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Posted

The standard proof that a polynomial cannot be irreducible if it has repeated roots uses calculus (differentation). I would like to find a proof without calculus . . .

 

Posted

If [latex]a[/latex] is a repeated root of a polynomial [latex]p(x)[/latex] then [latex]p(x)=(x-a)^kq(x)[/latex] for some polynomial [latex]q(x)[/latex] and integer [latex]k\geqslant2[/latex].

 

Thus [latex]p(x)=(x-a)r(x)[/latex] where [latex]r(x)=(x-a)^{k-1}q(x)[/latex] and both [latex]x-a[/latex] and [latex]r(x)[/latex] are not units.

Posted

in the first line above, neither (xa)k nor q(x) will generally be "polynimials" inasmuch as they do not have rational coefficients.

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