If $x=2+i$, $gcd(a,b,c)=1$, and $ax^4+bx^3+cx^2+bx+a=0$, then what is $|c|$? 
Suppose
  $$a(2+i)^4 + b(2+i)^3 + c(2+i)^2 + b(2+i) + a = 0,$$
  where $a,b,c$ are integers whose greatest common divisor is $1$.
  Determine $|c|$.

So I first simplified the exponents and combined like terms.
I received $$a(-6+24i)+b(4+12i)+c(3+4i)=0.$$
I don't really know how to progress. What is the answer $|c|$?
 A: We use your calculation. Note that as the other answers show there would have been nicer ways to proceed.
From the equation that you obtained we get $-6a+4b+3c=0$ and $24a+12b+4c=0$. Divide the second expression through by $4$, and add the first expression. We get $7b+4c=0$.
Thus $4c=-7b$ and therefore $c=7k$, $b=4k$ for some integer $k$. Substituting in the first equation we get $6a=5k$. It follows that $k=6m$ for some integer $m$. Thus $c=42m$, $b=-24m$, and $a=5m$. 
By relative primality we get $|m|=1$, and therefore $|c|=42$.
A: Hint: since $a,b,c \in \mathbb Z$ the conjugate of any root will also be a root.
Because the coefficients are symmetric, there is another easy way (applicable to such symmetric polynomials) to obtain another root. I think you wouldn't have been given this question without knowing, or having in your notes, how that works in the case where roots are rational. It works for complex roots too (easy to prove).
That gets you all the roots easily and you should be able to find an efficient way of finding the polynomial.
You are being asked to notice the features of the polynomial you are given and use them in an unusual situation. Give it a go and see what you come up with. 
A: A polynomial of this symmetric form can be reduced in degree by the following trick. Divide by $x^2$ and get:
$\begin{align}
a (x^2 + x^{-2}) + b (x + x^{-1}) + c
  &= a (x + 1/x)^2 + b (x + 1/x) + c - 2 a
\end{align}$
The substitution $y = x + 1/x$ finishes this off.
This works because the powers of $x + 1/x$ turn out to be combinations of $x^k + x^{-k}$.
