proof of factor theorem due to Fermat

Lemma (cf. factor theorem).  If the polynomial

$$f(x):=a_0{x}^n+a_1{x}^{n-1}+\mathrm{\cdots }+a_{n-1}x+a_n$$

vanishes at  $x=c$,  then it is divisible by the difference $x-c$, i.e. there is valid the identic equation

$f(x)\equiv (x-c)q(x)$(1)

where $q(x)$ is a polynomial of degree $n-1$, beginning with the $a_0{x}^{n-1}$.

The lemma is here proved by using only the properties of the multiplication and addition, not the division.

Proof.  If we denote  $x-c=y$,  we may write the given polynomial in the form

$$f(x)=a_0{(y+c)}^n+a_1{(y+c)}^{n-1}+\mathrm{\cdots }+a_{n-1}(y+c)+a_n.$$

It’s clear that every ${(y+c)}^k$ is a polynomial of degree $k$ with respect to $y$, where $y^k$ has the coefficient 1 and the is $c^k$.  This implies that $f(x)$ may be written as a polynomial of degree $n$ with respect to $y$, where $y^n$ has the coefficient $a_0$ and the on $y$ is equal to  $a_0{c}^n+a_1{c}^{n-1}+\mathrm{\cdots }+a_{n-1}c+a_n$, i.e. $f(c)$.  So we have

$$f(x)=a_0{y}^n+b_1{y}^{n-1}+b_2{y}^{n-2}+\mathrm{\cdots }+b_{n-1}y+f(c)=f(c)+y\cdot (a_0{y}^{n-1}+b_1{y}^{n-2}+\mathrm{\cdots }+b_{n-1}+a_n),$$

where  $b_1,b_2,\mathrm{\dots },b_{n-1}$  are certain coefficients.  If we return to the indeterminate $x$ by substituting in the last identic equation $x-c$ for $y$, we get

$$f(x)\equiv f(c)+(x-c)[a_0{(x-c)}^{n-1}+b_1{(x-c)}^{n-2}+\mathrm{\cdots }+b_{n-1}].$$

When the powers ${(x-c)}^k$ are expanded to polynomials, we see that the expression in the brackets is a polynomial $q(x)$ of degree  $n-1$  with respect to $x$ and with the coefficient $a_0$ of $x^{n-1}$.  Thus we obtain

$f(x)\equiv f(c)+(x-c)q(x).$(2)

This result is true independently on the value of $c$.  If this value is chosen such that  $f(c)=0$,  then (2) reduces to (1), Q. E. D.