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I have to find rank of the matrix from the given characteristic polynomial. Here is my attempt.

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Now how to check other possibilities of rank ?

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    Do you know the Jordan normal form of a matrix? – 2017-01-29
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    Jordan form is probably overkill here. – 2017-01-29

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Recall that the geometric multiplicity of an eigenvalue is less than or equal to the algebraic multiplicity. In your case, since $\beta \neq 0$, the algebraic multiplicity of the eigenvalue $0$ is $3$ and the geometric multiplicity of $0$ is given by $\dim \ker A$ so we have $1 \leq \dim \ker A \leq 3$ (or equivalently, $2 \leq \operatorname{rank} A \leq 4$). By playing with examples of block diagonal matrices, you can see that indeed all three cases are possible.

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    You mean the geometric multiplicity of eigenvalue 0 is numerically equal to the nulity of matrix ? – 2017-01-30
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    @AbhishekChandra: Yep. – 2017-01-30
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Hint: The eigenspace associated with the eigenvalue $0$ is also the kernel (AKA nullspace) of $A$. How does this relate to the rank? What are the possible dimensions of this eigenspace?

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    I feel like this should be a comment, not an answer. – 2017-01-29
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    @JEFFYELTON I respectfully disagree, pending feedback from the asker. – 2017-01-29
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    @JEFFYELTON I think this is the best answer possible (provided one doesn't do all the work for the OP). – 2017-01-30