Let $p\in\mathbb{Z}[x]$, set $d = \textbf{Deg}(p)$, and write $$p(x) = \sum_{i=0}^{d}{a_ix^i}$$ for some sequence $\{a_0,a_1,a_2,...., a_{d-1}\}$. Is there a mapping $\mathcal{F}$ so that $$\mathcal{F}\biggl(\{a_0,a_1,a_2,...., a_{d-1}\}\biggr) = \{x:~ \sum_{i=0}^{d}{a_ix^i} = 0 \}.$$ That is does a general formula for the roots of every polynomial exist. The Quadratic formula is known and we can write the roots of $a_3x^2+a_2x+a_1$ as $$\dfrac{-a_2\pm \sqrt{a_2^2-4a_3a_1}}{2a_3}.$$ So we know when $d=3$ the function $\mathcal{F}$ at $\{a_1,a_2,a_3\}$ is $\dfrac{-a_2\pm \sqrt{a_2^2-4a_3a_1}}{2a_3}$. There might exist a function in the form of a quotient $$\mathcal{F}(\{a_0,a_1,a_2,...,a_{d-1}\})= \dfrac{f(a_0,a_1,a_2,...,a_{d-1})}{g(a_0,a_1,a_2,... ,a_{d-1})}.$$ When $d=2$ we have $f:= -a_2\pm \sqrt{a_2^2-4a_3a_1}$ and $g:= 2a_3$ is a solution.
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2$\begingroup$ Of course there is such a mapping, you have just defined it (though for $d\ge5$ it is has no representation by sums, products, quotients, radicals, according to the Ruffini-Abel-Galois Thm) $\endgroup$– Pietro MajerCommented 22 hours ago
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$\begingroup$ You use $f$ to mean two different things (a function, and its numerator). You also use $\{a_0, \dotsc, a_d\}$, which looks like a set, but you call it a sequence, and certainly you seem to treat it as a sequence (since the set of coefficients of a polynomial certainly isn't enouhg to recover its set of roots). $\endgroup$– LSpiceCommented 22 hours ago
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$\begingroup$ For d=3 the formula is more complicated. $\endgroup$– AnixxCommented 20 hours ago
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$\begingroup$ The formulas for degree 3 and 4 are not of the form of quotinent. Here is the formula for degree 4: math.stackexchange.com/a/1135224/2513 $\endgroup$– AnixxCommented 20 hours ago
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$\begingroup$ Why you close question? I'm Blocked? $\endgroup$– Wuu tang clanCommented 18 hours ago
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