Polynomial ring/Field/Zero/Linear factor/Fact/Proof

Proof

If ${}P$ is a multiple of ${}X-a$, then we can write

${}P=(X-a)Q\,$

with another polynomial ${}Q$. Inserting ${}a$ yields

${}P(a)=(a-a)Q(a)=0\,.$

In general, there exists, due to fact, a representation

${}P=(X-a)Q+R\,,$

where either ${}R=0$ or the degree of ${}R$ is ${}0$, so in both cases ${}R$ is a constant. Inserting ${}a$ yields

${}P(a)=R\,.$

So if ${}P(a)=0$ holds, then the remainder must be ${}R=0$, and this means ${}P=(X-a)Q$.

To fact