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Mathematics
General Math
Prove an nth-degree polynomial has exactly n roots
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[QUOTE="jbunniii, post: 5485674, member: 81553"] The fundamental theorem of algebra has no really elementary proof. It says that any nonconstant polynomial (with real or complex coefficients) has at least one complex root. So it is really a statement involving complex analysis, and indeed the most straightforward proof uses complex analysis. However, there are many other ways to prove it. Here is a book which proves the theorem six different ways and is intended as a semester-long course at the senior undergraduate level: [URL]https://www.amazon.com/dp/0387946578/?tag=pfamazon01-20[/URL] by Benjamin Fine and Gerhard Rosenberger. So that gives you some idea of the level of mathematics required for a rigorous proof. However, note that if we assume the fundamental theorem of algebra, then it is not hard to show that your ##f(x)## can be written in the form ##A(x-x_1)(x-x_2)\cdots(x-x_n)##. We can do this by induction on ##n##, the degree of ##f(x)##. First, it is clear that if ##n=0## or ##n=1## then ##f(x)## is already in that form. So suppose that ##n \geq 2## and that the result is true for all polynomials of degree smaller than ##n##, and that ##f(x)## has degree ##n##. By the fundamental theorem of algebra, ##f(x)## has at least one complex root, call it ##x_1##. By the division algorithm, we may write ##f(x) = g(x)(x-x_1) + r(x)##, where ##g(x)## is a polynomial and ##r(x)## is zero or a polynomial of degree less than ##1##, i.e., ##r(x)## is a constant, so just write it as ##r##. Thus ##f(x) = g(x)(x-x_1) + r##. Plugging in ##x=x_1## shows that ##r=0##. Consequently, ##f(x) = g(x)(x-x_1)##. Comparing degrees, we see that ##g(x)## must have degree ##n-1##, so the induction hypothesis applies, and we can write ##g(x) = A(x-x_2)(x-x_3)\cdots(x-x_n)##, and therefore ##f(x) = A(x-x_1)(x-x_2)\cdots(x-x_n)## as desired. [/QUOTE]
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Mathematics
General Math
Prove an nth-degree polynomial has exactly n roots
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