I am going to use (x-c) instead of alpha. OK?
Let's use the remainder theorem.
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?.