By expanding,

\(\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}−{4}\right)}{\left({x}^{{2}}−{2}{i}{x}+{2}{i}{x}+{4}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}−{4}\right)}{\left({x}^{{2}}+{4}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={a}{\left[{x}^{{2}}{\left({x}−{4}\right)}+{4}{\left({x}−{4}\right)}\right]}\)

\(\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}^{{3}}−{4}{x}^{{2}}+{4}{x}−{16}\right)}\)

For simplicity, use a=1. So, a possible third-degree polynomial is:

\(\displaystyle{f{{\left({x}\right)}}}={1}{\left({x}^{{3}}−{4}{x}^{{2}}+{4}{x}−{16}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}−{4}{x}^{{2}}+{4}{x}−{16}\)