Condition for Linear Divisor of Polynomial

Theorem
Let $\map P x$ be a polynomial in $x$.

Let $a$ be a constant.

Then $x - a$ is a divisor of $\map P x$ $a$ is a root of $P$.

Proof
From the Little Bézout Theorem, the remainder of $\map P x$ when divided by $x - a$ is equal to $\map P a$.

Sufficient Condition
Let $x - a$ be a divisor of $\map P x$.

From the Little Bézout Theorem, the remainder of $\map P x$ when divided by $x - a$ is equal to $\map P a$.

By definition of divisor, the remainder of $\map P x$ when divided by $x - a$ equals $0$.

That is:
 * $\map P a = 0$

It follows by definition that $a$ is a root of $P$.

Necessary Condition
Let $a$ be a root of $P$.

From the Little Bézout Theorem, and by Polynomial Long Division, we have:


 * $\map P x = \paren {x - a} \map Q x + \map P a$

where $\map Q x$ is a polynomial in $x$ of degree one less than $\map P x$.

By definition of root of polynomial:
 * $\map P a = 0$

So we have:


 * $\map P x = \paren {x - a} \map Q x$

and the result follows by definition of divisor.