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Why does $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots a_1x+a_0=a_n(x-r_1)(x-r_2)(x-r_3)\dots(x-r_n)?$ If $r$ are the roots of $f(x)?$

I believe this has something to do with the Fundamental Theorem of arithmetic - which states that any degree $n$ polynomial has exactly $n$ roots - but why do we have the factorization above? Is there some kind of proof stemming from the Fundamental Theorem??

Since I'm not familiar with polynomials, a detailed explanation along with an answer would be greatly appreciated.

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If you are assuming that $f$ has $n$ roots, which are $r_1,\ldots,r_n$, then all you need is to observe that, if$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0-a_n(x-r_1)\ldots(x-r_n),$$then $p(x)$ has $n$ roots ($r_1,\ldots,r_n$), but its degree is smaller than $n$. Therefore, $p(x)$ is the null polynomial.

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  • $\begingroup$ How will we know that its degree is smaller than $n$ and that p(x) is a polynomial? $\endgroup$ – Mathematrix Mar 3 '18 at 7:48
  • $\begingroup$ @Mathematrix $p(x)$ is a polynomial because it is one polynomial minus another one. The monomials with highest degree in both of them are $a_nx^n$ and therefore, when you take the first one and subtract the second one, you get a polynomial whose degree is, at most, $n-1$. $\endgroup$ – José Carlos Santos Mar 3 '18 at 7:54
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(See This link to learn about polynomial division)

First, take your polynomial $f(x)$ and divide it (as described in the link) by one of the factors, say $x-r_1$. Then you will get that $f(x) = (x-r_1)g(x) + c$ for some polynomial $g(x)$ and some constant polynomial $c$. But $c$ actually has to be $0$ since $0 = f(r_1) = (r_1-r_1)g(r_1) + c = c$. This means that $f(x) = (x-r_1)g(x)$. Note that the degree of $g(x)$ is less than the degree of $f(x)$. Repeat this process for $g(x)$ and by induction, you can factor $f(x)$ into a product of linear factors assuming that it has such roots.

(There are issues about whether the polynomial even has roots. This may not be the case over many fields. But over $\mathbb{C}$ every polynomial has a root, although that is difficult to prove)

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