I am going to use (x-c) instead of alpha. OK?

Let's use the remainder theorem.

Then,

For some quotient q(x).

If P(c)=0, the P(x)=(x-c)q(x); that is, x-c is a factor of P(x).

Also, if x-c is a factor of P(x), then the remainder when we divide P(x) by

x-c must be 0. Therefore, by the remainder theorem P(c)=0.

Is that what you were getting at?. Unless you want to be a purist and prove the remainder theorem as well.

Is that want you wanna do?.