Finding Repeated Eigenvalues for $\begin{bmatrix} 3 & 0 & 0 \\ 0 & x & 2 \\ 0 & 2 & x \end{bmatrix}$ I'm reviewing some linear algebra. I'm having trouble with this problem:

For what values of $x$ does the matrix 
  \begin{bmatrix}
    3 & 0 & 0 \\
    0 & x & 2 \\
    0 & 2 & x 
\end{bmatrix}
  Have at least one repeated eigenvalue?

I'm not sure how to start this (aside from writing down the characteristic polynomial).
Edited: As per Shaun's direction.
 A: \begin{equation}
 p(\lambda) =\det 
 \begin{bmatrix}
  3 - \lambda& 0 & 0 \\
  0 & x - \lambda& 2 \\
  0 & 2 & x - \lambda
 \end{bmatrix}
 =
 0
\end{equation}
So
\begin{equation}
 (3-\lambda)( (x-\lambda)^2 - 4) = (\lambda - 3)(x - \lambda - 2)(x - \lambda + 2) = 0
\end{equation}
So no matter what there is $\lambda_1 = 3$. Arguement goes on the other two eigenvalues $\lambda_2 = x - 2$ and $\lambda_3 = x + 2$. You've got the following cases:


*

*$\lambda_2 \neq \lambda_3$ no matter what. 

*$\lambda_1 = \lambda_2$ gives $ x = 3 + 2 = 5$

*$\lambda_1 = \lambda_3$ gives $x = 3-2 = 1$


So two values of $x$ which are $1,5$.
A: Suppose that the matrix $A$ is given by the following matrix
$$ A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & x & 2 \\ 0 & 2 & x \end{bmatrix}  \tag{1} $$
in order find the eigenvalues we do the following
$$ \det(A - I \lambda)  = \det\left(\begin{bmatrix} 3 - \lambda & 0 & 0 \\ 0 & x - \lambda  & 2 \\ 0 & 2 & x - \lambda \end{bmatrix}\right)  \tag{2}$$
then we get 
$$ \det(A - I \lambda)  = (3-\lambda) \cdot \begin{vmatrix} x -\lambda & 2 \\ 2 & x-\lambda  \end{vmatrix} - 0 \cdot \begin{vmatrix} 0 & 2 \\ 0 & x -\lambda  \end{vmatrix} +0 \cdot \begin{vmatrix}  0 & x- \lambda \\ 0 & 2 \end{vmatrix}   \tag{3} $$
$$  \det(A - I \lambda)  = (3-\lambda) \cdot  \left( ( x-\lambda)^{2} - 2^{2}\right) \tag{4} $$ 
$$  \det(A - I \lambda)  = (3-\lambda) \cdot  \left( \lambda^{2}  -2\lambda x + x^{2} -4\right) \tag{5} $$ 
So you get repeated eigenvalues when this expression $ \lambda^{2}  -2\lambda x + x^{2} -4 $ generates them.
We want for the expression to generate eigenvalues that are equal to $3$
Note then for the expression
$$  \lambda^{2}  -2\lambda x + x^{2} -4 \tag{6}$$
if we insert $ x=1$  we get
$$  \lambda^{2}  -2\lambda  + 1 -4 = \lambda^{2} - 2\lambda -3 \tag{7} $$
$$ \lambda^{2} -2\lambda -3 = (\lambda-3)(\lambda +1) \tag{8} $$
So we have
$$ \det(A-I\lambda) =(3-\lambda) (\lambda-3)(\lambda+1) \tag{9}$$
letting $x=5$
$$ \lambda^{2}  -2\lambda 5 + 25 -4 = \lambda^{2} -10\lambda + 21  
\tag{10}$$
$$ \lambda^{2} -10 \lambda +21 = (\lambda -7)(\lambda-3) \tag{11} $$ 
$$ \det(A-I\lambda) = (3-\lambda) (\lambda-7)(\lambda-3) \tag{12}$$
