一條F.4的數學表題目

f(x) = x^2006 + 2006x^2 – 2007

a.
Consider
f(1)=(1)^2006 + 2006(1)^2 – 2007=0
f(-1)=(-1)^2006 + 2006(-1)^2 – 2007=0
By Reminder theorem,
f(x) is divisible by (x+1) and (x-1)

b.
f(x)=(x^2-1)Q(x) + ax+b
f(x)=(x+1)(x-1)Q(x) + ax+b
f(1)=a+b
=>a+b=0 ----------------(1)
f(-1)=-a+b
b-a=0
a=b------------------(2)
From (1)
b+b=0
b=0
a=b=0

c.
By a,
f(x) = x^2006 + 2006x^2 – 2007 is divisible by (x+1) and (x-1)
f(4) = 4^2006+2006(4)^2-2007 is divisible by (4+1) and (4-1)
4^2006+2006(4)^2-2007 is divisible by 5 and 3
4^2006+2006(4)^2-2007 is not a prime number.

+10000000000000000000000000
其實I don't know