proof of factor theorem

Suppose that $f(x)$ is a polynomial with real or complex coefficients of degree $n-1$ . Since $f$ is a polynomial, it is infinitely differentiable. Therefore, $f$ has a Taylor expansion about $a$ . Since $f^{(n)}(x)=0$ , the terminates after the $n-1^{\\text{th}}$ term. Also, the $n^{\\text{th}}$ remainder of the Taylor series vanishes; i.e. (http://planetmath.org/Ie), $\\displaystyle R_{n}(x)=\\frac{f^{(n)}(y)}{n!}x^{n}=0$ . Thus, the function is equal to its Taylor series. Hence,

$\\begin{array}[]{rl}f(x)&\\displaystyle=\\sum_{k=0}^{n-1}\\frac{f^{(k)}(a)}{k!}(x-%a)^{k}\\\\&\\\\&\\displaystyle=f(a)+\\sum_{k=1}^{n-1}\\frac{f^{(k)}(a)}{k!}(x-a)^{k}\\\\&\\\\&\\displaystyle=f(a)+(x-a)\\sum_{k=1}^{n-1}\\frac{f^{(k)}(a)}{k!}(x-a)^{k-1}\\\\&\\\\&\\displaystyle=f(a)+(x-a)\\sum_{k=0}^{n-2}\\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^{k}.%\\end{array}$

If $f(a)=0$ , then $\\displaystyle f(x)=(x-a)\\sum_{k=0}^{n-2}\\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^{k}$ . Thus, $f(x)=(x-a)g(x)$ , where $g(x)$ is the polynomial $\\displaystyle\\sum_{k=0}^{n-2}\\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^{k}$ . Hence, $x-a$ is a factor of $f(x)$ .

Conversely, if $x-a$ is a factor of $f(x)$ , then $f(x)=(x-a)g(x)$ for some polynomial $g(x)$ . Hence, $f(a)=(a-a)g(a)=0$ .

It follows that $x-a$ is a factor of $f(x)$ if and only if $f(a)=0$ .

单词	ProofOfFactorTheorem
释义	proof of factor theorem Suppose that $f(x)$ is a polynomial with real or complex coefficients of degree $n-1$ . Since $f$ is a polynomial, it is infinitely differentiable. Therefore, $f$ has a Taylor expansion about $a$ . Since $f^{(n)}(x)=0$ , the terminates after the $n-1^{\\text{th}}$ term. Also, the $n^{\\text{th}}$ remainder of the Taylor series vanishes; i.e. (http://planetmath.org/Ie), $\\displaystyle R_{n}(x)=\\frac{f^{(n)}(y)}{n!}x^{n}=0$ . Thus, the function is equal to its Taylor series. Hence, $\\begin{array}[]{rl}f(x)&\\displaystyle=\\sum_{k=0}^{n-1}\\frac{f^{(k)}(a)}{k!}(x-%a)^{k}\\\\&\\\\&\\displaystyle=f(a)+\\sum_{k=1}^{n-1}\\frac{f^{(k)}(a)}{k!}(x-a)^{k}\\\\&\\\\&\\displaystyle=f(a)+(x-a)\\sum_{k=1}^{n-1}\\frac{f^{(k)}(a)}{k!}(x-a)^{k-1}\\\\&\\\\&\\displaystyle=f(a)+(x-a)\\sum_{k=0}^{n-2}\\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^{k}.%\\end{array}$ If $f(a)=0$ , then $\\displaystyle f(x)=(x-a)\\sum_{k=0}^{n-2}\\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^{k}$ . Thus, $f(x)=(x-a)g(x)$ , where $g(x)$ is the polynomial $\\displaystyle\\sum_{k=0}^{n-2}\\frac{f^{(k+1)}(a)}{(k+1)!}(x-a)^{k}$ . Hence, $x-a$ is a factor of $f(x)$ . Conversely, if $x-a$ is a factor of $f(x)$ , then $f(x)=(x-a)g(x)$ for some polynomial $g(x)$ . Hence, $f(a)=(a-a)g(a)=0$ . It follows that $x-a$ is a factor of $f(x)$ if and only if $f(a)=0$ .
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