Given \(\displaystyle{f{{\left({x}\right)}}}={x}^{{4}}-{8}{x}^{{2}}-{8}{x}+{15}\)

The all the roots of the above expression is

X=1

X=3

X=-2+i

X=-2-i

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

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

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

\(\displaystyle={\left({x}^{{4}}+{4}{x}^{{3}}+{5}{x}^{{2}}-{4}{x}^{{3}}-{16}{x}^{{2}}-{20}{x}+{3}{x}^{{2}}+{12}{x}+{15}\right)}\)

\(\displaystyle={x}^{{4}}-{8}{x}^{{2}}-{8}{x}+{15}\)

The all the roots of the above expression is

X=1

X=3

X=-2+i

X=-2-i

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

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

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

\(\displaystyle={\left({x}^{{4}}+{4}{x}^{{3}}+{5}{x}^{{2}}-{4}{x}^{{3}}-{16}{x}^{{2}}-{20}{x}+{3}{x}^{{2}}+{12}{x}+{15}\right)}\)

\(\displaystyle={x}^{{4}}-{8}{x}^{{2}}-{8}{x}+{15}\)