If this problem is really D2, Im probably doing something wrong but heres what I found.
Let x=0 and substitute into the given equation:
0*p(1)=(15)*p(0)
so p(0)=0
Let x=1 and substitute again
1*p(0)=(14)*p(1)
so P(1)=0
this continues to p(14)=0
now for p(15)
15*p(14) = 0*p(15)
0 = 0*p(15)
so p(15) is indeterminate.
Ok lets define p(x) = q(x)*x*(x1)*...*(x14) and try to find a formula for q(x) by going further. Try x=16
16*p(15) = 1*p(16)
16*q(15)*15! = q(16)*16!
q(15)=q(16)
Try x=17
17*p(16) = 2*p(17)
17*q(16)*16! = 2*q(17)*17!/2!
q(16)=q(17)
Try x=18
18*p(17) = 3*p(18)
18*q(17)*17!/2! = 3*q(18)*18!/3!
q(17)=q(18)
etc... so q(n)=q(n+1) [for all integer n>14]
This means q(n)= C where C is a constant.
So the polynomial p(x)=C*x*(x1)*(x2)*...*(x14)

Posted by Jer
on 20101126 17:27:14 