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

Given information:

The polynomial P(x) is,

\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

Concept used:

Linear and Quadratic Factors Theorem:

Any polynomial that has real coefficients can be factored into the product of linear and irreducible quadratic factors.

Calculation:

The given polynomial P(x) is,

\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

Rewrite the above polynomial as,

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

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

Use identity \(\displaystyle{a}^{{{2}}}-{b}^{{{2}}}={\left({a}-{b}\right)}{\left({a}+{b}\right)}\) to factor the above equation as,

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

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

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

The factors x, \(\displaystyle{\left({x}-{2}\right)}\) and \(\displaystyle{\left({x}+{2}\right)}\) are linear factors.

The factor \(\displaystyle{\left({x}^{{{2}}}+{4}\right)}\) is irreducible, since it has no real zeros.

Conclusion:

Thus, the factored form of the polynomial P(x) that has linear and irreducible quadratic factors is \(\displaystyle{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{2}}}+{4}\right)}\)

b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients.

Given: The polynomial P(x) is,

\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

Calculation:

From part (a) the factored form of the polynomial P(x) is,

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

Now, factor the remaining quadratic factor to obtain the complete factorization as,

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

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

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

The above factors are linear factors with complex coefficients.

Conclusion:

Thus, the factored form of the polynomial P(x) that has linear factors is \(\displaystyle{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}\).

Given information:

The polynomial P(x) is,

\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

Concept used:

Linear and Quadratic Factors Theorem:

Any polynomial that has real coefficients can be factored into the product of linear and irreducible quadratic factors.

Calculation:

The given polynomial P(x) is,

\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

Rewrite the above polynomial as,

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

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

Use identity \(\displaystyle{a}^{{{2}}}-{b}^{{{2}}}={\left({a}-{b}\right)}{\left({a}+{b}\right)}\) to factor the above equation as,

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

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

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

The factors x, \(\displaystyle{\left({x}-{2}\right)}\) and \(\displaystyle{\left({x}+{2}\right)}\) are linear factors.

The factor \(\displaystyle{\left({x}^{{{2}}}+{4}\right)}\) is irreducible, since it has no real zeros.

Conclusion:

Thus, the factored form of the polynomial P(x) that has linear and irreducible quadratic factors is \(\displaystyle{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}^{{{2}}}+{4}\right)}\)

b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients.

Given: The polynomial P(x) is,

\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

Calculation:

From part (a) the factored form of the polynomial P(x) is,

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

Now, factor the remaining quadratic factor to obtain the complete factorization as,

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

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

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

The above factors are linear factors with complex coefficients.

Conclusion:

Thus, the factored form of the polynomial P(x) that has linear factors is \(\displaystyle{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}\).