a) To find: The factor of the polynomial into linear and irreducible quadratic factor with real coefficients.

Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

Using the Rational Zeros Theorem, possible rational zeros are \(\displaystyle\pm{1},\pm{2},\pm{4},\pm{8},\pm{16},\pm{32},\pm{64}\).

Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{6}}}-{64}\), we get

\(\displaystyle{P}{\left({2}\right)}={2}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={64}-{64}\)

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

Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\).

Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

Substitute \(\displaystyle{x}=-{2}\in{\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\), we get

\(\displaystyle{\left({\left(-{2}\right)}^{{{5}}}+{2}{\left(-{2}\right)}^{{{4}}}+{4}{\left(-{2}\right)}^{{{3}}}+{8}{\left(-{2}\right)}^{{{2}}}+{16}{\left(-{2}\right)}+{32}\right)}\)

\(\displaystyle=-{32}+{32}-{32}+{32}-{32}+{32}\)

\(\displaystyle={0}\)

Therefore, \(\displaystyle{x}=-{2}\) is a factor of \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\).

Factorize the polynomial, \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

\(\displaystyle\Rightarrow{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}={\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

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

Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{4}\right)}{\left({x}^{{{2}}}-{2}{x}+{4}\right)}\).

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

Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

Using the Rational Zeros Theorem, possible rational zeros are \(\displaystyle\pm{1},\pm{2},\pm{4},\pm{8},\pm{16},\pm{32},\pm{64}\).

Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{6}}}-{64}\), we get

\(\displaystyle{P}{\left({2}\right)}={2}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={64}-{64}\)

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

Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\).

Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

Substitute \(\displaystyle{x}=-{2}\in{\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\), we get

\(\displaystyle{\left({\left(-{2}\right)}^{{{5}}}+{2}{\left(-{2}\right)}^{{{4}}}+{4}{\left(-{2}\right)}^{{{3}}}+{8}{\left(-{2}\right)}^{{{2}}}+{16}{\left(-{2}\right)}+{32}\right)}\)

\(\displaystyle=-{32}+{32}-{32}+{32}-{32}+{32}\)

\(\displaystyle={0}\)

Therefore, \(\displaystyle{x}=-{2}\) is a factor of \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\).

Factorize the polynomial, \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

\(\displaystyle\Rightarrow{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}={\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

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

\(\displaystyle\Rightarrow{P}{\left({x}\right)}\)

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

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

Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

Using the Rational Zeros Theorem, possible rational zeros are \(\displaystyle\pm{1},\pm{2},\pm{4},\pm{8},\pm{16},\pm{32},\pm{64}\).

Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{6}}}-{64}\), we get

\(\displaystyle{P}{\left({2}\right)}={2}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={64}-{64}\)

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

Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\).

Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

Substitute \(\displaystyle{x}=-{2}\in{\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\), we get

\(\displaystyle{\left({\left(-{2}\right)}^{{{5}}}+{2}{\left(-{2}\right)}^{{{4}}}+{4}{\left(-{2}\right)}^{{{3}}}+{8}{\left(-{2}\right)}^{{{2}}}+{16}{\left(-{2}\right)}+{32}\right)}\)

\(\displaystyle=-{32}+{32}-{32}+{32}-{32}+{32}\)

\(\displaystyle={0}\)

Therefore, \(\displaystyle{x}=-{2}\) is a factor of \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\).

Factorize the polynomial, \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

\(\displaystyle\Rightarrow{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}={\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

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

Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{4}\right)}{\left({x}^{{{2}}}-{2}{x}+{4}\right)}\).

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

Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

Using the Rational Zeros Theorem, possible rational zeros are \(\displaystyle\pm{1},\pm{2},\pm{4},\pm{8},\pm{16},\pm{32},\pm{64}\).

Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{6}}}-{64}\), we get

\(\displaystyle{P}{\left({2}\right)}={2}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({2}\right)}={64}-{64}\)

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

Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\).

Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

Substitute \(\displaystyle{x}=-{2}\in{\left({x}-{2}\right)}{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\), we get

\(\displaystyle{\left({\left(-{2}\right)}^{{{5}}}+{2}{\left(-{2}\right)}^{{{4}}}+{4}{\left(-{2}\right)}^{{{3}}}+{8}{\left(-{2}\right)}^{{{2}}}+{16}{\left(-{2}\right)}+{32}\right)}\)

\(\displaystyle=-{32}+{32}-{32}+{32}-{32}+{32}\)

\(\displaystyle={0}\)

Therefore, \(\displaystyle{x}=-{2}\) is a factor of \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\).

Factorize the polynomial, \(\displaystyle{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}\)

\(\displaystyle\Rightarrow{\left({x}^{{{5}}}+{2}{x}^{{{4}}}+{4}{x}^{{{3}}}+{8}{x}^{{{2}}}+{16}{x}+{32}\right)}={\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{4}}}+{4}{x}^{{{2}}}+{16}\right)}\)

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

\(\displaystyle\Rightarrow{P}{\left({x}\right)}\)

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

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