Question

# Given P(x)=3x^{5}-4x^{4}+9x^{3}-12x^{2}-12x+16, and that 2i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including P(x)=.

Factors and multiples
Given $$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}$$, and that 2i is a zero, write P in factored form (as a product of linear factors). Be sure to write the full equation, including $$\displaystyle{P}{\left({x}\right)}=$$.

2021-08-07

$$\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}$$
Given that,
2i is a zero of P(x)
$$\displaystyle\therefore-{2}{i}$$ is also a zero of P(x)
$$\displaystyle\therefore{\left({x}-{2}{i}\right)},{\left({x}+{2}{i}\right)}$$ are two factors of P(x)
Now $$\displaystyle{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}={x}^{{{2}}}-{\left({2}{i}\right)}^{{{2}}}$$
$$\displaystyle={x}^{{{2}}}-{4}{i}^{{{2}}}$$
$$\displaystyle{x}^{{{2}}}+{4}$$
$$\displaystyle\therefore{P}{\left({x}\right)}={3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}$$
$$\displaystyle={3}{x}^{{{3}}}{\left({x}^{{{2}}}+{4}\right)}-{4}{x}^{{{2}}}{\left({x}^{{{2}}}+{4}\right)}-{3}{x}{\left({x}^{{{2}}}+{4}\right)}+{4}{\left({x}^{{{2}}}+{4}\right)}$$
$$\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left({3}{x}^{{{3}}}-{4}{x}^{{{2}}}-{3}{x}+{4}\right)}$$
$$\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left[{3}{x}^{{{2}}}{\left({3}{x}-{4}\right)}-{1}{\left({3}{x}-{4}\right)}\right]}$$
$$\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left({3}{x}-{4}\right)}{\left({x}^{{{2}}}-{1}\right)}$$
$$\displaystyle={\left({x}^{{{2}}}+{4}\right)}{\left({3}{x}-{4}\right)}{\left({x}+{1}\right)}{\left({x}-{1}\right)}$$
$$\displaystyle={\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({3}{x}-{4}\right)}{\left({x}+{1}\right)}{\left({x}-{1}\right)}$$
For zeros,
$$\displaystyle{P}{\left({x}\right)}={0}$$
$$\displaystyle\Rightarrow{3}{x}^{{{5}}}-{4}{x}^{{{4}}}+{9}{x}^{{{3}}}-{12}{x}^{{{2}}}-{12}{x}+{16}={0}$$
$$\displaystyle\Rightarrow{\left({x}+{2}{i}\right)}{\left({x}-{2}{i}\right)}{\left({3}{x}-{4}\right)}{\left({x}+{1}\right)}{\left({x}-{1}\right)}={0}$$
$$\therefore$$ Zeros of P(x) are
$$\displaystyle-{2}{i},{2}{i},{\frac{{{4}}}{{{3}}}},-{1},{1}$$.