# Given: n=3. 4 and 2i are zeros. f(-1)=75 . Find an nth-degree polynomial function with real coefficients

Given:
$$n=3$$.
4 and 2i are zeros.
$$f(-1)=75$$
Find an nth-degree polynomial function with real coefficients

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2abehn

$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{4}\right)}{\left({x}-{2}{i}\right)}{\left({x}+{2}{i}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{4}\right)}{\left({x}^{{2}}-{2}{i}{x}+{2}{i}{x}+{4}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}-{4}\right)}{\left({x}^{{2}}+{4}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left[{x}^{{2}}{\left({x}-{4}\right)}+{4}{\left({x}-{4}\right)}\right]}$$
$$\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}^{{3}}-{4}{x}^{{2}}+{4}{x}-{16}\right)}$$
Use $$\displaystyle{f{{\left(-{1}\right)}}}={75}$$ to solve for a:
$$\displaystyle-{75}={a}{\left[{\left(-{1}\right)}^{{3}}-{4}{\left(-{1}\right)}^{{2}}+{4}{\left(-{1}\right)}-{16}\right]}$$
$$-75=-25a$$
$$a=3$$
$$\displaystyle{f{{\left({x}\right)}}}={3}{x}^{{3}}-{12}{x}^{{2}}+{12}{x}-{48}$$