2022 NYJC P1 Q5

It is given that \(-1+2\mathbf{i}\) is a root of the equation \({{z}^{3}}+2\left( 1+\mathbf{i} \right){{z}^{2}}+\left( 5+4\mathbf{i} \right)z+10\mathbf{i}=0\).

Explain why \(-1-2\mathbf{i}\) may not be a root.[1]

Without using a calculator, solve the equation \({{z}^{3}}+2\left( 1+\mathbf{i} \right){{z}^{2}}+\left( 5+4\mathbf{i} \right)z+10\mathbf{i}=0\), giving your answers in the form \(a+\mathbf{i}\,b\), where \(a\) and \(b\) are exact values.[5]

Hence solve \(\mathbf{i}{{z}^{3}}+2\left( 1+\mathbf{i} \right){{z}^{2}}+\left( 4-5\mathbf{i} \right)z-10\mathbf{i}=0\).[2]