proof of factor theorem using division
ProofOfFactorTheoremUsingDivision
2005-03-22 12:58:30
2006-06-20 15:20:51
Proof
factor theorem
0
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\begin{lemma}[cf. factor theorem]
Let $R$ be a commutative ring with identity and let $p(x)\in R[x]$ be a polynomial with coefficients in $R$. The element $a\in R$ is a root of $p(x)$ if and only if $(x-a)$ divides $p(x)$.
\end{lemma}
\begin{proof}
Let $p(x)$ be a polynomial in $R[x]$ and let $a$ be an element of $R$.
\begin{enumerate}
\item First we assume that $(x-a)$ divides $p(x)$. Therefore, there is a polynomial $q(x)\in R[x]$ such that $p(x)=(x-a)\cdot q(x)$. Hence, $p(a)=(a-a)\cdot q(a)=0$ and $a$ is a root of $p(x)$.\\
\item Assume that $a$ is a root of $p(x)$, i.e. $p(a)=0$. Since $x-a$ is a monic polynomial, we can perform the \PMlinkname{polynomial long division}{LongDivision} of $p(x)$ by $(x-a)$. Thus, there exist polynomials $q(x)$ and $r(x)$ such that:
$$p(x)=(x-a)\cdot q(x) + r(x)$$
and the degree of $r(x)$ is less than the degree of $x-a$ (so $r(x)$ is just a constant). Moreover, $0=p(a)=0+r(a)=r(a)=r(x)$. Therefore $p(x)=(x-a)\cdot q(x)$ and $(x-a)$ divides $p(x)$.
\end{enumerate}
\end{proof}