How can I tell if a matrix is diagonalizable knowing only the trace, one eigenvalue, and a result of the characteristic polynomial?

You have two missing eigenvalues, other than the known eigenvalue $2$. The first equation tells you that their sum is $-4$. The last equation gives you another equation: since $\chi_A(t) = (t-2)(t-\lambda)(t-\mu)$, plugging in $t=1$ allows you to solve for $\lambda$ and $\mu$.

The characteristic polynomial is $(X-2)(X^2+bX+c)$. The trace is the coefficient of $X^2$ that is:$b-2=-2$.

The last equation enables you to find $c$.