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# Write a third-degree polynomial in standard form with roots x = 4 and x - 2i.f(x)=a(x−4)(x−2i)(x+2i)

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asked 2021-06-09

Write a third-degree polynomial in standard form with roots $$x = 4$$ and $$x - 2i$$.

$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}−{4}\right)}{\left({x}−{2}{i}\right)}{\left({x}+{2}{i}\right)}$$

## Answers (1)

2021-06-10

By expanding,
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}−{4}\right)}{\left({x}^{{2}}−{2}{i}{x}+{2}{i}{x}+{4}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}−{4}\right)}{\left({x}^{{2}}+{4}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left[{x}^{{2}}{\left({x}−{4}\right)}+{4}{\left({x}−{4}\right)}\right]}$$

$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}^{{3}}−{4}{x}^{{2}}+{4}{x}−{16}\right)}$$

For simplicity, use a=1. So, a possible third-degree polynomial is:
$$\displaystyle{f{{\left({x}\right)}}}={1}{\left({x}^{{3}}−{4}{x}^{{2}}+{4}{x}−{16}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}−{4}{x}^{{2}}+{4}{x}−{16}$$

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