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avec_holl
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Homework Statement
Prove that for any polynomial function [tex]f[/tex] and number [tex]a[/tex], there exists a polynomial function [tex]g[/tex] and number [tex]b[/tex] such that: [tex]f(x) = (x-a)g(x) + b[/tex]
Homework Equations
N/A
The Attempt at a Solution
Proof: Let [tex]P(n)[/tex] be the statement that for some natural number [tex]n[/tex],
[tex]f(x) = a_nx^n + \dots + a_0 = (x-\alpha)g(x) + \beta[/tex]
Clearly [tex]P(1)[/tex] is true since we have that [tex]a_1x + a_0 = (x-\alpha)(a_1) + \beta[/tex]. Now suppose that [tex]P(k)[/tex] holds. To complete the proof, we need only show that if [tex]P(k)[/tex] holds then [tex]P(k+1)[/tex] also holds. Considering the polynomial function [tex]f[/tex] defined such that,
[tex]f(x) = a_{k+1}x^{k+1} + \dots + a_0 = a_{k+1}x^{k+1}\; + \;(x-\alpha)g(x)\; + \;b_1[/tex]
[tex]f(x) = a_{k+1}(x)(x^k)\; + \;(x-\alpha)g(x)\; + \;b_1[/tex]
Applying the Remainder Theorem to the polynomials [tex]a_{k+1}x[/tex] and [tex]x^k[/tex] - we've already proved this when [tex]n=1[/tex] and assumed that it holds for [tex]n=k[/tex] so we can apply it to those polynomials. This yields,
[tex]f(x) = [(x-\alpha)(a_{k+1}) + b_2][(x-\alpha)h(x) + b_3] + (x-\alpha)g(x) + b_1[/tex]
[tex]f(x) = a_{k+1}(x-\alpha)^2h(x) + b_2(x-\alpha)h(x) + a_{k+1}b_3(x-\alpha) + (x-\alpha)g(x) + b_1 + b_2b_3[/tex]
[tex]f(x) = (x-\alpha)[a_{k+1}(x-\alpha)h(x) + b_2h(x) + g(x) + a_{k+1}b_3] + b_1+b_2b_3[/tex]
Letting [tex]L(x) = a_{k+1}(x-\alpha)h(x) + b_2h(x) + g(x) + a_{k+1}b_3[/tex] and [tex]\beta = b_1 + b_2b_3[/tex] we find that,
[tex]f(x) = (x-\alpha)L(x) + \beta[/tex]
Is this a valid proof? It seems awfully fishy . . . if anyone could point out mistakes and make suggestions it would be much appreciated. Thanks!