(2/7,-1) and 9+1/(3i) are zeros. Find a polynomial function with real

ossidianaZ 2021-08-14 Answered
\(\displaystyle{\left(\frac{{2}}{{7}},-{1}\right)}{\quad\text{and}\quad}{9}+\frac{{1}}{{{3}{i}}}\) are zeros.
Find a polynomial function with real coefficients.

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Expert Answer

liingliing8
Answered 2021-08-15 Author has 12132 answers
\(\displaystyle{\left({x}-\frac{{2}}{{7}}\right)}{\left({x}-{\left(-{1}\right)}\right)}{\left({x}-{\left({9}-\sqrt{{{3}{i}}}\right)}\right)}{\left({x}-{\left({9}+\sqrt{{{3}{i}}}\right)}\right)}={\left({x}-\frac{{2}}{{7}}\right)}{\left({x}+{1}\right)}{\left({x}-{\left({9}-\sqrt{{{3}{i}}}\right)}\right)}{\left({x}-{\left({9}+\sqrt{{{3}{i}}}\right)}\right)}\)
\(\displaystyle={\left({x}^{{2}}+{x}-\frac{{2}}{{7}}{x}-\frac{{2}}{{7}}\right)}{\left({\left({x}-{9}\right)}^{{2}}-{\left(\sqrt{{{3}{i}}}\right)}^{{2}}\right)}\)
\(\displaystyle={\left({x}^{{2}}+\frac{{5}}{{7}}{x}-\frac{{2}}{{7}}\right)}{\left({x}^{{2}}-{18}{x}+{81}+{3}\right)}\)
\(\displaystyle=\frac{{1}}{{7}}{\left({x}^{{2}}+{5}{x}-{2}\right)}{\left({x}^{{2}}-{18}{x}+{84}\right)}\)
\(\displaystyle=\frac{{1}}{{7}}{\left({x}^{{4}}-{13}{x}^{{3}}-{8}{x}^{{2}}+{456}{x}-{168}\right)}\)
\(\displaystyle\frac{{1}}{{7}}{\left({x}^{{4}}-{13}{x}^{{3}}-{8}{x}^{{2}}+{456}{x}-{168}\right)}\) is the polynomial function with real coefficients for the given zeros.
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