Show that every integer eigenvalue of $A$ divides the determinant of $A$. 
Let $A$ be a square matrix of order $n$ whose entries are all integers.
  Show that every integer eigenvalue of $A$ divides the determinant of $A$.

I am not able to understand how to show this.
We know that $\det A$ is the product of eigen values and so every eigen value must divide $\det A$.
But if a matrix has eigen values $3$ and $\frac{4}{3}$ then if I do the product then the factor $3$  gets neutralized if I do the product then how does it appear as a factor of $\det A$?
Please help.

Let $P(x)=\sum_{k=0}^na_kx^k$ be the charasteristic polynomial. Then $|P(0)|=|\det A|$. Moreover, the coefficients of $P$ are integer. If $\lambda$ is an integer eigenvalue, then 
$$0=P(\lambda)=P(0)+\sum_{k=1}^na_k\lambda^k$$
Therefore
$$|\det A|=|P(0)|=|\lambda|\cdot\left|\sum_{k=1}^na_k\lambda^{k-1}\right|$$
Hence $\lambda$ divides $\det A$.
