# Thread: Polynomial division and dividing by zero

1. ## Polynomial division and dividing by zero

Suppose I define $\displaystyle g(x,y) = \frac {f(x)-f(y)}{x-y}$ and take $\displaystyle f(z) = z^{n}$. So $\displaystyle g(x,y) = x^{n-1}+ x^{n-2}y + \cdots + xy^{n-2} + x^{k-1}$. If I take $\displaystyle y=x$, then $\displaystyle g(x,x) = nx^{n-1}$. Is this valid? If I look at the original equation, does this also mean $\displaystyle g(x,x) = \frac {0}{0}$? I'm not sure how to understand division by zero.

2. $\displaystyle g(x,x) \neq nx^{n-1}$ because deriving $\displaystyle g(x,y) = x^{n-1}+ x^{n-2}y + \cdots + xy^{n-2} + x^{k-1}$ from $\displaystyle g(x,y) = \frac {f(x)-f(y)}{x-y}$ is valid only when $\displaystyle x \neq y$.

Again, if $\displaystyle x=y$ then $\displaystyle g$ is not defined, therefore you can't say that $\displaystyle g(x,x) = nx^{n-1}$

3. OK, the problem I'm doing says $\displaystyle g(x,y)$ is always divisible by $\displaystyle x - y$, so instead, can I say because $\displaystyle (x-y)g(x,x) = (x-y)(nx^{n-1})$, and then using the cancellation law, $\displaystyle g(x,x)$ is the same polynomial as $\displaystyle nx^{n-1}$?