Lemma (cf. factor theorem). If the polynomial $$f(x) := a_0x^n\!+\!a_1x^{n-1}\!+\cdots+\!a_{n-1}x\!+\!a_n$$ vanishes at $x = c$ , then it is divisible by the difference $x\!-\!c$ , i.e. there is valid the identic equation

(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 $$f(x) = a_0(y\!+\!c)^n\!+\!a_1(y\!+\!c)^{n-1}\!+\cdots+\!a_{n-1}(y\!+\!c)\!+\!a_n.$$ 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 constant term 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 term independent on $y$ is equal to $ a_0c^n\!+\!a_1c^{n-1}\!+\cdots+\!a_{n-1}c\!+\!a_n$ , i.e. $f(c)$ . So we have $$f(x) = a_0y^n\!+\!b_1y^{n-1}\!+\!b_2y^{n-2}\!+\cdots+\!b_{n-1}y\!+f(c) = f(c)+y\cdot(a_0y^{n-1}\!+\!b_1y^{n-2}\!+\cdots+\!b_{n-1}\!+\!a_n),$$ where $b_1,\,b_2,\,\ldots,\,b_{n-1}$ are certain coefficients. If we return to the indeterminate $x$ by substituting in the last identic equation $x\!-\!c$ for $y$ , we get $$f(x) \equiv f(c)+(x\!-\!c)[a_0(x\!-\!c)^{n-1}\!+\!b_1(x\!-\!c)^{n-2}\!+\cdots+\!b_{n-1}].$$ 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

(2)

Bibliography

1

ERNST LINDELÖF: Johdatus korkeampaan analyysiin (`Introduction to Higher Analysis'). Fourth edition. WSOY, Helsinki (1956).