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I have a tough time understanding on the factorization of polynomials completely i.e. factoring a polynomial, say $f(x)$ to linear factors such that $a(x-r_1)\cdots (x-r_d)$ for some $d\in \mathbb{N}$ and $a,r_1,...,r_d\in \mathbb{F}$/

If we are able to factor polynomial $f(x)$ (with $\text{deg}(f(x))=d$) completely over the field $\mathbb{F}$, does this imply that $f(x)$ has the total of $d$ roots in $\mathbb{F}$?

For the contrary, if the polynomial $f(x)$ has the total of $t$ roots in $\mathbb{F}$ where $t<d$, then can we conclude that $f(x)$ may not be factored completely over the field $\mathbb{F}$?

  • Yes: the number of roots of a polynomial (counting multiplicity) is, by definition, the number of linear factors of the factored version of the polynomial (counting multiplicity). – Greg Martin Apr 23 '21 at 05:34
  • Beware that a polynomial can have more roots than its degree over non-domains, e.g. $,x^2-1,$ has four roots $,\pm1,\pm3,$ over $\Bbb Z_8 =$ integers $!\bmod 8.,$ Over a field (or domain) the roots correspond to linear factors - which are (cancellable) primes, so occur uniquely in factorizations, so we can't have ave nonunique factorizations like $,(x-1)(x+1) = (x-3)(x+3),$ in the prior example over $\Bbb Z_8.,$ See here for more. – Bill Dubuque Apr 23 '21 at 16:51

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