proof of factor theorem due to Fermat
where is a polynomial of degree , beginning with the .
Proof. If we denote , we may write the given polynomial in the form
It’s clear that every is a polynomial of degree with respect to , where has the coefficient 1 and the is . This implies that may be written as a polynomial of degree with respect to , where has the coefficient and the on is equal to , i.e. . So we have
where are certain coefficients. If we return to the indeterminate by substituting in the last identic equation for , we get
When the powers are expanded to polynomials, we see that the expression in the brackets is a polynomial of degree with respect to and with the coefficient of . Thus we obtain
This result is true independently on the value of . If this value is chosen such that , then (2) reduces to (1), Q. E. D.
|Title||proof of factor theorem due to Fermat|
|Date of creation||2013-03-22 15:40:12|
|Last modified on||2013-03-22 15:40:12|
|Last modified by||pahio (2872)|
|Synonym||proof of factor theorem without division|