When writing a polynomial as a product of linear factors, why is the leading coefficient a factor?

Question

I get that if a polynomial has roots $r_1,...,r_n$, where $n$ is the degree of the polynomial, then $(x-r_1),...,(x-r_n)$ are linear factors. But a polynomial $a_1x^n+...+a_nx+a_{n+1}$ is written as $a_1(x-r_1)...(x-r_n)$ with the extra leading coefficient $a_1$ as a constant factor. Why is this? I don't see how, in the general sense, successive polynomial division by said linear factors end up with specifically $a_1$ as the quotient.

Answer 1

Let's check what we get if we omit the factor $a_1$ :

$$ \prod_{i=1}^{n} (x - r_i) = \sum_{d=0}^n c_d x^d$$ where $$c_d = \sum_{I \in\mathcal{P}_{n-d}} \prod_{i \in I} (-r_i) $$ $\mathcal{P}_{n-d}$ denotes the set of all subsets of $\{1, \dots, n\}$ of size $n-d$.

Now for $d=n$, there is only one subset of size 0, namely the empty set. Hence, $c_n$ is the empty product : namely $c_n = 1$.

So, to have the equality you need to multiply by $a_1$.

By the way, if $k$ is a field, then $k[x]$ is a unique factorization domain, its units are the elements of $k^*$. $a_1$ corresponds to the "unit part" of the factorization of a non constant polynomial $P \in k[x]$ whereas each $(x-r)$ is prime in $k[x]$.