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

Tabansi 2021-08-14 Answered
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.
\(\displaystyle{P}{\left({x}\right)}={x}^{{{5}}}-{16}{x}\)

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Expert Answer

falhiblesw
Answered 2021-08-15 Author has 18601 answers
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 \(\displaystyle{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}\).
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