Let f(x)=x^4+ax^3+bx^2+cx+d then using the given info:
(1) 10=1+a+b+c+d (because f(1)=10)
(2) 20=16+8a+4b+2c+d (f(2)=20)
(3)=(2)-(1): 10=15+7a+3b+c and d=9-a-b-c from (1), 7a+3b+c=-5
(4) 30=81+27a+9b+3c+d=90+26a+8b+2c; 26a+8b+2c=-60, 13a+4b+c=-30 (f(3)=30)
(5)=(4)-(3): 6a+b=-25, b=-25-6a. c=-5-7a-3b from (3) so c=-5-7a+75+18a=70+11a.
Also, d=9-a-b-c=9-a+25+6a-70-11a=-36-6a.
We have three variables in terms of the fourth and f(x)=x^4+ax^3-(25+6a)x^2+(70+11a)x-36-6a.
f(12)=12^4+12^3a-12^2(25+6a)+12(70+11a)-36-6a;
f(-8)=8^4-8^3a-8^2(25+6a)-8(70+11a)-36-3a.
f(12)+f(-8)=12^4+8^4+1216a-5200-1248a+280+44a-72-12a=19840 (a's total to zero).
19840÷9920=2. ANSWER: 2