Matrix information that can be extracted from the characteristic polynomial 
Let $A\in M_{3}(\mathbb{R})$ with $f_{\lambda}(x)=x^3+4x^2+5x+2$
  Find the eigenvalues, the trace and the determinant of $A$

$f_{\lambda}(x)=x^3+4x^2+5x+2$ the dividers of $2$ are $-1,-2$ we get that 
$(-1)^3+4(-1)^2+5(-1)+2=0$ dividing the polynomial we get $f_{\lambda}(x)=x^3+4x^2+5x+2=(x+2)(x+1)^2$ so $
\lambda_{1}=-2$ and $\lambda_{2}=-1$
Now the trace is minus the coefficient of the element of the the $n-1$ power, in this case $-4$
The determinant is the value of the free variable which is $2$
Is this correct? where can I find more information about the coefficients of the characteristic polynomial?
 A: You are correct in determining the eigenvalues and the trace, but note that the constant term of the characteristic polynomial of a matrix $A$ is $\det (-A)=(-1)^n\det A$ (where $n$ is the degree of the characteristic polynomial), so, in this case $\det A=-2$.
In general the coeffcients of thecharacteristic polynomial are functions of the entries of the matrix and can be expressed as the traces of the exterior powers of $A$, as you can see here. But this cannot be used to find the entries of the matrix because all similar matrices have the same characteristic polynomial. 
A: One has for an $n\times n$ matrix:
$$ f_\lambda(x) = \sum_{k=0}^n x^{n-k} (-1)^{k} \ {\rm tr } \wedge^k A
 = \sum_{k=0}^n x^{n-k} (-1)^k \sum_{i_1<...<i_k} \lambda_{i_1} ... \lambda_{i_k} $$
where $\lambda_1,...,\lambda_n$ are the $n$ eigenvalues of $A$.
For the traces one has in terms of elements of $A$:
$$ {\rm tr} \wedge^k A= \sum_{i_1<i_2<...<i_k}  
   \left| \begin{matrix} a_{i_1 i_1} & ... & a_{i_1 i_k} \\
    .. & ... & .. \\
    a_{i_k i_1} & ... & a_{i_k i_k } \end{matrix} \right|$$
so e.g. in the case $n=3$
$$ {\rm tr} \wedge^2 A= 
   \left| \begin{matrix} a_{1 1} &  a_{1 2} \\
    a_{2 1 }  & a_{2 2 } \end{matrix} \right| +
 \left| \begin{matrix} a_{1 1} &  a_{1 3} \\
    a_{3 1 }  & a_{3 3 } \end{matrix} \right|+
 \left| \begin{matrix} a_{2 2} &  a_{2 3} \\
    a_{3 2 }  & a_{3 3 } \end{matrix} \right|
$$
One has ${\rm tr } \wedge^1 A = {\rm tr } A$ (usual trace) and
${\rm tr } \wedge^n A = {\rm det\ }A$. In your case $\det A=-2$.
