Let's take polynomial $p(x)$ and say that $p(x)/(x-3)=q(x)+2/(x-3)$.
Now let's say that we set the value of $x=3$. Does this prove that when $p(x)=3$, $p(x)=2$?
Part of me thinks it would, because $p(x)=q(x)(x-3) + 2$. And when $x=3$, we are left with $p(x)=2$.
But another part of me thinks that it wouldn't because if $x=3$:
$ - (x-3) * p(x)/(x-3) = (x-3) * q(x)+(x-3) * (2/(x-3))$ $ - 0 * p(x)/0 = 0+(x-3) * 0(2/0)$
So wouldn't the answer be undefined, because $0$ times $2/(x-3)$ when $x=3 = 0 $ times $2/0$ iss undefined?
So is the answer 2 or undefined? and why is the other interpretation incorrect?
Do not divide by zero.
Writing $\frac{p(x)}{(x-3)}$ means you are assuming the case when $x \ne 3$.
If we indeed have $\forall x\in \mathbb{R}, p(x)=q(x)(x-3)+2$ where $q$ is a polynomial, then $p(3)=2$.