Linear and Quadratic Factors. A polynomial P is given (a)

Aneeka Hunt 2021-08-14 Answered
Linear and Quadratic Factors. 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.
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}+{16}\)

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tabuordg
Answered 2021-08-15 Author has 17089 answers
a) To find: The factor of the polynomial into linear and irreducible quadratic factor with real coefficients.
Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}+{16}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={x}^{{{4}}}+{4}{x}^{{{2}}}+{4}{x}^{{{2}}}+{16}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={x}^{{{2}}}{\left({x}^{{{2}}}+{4}\right)}+{4}{\left({x}^{{{2}}}+{4}\right)}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}^{{{2}}}+{4}\right)}{\left({x}^{{{2}}}+{4}\right)}\)
Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}^{{{2}}}+{4}\right)}{\left({x}^{{{2}}}+{4}\right)}\).
b) To find: The factor completely into linear factors with complex coefficients.
Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}+{16}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={x}^{{{4}}}+{4}{x}^{{{2}}}+{4}{x}^{{{2}}}+{16}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={x}^{{{2}}}{\left({x}^{{{2}}}+{4}\right)}+{4}{\left({x}^{{{2}}}+{4}\right)}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}^{{{2}}}+{4}\right)}{\left({x}^{{{2}}}+{4}\right)}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}\)
Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}\).
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