Eigen values always occur in conjugate pairs.So all the 3 roots will be 2+i,2-i,3. Determinant of P= product of eigen values=15
EDIT: It is wrong to say that “Eigen values always occur in conjugate pairs”. What makes them appear in conjugate pair is the fact that given matrix is a real matrix. Why?
It is because determinant of real matrix is a real number and the determinant is equal to product of all the eigen values.
$\lambda _{1}$$\lambda _{2}$$\lambda _{3}$$\lambda _{4}$….. = determinant
If determinant is real number only then the product $\lambda _{1}$$\lambda _{2}$$\lambda _{3}$$\lambda _{4}$… is forced to be a real number. This implies the number of imaginary eigen values should be even i.e. they must occur in pairs. Now why conjugate pairs? you can see that from this equation :- $\lambda$$=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
Therefore, since the matrix is real, the eigen values occur in pairs...and moreover in conjugate pairs. Hence third eigen value must be 2-i
