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

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

falhiblesw
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:
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 $$\displaystyle{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}$$.