caledonia Posted May 29, 2015 Share Posted May 29, 2015 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 . . . Link to comment Share on other sites More sharing options...
Olinguito Posted June 3, 2015 Share Posted June 3, 2015 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. Link to comment Share on other sites More sharing options...
caledonia Posted June 4, 2015 Author Share Posted June 4, 2015 in the first line above, neither (x – a)k nor q(x) will generally be "polynimials" inasmuch as they do not have rational coefficients. Link to comment Share on other sites More sharing options...
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