\(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}\)

Given that,

2i is a zero of P(x)

\(\displaystyle\therefore-{2}{i}\) is also a zero of P(x)

\(\displaystyle\therefore{\left({x}-{2}{i}\right)},{\left({x}+{2}{i}\right)}\) are two factors of P(x)

Now \(\displaystyle{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}={x}^{{{2}}}-{\left({2}{i}\right)}^{{{2}}}\)

\(\displaystyle={x}^{{{2}}}-{4}{i}^{{{2}}}\)

\(\displaystyle{x}^{{{2}}}+{4}\)

\(\displaystyle\therefore{P}{\left({x}\right)}={3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}\)

\(\displaystyle={3}{x}^{{{3}}}{\left({x}^{{{2}}}+{4}\right)}-{4}{x}^{{{2}}}{\left({x}^{{{2}}}+{4}\right)}-{3}{x}{\left({x}^{{{2}}}+{4}\right)}+{4}{\left({x}^{{{2}}}+{4}\right)}\)

\(\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left({3}{x}^{{{3}}}-{4}{x}^{{{2}}}-{3}{x}+{4}\right)}\)

\(\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left[{3}{x}^{{{2}}}{\left({3}{x}-{4}\right)}-{1}{\left({3}{x}-{4}\right)}\right]}\)

\(\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left({3}{x}-{4}\right)}{\left({x}^{{{2}}}-{1}\right)}\)

\(\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left({3}{x}-{4}\right)}{\left({x}+{1}\right)}{\left({x}-{1}\right)}\)

\(\displaystyle={\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({3}{x}-{4}\right)}{\left({x}+{1}\right)}{\left({x}-{1}\right)}\)

For zeros,

\(\displaystyle{P}{\left({x}\right)}={0}\)

\(\displaystyle\Rightarrow{3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}={0}\)

\(\displaystyle\Rightarrow{\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({3}{x}-{4}\right)}{\left({x}+{1}\right)}{\left({x}-{1}\right)}={0}\)

\(\therefore\) Zeros of P(x) are

\(\displaystyle-{2}{i},{2}{i},{\frac{{{4}}}{{{3}}}},-{1},{1}\).