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

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.

• If you only multiply the factors $$(x-r_1)\cdots (x-r_n)$$ the leading coefficient would be $1$. – quasi May 11 '18 at 4:55
• 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 $(x-r_1)\ldots(x-r_n)$. – hardmath May 11 '18 at 4:56
• @quasi Why are you writing a perfectly good answer as a comment? – Arthur May 11 '18 at 5:43
• @hardmath Why are you writing a perfectly good answer as a comment? – Arthur May 11 '18 at 5:43
• @Arthur: I thought my comment would have to be fleshed out a little more to qualify as an answer. – quasi May 11 '18 at 5:50

## 1 Answer

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 $$k[x]$$.