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$P(x)$ 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.

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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 $2x$.

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  • $\begingroup$ Very clear explanation, thank you. $\endgroup$ – user574848 Sep 26 '18 at 11:48
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Hint: by euclidian division $p(x) = (x^2-9) q(x) + a x + b$.

Then $p(3)-p(-3)= 6a = 6 - (-6) = 12$ so $a=2$.

Since $p(x)$ is odd, all coefficients of even powers of $x$ are $0$, including the free term, so $b=0$.

So the remainder is $\cdots$

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Hint: the linked question is quite useful for your purpose actually. Let $ax+b$ be the remainder when $p(x)$ is divided by $x^2-9$: $$ p(x)=(x^2-9)q(x)+ax+b. $$ We know \begin{aligned} 6\color{blue}{=}p(3)=0q(3)+3a+b&\implies 3a+b=6,\\ -6=-p(3)\color{blue}{=}p(-3)=0q(-3)+(-3)a+b&\implies-3a+b=-6 \end{aligned} The $\color{blue}{\text{colored}}$ bits above are where you use the given information.

So, two equations, two unknowns ($a$ and $b$), can you continue?

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By the Division Algorithm for polynomials, the remainder has the form $ax+b$ for constants $a, b$. Hence, we have:$$p(x)=(x^2-9)q(x)+ax+b\implies 6=3a+b;-6=-3a+b$$, where $q(x)$ is a polynomial of degree $2$ less than $p(x)$ . Solving the two equations obtained for $a, b$; we obtain the remainder as $2x$.

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