a) To find: The factorization of the polynomial P(x) in linear and irreducible quadratic factors with real coefficients.

Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)

Using the Rational Zeros Theorem, possible rational zeros are \(\pm 1, \pm 2, \pm 4\).

Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\), we get

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

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={8}-{4}-{4}\)

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={0}\)

Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\).

Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\).

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{2}\right)}\)

Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{2}\right)}\).

b) To find: The factor completely into linear factors with complex coefficients.

Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)

Using the Rational Zeros Theorem, possible rational zeros are \(\displaystyle\pm{1},\pm{2},\pm{4}\).

Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\), we get

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

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={8}-{4}-{4}\)

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={0}\)

Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\).

Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{2}\right)}\)

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

Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}-{\left(-{1}+{i}\right)}\right)}{\left({x}-{\left(-{1}-{i}\right)}\right)}\).