TITLE: P(x) is an odd polynomial. When (x−3) is factored out, the remainder is 6.
QUESTION [9 upvotes]: $P(x)$ is an odd polynomial. When $(x−3)$ is factored out, the remainder is $6$. What is the remainder when $P(x)$ is divided by $(x^2-9)$?

Using the remainder theorem, when $x=3$, $P(x)=6$, and if the function is odd then when $x=-3$, $P(x)=-6$.
I would very much appreciate if someone can guide me through the solving process, since that is what I care about the most. Thanks in advance.

REPLY [4 votes]: Since you asked for a bit of guidance...
In general, if we divide a degree $n$ polynomial $p(x)$ by a degree $m<n$ polynomial $d(x)$, we will end up with a degree $n-m$ polynomial $q(x)$ and a degree $m-1$ (at most) remainder $r(x)$. (This follows from the division algorithm - it might help to perform a few long or synthetic divisions on some polynomials just to play around with this idea.)
We can now express $p$ in terms of $d$, $q$ and $r$ as:
$$\overbrace{p(x)}^{\text{degree }n}=\underbrace{q(x)}_{\text{degree } n-m}\overbrace{d(x)}^{\text{degree }m}+\underbrace{r(x)}_{\text{degree }m-1\text{ (at most)}}$$
In the problem that you present, they give us that when $p(x)$ is divided by $d(x)=x-3$, the remainder is $6$. By the remainder theorem, we can conclude $p(3)=6$.
Since we are told that $p$ is an odd function, $p(-3)=-p(3)=-6$.
We are asked for the remainder when $p(x)$ is divided by $x^2-9=(x+3)(x-3)$. Notice that this is a degree $2$ polynomial, so the remainder will be a degree $1$ (linear) polynomial. In other words, $r(x)$ will have the form $ax+b$.
Expressing $p(x)$ in terms of this new information, then:
$$p(x)=q(x)(x+3)(x-3)+(ax+b)$$
where $q$ is some polynomial.
From this, we can see that $p(3)=3a+b$, which we know is $6$, and $p(-3)=-3a+b$, which we know is $-6$. Solving these simultaneously gives us $a=2$ and $b=0$. Thus the remainder is the degree $1$ polynomial $2x$.