# P(x) is an odd polynomial. When (x−3) is factored out, the remainder is 6.

$$P(x)$$ 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.

Since you asked for a bit of guidance...

In general, if we divide a degree $$n$$ polynomial $$p(x)$$ by a degree $$m polynomial $$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 $$2x$$.

• Very clear explanation, thank you. Sep 26 '18 at 11:48

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 $\cdots$

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?

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 $2x$.