Question

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

Factors and multiples
ANSWERED
asked 2021-08-15
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.
\(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)

Expert Answers (1)

2021-08-16

a) To find: The factorization of the polynomial P(x) in linear and irreducible quadratic factors with real coefficients.
Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)
Using the Rational Zeros Theorem, possible rational zeros are \(\pm 1, \pm 2, \pm 4\).
Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\), we get
\(\displaystyle{P}{\left({2}\right)}={2}^{{{3}}}-{2}{\left({2}\right)}-{4}\)
\(\displaystyle\Rightarrow{P}{\left({2}\right)}={8}-{4}-{4}\)
\(\displaystyle\Rightarrow{P}{\left({2}\right)}={0}\)
Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\).
Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\).
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{2}\right)}\)
Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{2}\right)}\).
b) To find: The factor completely into linear factors with complex coefficients.
Consider, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)
Using the Rational Zeros Theorem, possible rational zeros are \(\displaystyle\pm{1},\pm{2},\pm{4}\).
Substitute \(\displaystyle{x}={2}\in{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\), we get
\(\displaystyle{P}{\left({2}\right)}={2}^{{{3}}}-{2}{\left({2}\right)}-{4}\)
\(\displaystyle\Rightarrow{P}{\left({2}\right)}={8}-{4}-{4}\)
\(\displaystyle\Rightarrow{P}{\left({2}\right)}={0}\)
Therefore, \(\displaystyle{x}={2}\) is a factor of \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\).
Factorize the polynomial, \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}^{{{2}}}+{2}{x}+{2}\right)}\)
\(\displaystyle\Rightarrow{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}-{\left(-{1}+{i}\right)}\right)}{\left({x}-{\left(-{1}-{i}\right)}\right)}\)
Therefore, \(\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}-{\left(-{1}+{i}\right)}\right)}{\left({x}-{\left(-{1}-{i}\right)}\right)}\).

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