Question

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

Factors and multiples
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.
$$\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}-{2}{x}-{4}$$

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, $$\displaystyle{P}{\left({x}\right)}={\left({x}-{2}\right)}{\left({x}-{\left(-{1}+{i}\right)}\right)}{\left({x}-{\left(-{1}-{i}\right)}\right)}$$.