1. ## Functions

If you have
$f(x)=x^{2}+3$
I know f(x) just means the function of x, or just y.

But take this:

If
$\frac{f(x)}{x-k}$
then
$r=f(k)$
Where r= remainder
This makes no since to me. some function divided by x-k yields the function of k.

How do you come up with a function?

Edit: normally f(x),f(c),f(g) are all put in the place of possible functions, but this one is not.

2. Originally Posted by integral
If
$\frac{f(x)}{x-k}$
then
$r=f(k)$
Where r= remainder
This makes no since to me. some function divided by x-k yields the function of k.
This statement is deduced based on the Polynomial Remainder Theorem. That is, if you have a function $f(x)$, and you divide it with another function $x - c$, then the remainder of the division is equals to $f(c)$. Proof:

Let's say you have a function $f(x) = (x - c)q(r) + r(x)$, where $q(r)$ is the quotient (the result of dividing with $x - c$) and $r(x)$ is the remainder (after dividing with $x - c$). According to the Polynomial Division Algorithm, the remainder of the division must have a smaller degree than the divisor, in this case, $x - c$. Since $x - c$ has a degree of 1, $r(x)$ must be a constant polynomial.

Now, if you substitute $x = c$ into the function, you'll end up with:

$f(c) = (c - c)q(r) + r(c)$
$f(c) = (0)q(r) + r(c)$
$f(c) = r(c)$

Therefore, the remainder is equals to $f(c)$.