How can I prove that if $\alpha$ is a root of the polynomial $S$ then $S(x) = Q(x)(x-\alpha)$, where the highest exponent $(Q)=n-1$ being $n$ the highest exponent of $S$.
This can by generalized as: $$K\prod_{i=1}^n (x-a_i)$$
I tried to prove this by generalizing $Q$ as $q _{n-1}x^{n+1}+ ... +q_1x+q_0$ and decomposing it with $\alpha$ as $(x-\alpha)\cdot(Q)=S$ so $S= \alpha q_{n-1}x^{n+2}+...+αq_1x^{n+1}+q_0$
But now I'm more confused and I can't find a way to prove this correctly because I feel this is not the right way of proceeding. However, I have no idea another way to start the proof.
Let $\alpha_i$ be the $n$ roots, for $i\in[1,n]$.
As $\alpha_1$ is a root, you can write
$$Q_n(x)=(x-\alpha_1)Q_{n-1}(x).$$
But clearly, $Q_{n-1}$ has the roots $\alpha_i$ for $i\in[2,n]$, and you can write
$$Q_{n-1}=(x-\alpha_2)Q_{n-2}(x)$$
and so on.
Finally,
$$Q_n(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)Q_0(x)$$ where $Q_0$ is of degree $0$.