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
ANSWERED
asked 2021-08-06
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 \(\displaystyle{P}{\left({x}\right)}=\).

Answers (1)

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
\(\displaystyle-{2}{i},{2}{i},{\frac{{{4}}}{{{3}}}},-{1},{1}\).

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