[9x^2+19x+56=A(x-8)(x-7)+Bx(x-7)+Cx(x-8)\] \[9x^2+19x+56=A(x^2-15x+56)+B(x^2-7x)+C(x^2-8x)\]

\[9x^2+19x+56=Ax^2+Bx^2+Cx^2-15Ax-7Bx-8Cx+56A\] \[9x^2+19x+56=x^2(A+B+C)+x(-15A-7B-8C)+56A\]

so A+B+C=9 -15A-7B-8C=19 56A=56

so 56A=56 implies A=1 so we have 1+B+C=9 and -15-7B-8C=19

B+C=8 and -7B-8C=34 B=8-C -7(8-C)-8C=34 -56+7C-8C=34 -C=90 C=-90 B=8-(-90)=8+90=98

A=1, B=98, C=-90 A/x+B/(x-8)+C/(x-7)=1/x+98/(x-8)-90/(x-7)

i didn't do on scratch paper there may be a mistake above but if you look at what i was trying to do you should be able to figure how to do this problem and get the right answer you can also check my if you want by combining the partial fractions