# Question 028d7

Mar 10, 2016

a) f(0)#

#### Explanation:

Supposing $f \left(X\right)$ is a polynomial of degree $n$
$f \left(X\right) = {\sum}_{0}^{n} {a}_{i} \cdot {x}^{n} = {a}_{0} + {a}_{1} \cdot x + \ldots + {a}_{n} . {x}^{n}$
with
$f \left(X = 0\right) = {a}_{0}$

For $X = x - a$
$f \left(x - a\right) = {\sum}_{0}^{n} {a}_{i} \cdot {\left(x - a\right)}^{n}$

Dividing $f \left(x - a\right)$ by $\left(a - x\right)$ or $- \left(x - a\right)$ we get

$f \frac{x - a}{- \left(x - a\right)} = \frac{\left({\sum}_{0}^{n} {a}_{i} \cdot {\left(x - a\right)}^{n}\right)}{- \left(x - a\right)}$
$= \frac{{a}_{0} + {\sum}_{1}^{n} {a}_{i} \cdot {\left(x - a\right)}^{n}}{- \left(x - a\right)}$
$= {a}_{0} / \left(a - x\right) - {\sum}_{1}^{n} {a}_{i} \cdot {\left(x - a\right)}^{n - 1} = {a}_{0} / \left(a - x\right) - {a}_{1} - {a}_{2} \cdot x - \ldots - {a}_{n} \cdot {x}^{n - 1}$
So ${a}_{0}$ is the remainder of the division by $\left(a - x\right)$.
But, as we saw above, ${a}_{0} = f \left(X = 0\right)$
=> alternative $\left(a\right)$