Characteristic polynomial proof

The trace of a matrix is the sum of the entries on its main diagonal. Prove that if $A$ is a $2 \times 2$ matrix, then the characteristic polynomial of $A$ is $x^2 − {c_1}x + c_2$ where $c_1$ is the trace of $A$ and $c_2$ is the determinant of $A$.

Can anyone explain this to me? So far, I only know that $C_a (x) = \operatorname{det}(A-xI)$, that the product of eigenvalues (counting multiplicity) is the $\operatorname{det}{A}$, and the sum of eigenvalues (counting multiplicity) equals the trace of $A$. I am just lost as to how to apply these.

Let $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

Than it's characteristic equation is given by $|A-xI|=0$

i.e. $\begin{vmatrix} a-x & b \\ c & d-x \end{vmatrix}=0$

On solving the determinant : $(a-x)(d-x)-cb=0 \implies x^2-\underbrace{(a+d)}_{ tr.(\text{A})}x+\underbrace{(ad-cb)}_{\det(A)}=0$

Here $a+d = tr.(\text{A}), ~ad-cb = \det{(A)}$

Let $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$. The the characteristic polynomial is \begin{eqnarray} \begin{vmatrix} a-x& b\\ c& d-x \end{vmatrix}&=&(a-x)(d-x)-cb\\&=&x^2-(a+d)x+ad-bc\\&=&x^2-\text{Tr}(A)x+\det(A).\end{eqnarray}

Eigenvalues are the roots of the characteristic polynomial, and the characteristic polynomial is monic. This is a $2 \times 2$ matrix, so the characteristic polynomial has degree $2$. By the Fundamental Theorem of Algebra, there are two eigenvalues, let them be $\alpha_1$ and $\alpha_2$, not necessarily distinct. Then by the Factor Theorem, since the characteristic polynomial is monic it can be written in the form $$(x-\alpha_1)(x-\alpha_2) = x^2 - (\alpha_1+\alpha_2)x + \alpha_1\alpha_2,$$ and then you can use the results you know to relate these to $c_1$ and $c_2$.