# Finding an nth degree polynomial.

First post here. So I'm having a bit of trouble with the eponymous question type.

It's a bit embarrassing, as the problem is almost purely conceptual in nature, and I thought I had basically mastered this concept (this is the first time I've actually been "stuck" whilst doing this kind of problem). It's a two part question; I'm having difficulty with the second "part". My answers are in bold. Without further ado:

Find an $n^{th}$ degree polynomial function with real coefficients satisfying the given conditions. Suppose the division of $f(x)$ by $x-5$ gives a quotient $Q(x)$ and remainder $R$ of $23$.

(a) Find $f(5)$.

This part was simple enough:

By the Remainder Theorem, if $f(x)$ is divided by $(x - c)$, then the remainder is $f(c)$, hence if $f(x)$ is divided by $(x - 5)$, the remainder is $f(5)$, and

$f(5) = 23$

No trouble here, I think. This, though:

(b) Suppose $Q(9) = 4$; find $f(9)$

For some reason, I can't quite parse this part in my head. I tried to solve it logically:

By the Remainder Theorem: if $f(x)$ is divided by $(x - c)$, then the remainder is $f(c)$; therefore, $f(9)$ is the remainder of the division of $f(x)$ by $(x - 9)$, such that, by the Division Algorithm for Polynomials:

$f(x) = Q(9)\cdot(x-9) + R(x)$

and, since $Q(9) = 4$,

$f(x) = 4(x-9) + R(x)$

Here's where I get stuck. I'm not actually certain what to do at this point, or even if I'm on the right track. I've been sitting staring at my paper and trying out various "leads" that have occurred to me over the course of something like an hour, and I've reached "the point" where I'm too confused to even know what I'm doing anymore.

The worst part is, I can't shake the feeling that I'm missing something completely obvious. It's quite frustrating. Anyhow, that's the state of things. Any help you guys could give me would be greatly appreciated!

You should just write explicitely what is your expression of $f(x)$.
$f(x)=(x-5)Q(x)+23$
Obviously $f(5)=23$
But almost as obviously $f(9)=$...? knowing that $Q(9)=4$....