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[parent] proof of factor theorem (Proof)

Suppose that $ f(x)$ is a polynomial 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 expansion terminates after the $ n-1^{\text{th}}$ term. Also, the $ n^{\text{th}}$ remainder of the Taylor series vanishes; i.e., $ \displaystyle R_n(x)=\frac{f^{(n)}(y)}{n!}x^n=0$. Thus, the function is equal to its Taylor series. Hence,

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

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$.



"proof of factor theorem" is owned by Wkbj79. [ full author list (2) | owner history (1) ]
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Cross-references: factor, function, vanishes, Taylor series, remainder, term, Taylor expansion, differentiable, degree, polynomial

This is version 4 of proof of factor theorem, born on 2002-05-25, modified 2007-05-30.
Object id is 2937, canonical name is ProofOfFactorTheorem.
Accessed 3954 times total.

Classification:
AMS MSC: 12D05 (Field theory and polynomials :: Real and complex fields :: Polynomials: factorization)
 12D10 (Field theory and polynomials :: Real and complex fields :: Polynomials: location of zeros )
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