Question

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

Factors and multiples
ANSWERED
asked 2021-08-14
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.
\(\displaystyle{P}{\left({x}\right)}={x}^{{{4}}}+{8}{x}^{{{2}}}+{16}\)

Answers (1)

2021-08-15
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, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}\).
0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...