Question

设f(x)=x^4+ax^3+bx^2+cx+d,其中a、b、c、d是常数.如果f(1)=10,f(2)=20,f(3)

Answer 1

设F(x)=f(x)-10x

∴F(1)=F(2)=F(3)=0

∴F(x)=(x-1)(x-2)(x-3)(x+h)

f(10)=F(10)+100=9*8*7*(10+h)+100 f(x)=F(x)+10x

f(-6)=F(-6)-60=-7*8*9(-6+h)-60

f(10)+f(-6)=10*9*8*7+9*8*7h+100+6*7*8*9-7*8*9h-60

=5040+3024+100-60

=8104