It is given that, \(\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}-{4}{x}^{{3}}+{7}^{{2}}-{16}{x}+{12}\)

We have to find all the real and complex zeros of the polynomial.

We have, \(\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}-{4}{x}^{{3}}+{7}{x}^{{2}}-{16}{x}+{12}\)

For the zeros of polynomial,

\(\displaystyle\rightarrow{x}^{{4}}-{4}{x}^{{3}}+{7}{x}^{{2}}-{16}{x}+{12}={0}\)

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

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

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

\(\displaystyle\Rightarrow{x}={3}\) or \(\displaystyle{x}={1}\) or \(\displaystyle{x}^{{2}}=-{4}\)

\(\displaystyle\Rightarrow{x}={3}\) or \(\displaystyle{x}={1}\) or \(\displaystyle{x}=\pm{2}{i}\)

There are four zeros of the polynomial

Two real numbern \(x=1,3\)

Two complex number \(\displaystyle={2}{i},-{2}{i}\)