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 $P=(X-a)Q$.

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