Quadratic Mania!

Since, $\color{#D61F06}{m}$ and $\color{#3D99F6}{n}$ are the roots to the polynomial $\color{#20A900}{x^{2}-2x+3}$ , therefore $\color{#D61F06}{m^{2}-2m+3=0}$ and $\color{#3D99F6}{n^{2}-2n+3=0}$ .

By synthetic division, it can be seen that :- $(i)\,\,\, \color{#624F41}{m^{3}-3m^{2}+5m-2}=(\color{#D61F06}{\underbrace{m^{2}-2m+3}_\text{0}})(m-1)+1 \,\,\,\,\,\,\,\,\,\,\,\ (ii)\,\,\, \color{#624F41}{n^{3}-n^{2}+n+5}=(\color{#3D99F6}{\underbrace{n^{2}-2n+3}_\text{0}})(n+1)+2$ $\implies \color{#624F41}{m^{3}-3m^{2}+5m-2}=1$ and $\color{#624F41}{n^{3}-n^{2}+n+5}=2$ .

So, the required polynomial is $\color{#BA33D6}{x^{2}}-(\color{#624F41}{1+2})\color{#BA33D6}{x}+(\color{#624F41}{1 \cdot 2})\,=\, x^{2}-3x+2$ .