# A polynomial P is given (a) Factor P into linear

A polynomial P is given (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients.
$$\displaystyle{P}{\left({x}\right)}={x}^{{{6}}}-{64}$$

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