Given the roots,
[tex]1,2i,i[/tex]To have a polynomial with real coefficients, the complex conjugates of 2i and i must be roots of the polynomial as well
The conjugate of 2i is -2i while the conjugate of i is -i
The roots of the polynomial will be
[tex]1,2i,-2i,i,-i[/tex]The factors of the polynomial will be
[tex](x-1),(x-2i),(x+2i),(x-i),(x+i)[/tex]The factored form of the polynomial will be
[tex]\begin{gathered} f(x)=(x-1)(x-2i)(x+2i)(x-i)(x+i) \\ f(x)=(x^2+1)(x^2+4)(x-1) \end{gathered}[/tex]Hence, answer is option A
