Question #028d7

1 Answer
Mar 10, 2016

a) f(0)

Explanation:

Supposing f(X) is a polynomial of degree n
f(X)=sum_0^n a_i*x^n=a_0+a_1*x+...+a_n.x^n
with
f(X=0)=a_0

For X=x-a
f(x-a)=sum_0^n a_i*(x-a)^n

Dividing f(x-a) by (a-x) or -(x-a) we get

f(x-a)/(-(x-a))=((sum_0^n a_i*(x-a)^n))/(-(x-a))
=(a_0+sum_1^n a_i*(x-a)^n)/(-(x-a))
=a_0/(a-x)-sum_1^n a_i*(x-a)^(n-1)=a_0/(a-x)-a_1-a_2*x-...-a_n*x^(n-1)
So a_0 is the remainder of the division by (a-x).
But, as we saw above, a_0=f(X=0)
=> alternative (a)