<s>[INST] Let $A$ be a real $2×2$ matrix such that $A^6=I$. The total number of possibilities for the characteristic polynomial of $A$ is: Let $A$ be a real $2×2$ matrix such that $A^6 = I$ (where $I$ denote the identity $2×2$ matrix). The total number
of possibilities for the characteristic polynomial of $A$ is:
Annihilating polynomial is $x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1).$ Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial.
So, Possible CHaracteristic polynomials are $(x-1)(x+1),(x^2+x+1),(x^2-x+1)$.We can not take any 2-degree polynomial factor from the annihilating polynomial. Since complex roots occur in pairs. But the answer given is $5$. Can you help me where is my mistake?
 [/INST] This statement is false:

Characteristic polynomial divides any annihilating polynomial of degree greater than or equal to the degree of the characteristic polynomial. 

$$f(t) = (t-1)(t^4+7t^3+105t-999)$$
Annihilates $A=I$. The characteristic polynomial of $A$ does not divide $f(t)$.
It seems like you may be confusing characteristic polynomial with minimal polynomial.
Without loss of generality, you may assume $A$ is diagonal (it is diagonalizable, and similar matrices have the same characteristic polynomial). Your matrix, up to similarity will be
$$\pmatrix{1 & 0 \\ 0 & 1}, \pmatrix{1 & 0 \\ 0 & -1}, \pmatrix{-1 & 0 \\ 0 & -1},$$
$$ \pmatrix{-\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i & 0 \\ 0 & -\dfrac{1}{2} - \dfrac{ \sqrt{3}}{2}i}, \pmatrix{\dfrac{1}{2} + \dfrac{ \sqrt{3}}{2}i & 0 \\ 0 & \dfrac{1}{2} - \dfrac{ \sqrt{3}}{2}i}$$
And so it's characteristic polynomial (respectively) will be one of
$$(x-1)^2, (x-1)(x+1), (x+1)^2, x^2+x+1, x^2-x+1$$
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