Question #028d7

1 Answer
Mar 10, 2016

Answer:

#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)#