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.

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    $\begingroup$ If you only multiply the factors $$(x-r_1)\cdots (x-r_n)$$ the leading coefficient would be $1$. $\endgroup$ – quasi May 11 '18 at 4:55
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    $\begingroup$ The factors $x-r_k$ for respective roots are monic (first degree) polynomials, and the product of monic polynomials is always monic. So the only way the leading coefficient $a_1$ of $x^n$ can arise in the factorization of a polynomial with $n$ (not necessarily distinct) roots is by having a constant factor times $(x-r_1)\ldots(x-r_n)$. $\endgroup$ – hardmath May 11 '18 at 4:56
  • $\begingroup$ @quasi Why are you writing a perfectly good answer as a comment? $\endgroup$ – Arthur May 11 '18 at 5:43
  • $\begingroup$ @hardmath Why are you writing a perfectly good answer as a comment? $\endgroup$ – Arthur May 11 '18 at 5:43
  • $\begingroup$ @Arthur: I thought my comment would have to be fleshed out a little more to qualify as an answer. $\endgroup$ – quasi May 11 '18 at 5:50

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 $k[x]$.


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