If a polynomial [tex]p(x)=a_0+a_1x+a_2x^2+ \ldots +a_{n-1}x^{n-1}[/tex] is zero for more than [tex]n-1[/tex] x-values, then [tex]a_0=a_1= \ldots =0[/tex]. Use this result to prove that there is at most one polynomial of degree [tex]n-1[/tex] or less whose graph passes through n points in the plane with distinct x-coordinates.(adsbygoogle = window.adsbygoogle || []).push({});

Let p(x) be a polynomial of degree n-1 or less such that [tex]p(x_1)=y_1, p(x_2)=y_2, \ldots,p(x_n)=y_n[/tex]. Thus if there are n distinct points, then there will be n equations. If for each of these equations we subtract [tex]y_i[/tex] then we have n root to a polynomial of degree [tex]n-1[/tex] or less. So for any given root we have [tex]a_jx_i^{j}[/tex] where i denotes the coordiante of [tex](x_i,y_i)[/tex] and j denote the subscript for the coefficents of the terms of p(x). Since we can subtract [tex]y_i[/tex] from any of the n terms of the polynomial we have in general [tex]a_jx_i^{j}=y_j[/tex] where [tex]a_j[/tex] is the only variable. Since for any given point this equation has a unique solution for [tex]a_j[/tex], and since the coefficents make up the polynomial p(x), p(x) is unique.

Does anyone see a problem with the above argument?

Thanks in advance.

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# Homework Help: Proving the uniqueness of a polynomial

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