Factoring a cubic polynomial? So I have a matrix
$$A =
\begin{pmatrix}
-5 & -6 &  3 \\
3  &  4 & -3 \\
0  &  0 & -2
\end{pmatrix}
$$
I'm to find the characteristic polynomial and all the eigenvalues of $A$.
I have found $\det (A - \lambda I) = -\lambda^3 - 3\lambda^2 + 4$.
But I can't understand how to factor it to find the eigenvalues. The cubic factorization methods I've found don't seem to work on this one?
Any ideas?
 A: $$-λ^3 - 3λ^2 + 4=-(λ^3 -1)-( 3λ^2 -3)=$$
$$=-(λ-1)(λ^2+λ+1)-3(λ-1)(λ+1)$$
$$=(λ-1)(-λ^2-λ-1-3λ-3)=$$
$$=(λ-1)(-λ^2-4λ-4)=$$
$$=(1-λ)(λ^2+4λ+4)=$$
$$=(1-λ)(λ^2+2\cdot2λ+2^2)=$$
$$=(1-λ)(λ+2)^2$$
A: The last row of $A$ contains all zeroes except the last element, so that element $\lambda =−2$ must be an eigenvalue. Then, divide:
$$
\frac{-x^3-3x^2+4}{x+2}
  = -\frac{x^3+3x^2-4}{x+2}
  = -(x^2+x-2)
$$
A: Several good answers have been given already. Let me provide a quicker way for this particular matrix. By using a bit of theory and some observations, there's no need to explicitly calculate the characteristic polynomial at all.


*

*The matrix is block upper triangular. This immediately tells us that the eigenvalues are the union of the blocks on the diagonal. In this case, the eigenvalues come from
$$\left\{A'=\begin{pmatrix}-5 & -6\\3 & 4\end{pmatrix},\ (-2)\right\}$$
Therefore one of the eigenvalues is $-2$.

*Looking at the diagonal block $A'$, notice that each column sums to $-2$. This tells us that $(1\ \ 1)^\mathrm{T}$ is an eigenvector for $(A')^\mathrm{T}$ with eigenvalue $-2$. Since transposition does not change eigenvalues, this tells us that $A'$ also has eigenvalue $-2$.

*The last eigenvalue can be obtained from the trace of $A'$. The sum of the eigenvalues must equal the trace. The trace of $A'$ is $-1$ and we already know one of its eigenvalues to be $-2$. This tells us the last eigenvalue is $1$.
All together, we have the eigenvalues of $A$ as $\{-2,\ -2,\ 1\}$, all without calculating the characteristic polynomial.
A: The best way to factor a cubic is in my opinion by using the rational root theorem.  This states that a rational number $\frac{p}{q}$ is a root of a polynomial $\sum_{k=0}^na_kx^k,$ then $p$ is a factor of $a_0$ and q is a factor of $a_n$.  Since you have a cubic, you know that the end behavior tends towards opposite infinities, and also, irrational roots and complex roots come in pairs, so you must have 1 rational root with a cubic.  So for your problem, the factors of $4$ are $\pm{1}, \pm{2},$ and $\pm{4}$ and the factors of $-1$ are obviously $\pm{1}$.  So your rational roots can be just $\pm{1}, \pm{2}, \pm{4}$.  
