Let p(x)=x^4+ax^3+bx^2+cx+d, where a, b, c, and d are constants.
If p(1)=9, p(2)=18, and p(3)=27, find the value of
¼(p(10)+p(6)).
(In reply to
If there's only one answer... by Federico Kereki)
Let a be the unknown root, then p(x)=9x+(x1)(x2)(x3)(xa)
p(10)=9*8*7*(10a), p(6)=7*8*9*(6a)=9*8*7*(6+a)
So (p(10)+p(6))/4 = (7*8*9*(10a+6+a)90+54)/4=4*7*8*99=2007
Edited on March 16, 2007, 2:11 pm

Posted by Gamer
on 20070316 14:08:39 