Question

Question #028d7

Answer 1

`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)`