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

chillywilly12a 2021-08-11 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.
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}-{9}\)

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Answered 2021-08-12 Author has 17734 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,
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}-{9}\)
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,
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}-{9}\)
Rewrite the above polynomial as,
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{9}{x}^{{{2}}}-{x}^{{{2}}}-{9}\)
Now, factor the polynomial P(x) as,
\(\displaystyle{P}{\left({x}\right)}={x}^{{{2}}}{\left({x}^{{{2}}}+{9}\right)}-{1}{\left({x}^{{{2}}}+{9}\right)}\)
\(\displaystyle={\left({x}^{{{2}}}-{1}\right)}{\left({x}^{{{2}}}+{9}\right)}\)
\(\displaystyle={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}^{{{2}}}+{9}\right)}\)
The factor \(\displaystyle{\left({x}-{1}\right)}\) and \(\displaystyle{\left({x}+{1}\right)}\) are linear factors with real zeros.
The factor \(\displaystyle{\left({x}^{{{2}}}+{9}\right)}\) 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 \(\displaystyle{\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}^{{{2}}}+{9}\right)}\)
b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients.
Given: The polynomial P(x) is,
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}-{9}\)
Calculation:
From part (a) the factored form of the polynomial P(x) is,
\(\displaystyle{P}{\left({x}\right)}={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}^{{{2}}}+{9}\right)}\)
Now, factor the remaining quadratic factor to obtain the complete factorization as,
\(\displaystyle{P}{\left({x}\right)}={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}^{{{2}}}+{9}\right)}\)
\(\displaystyle={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}^{{{2}}}-{\left({3}{i}\right)}^{{{2}}}\right)}\)
\(\displaystyle={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}-{3}{i}\right)}{\left({x}+{3}{i}\right)}\)
The above factors are linear factors with complex coefficients.
Conclusion:
Thus, the factored form of the polynomial P(x) that has linear factors is \(\displaystyle{\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({x}-{3}{i}\right)}{\left({x}+{3}{i}\right)}\).
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