Mathematics, 05.08.2019 01:30, creeconnell1
How to find the simplest polynomial function with the given zeros 5,-3i
Answers: 1
![\bf \textit{difference of squares} \\\\ (a-b)(a+b) = a^2-b^2 \\\\\\ \textit{also recall that }i^2=-1\\\\ -------------------------------\\\\ \begin{cases} x=5\implies &x-5=0\\ x=0-3i\implies &x+3i=0\\ x=0+3i\implies &x-3i=0 \end{cases} \\\\\\ (x-5)(x+3i)(x-3i)=\stackrel{original~polynomial}{0} \\\\\\ (x-5)~~~[(x)^2~~-~~(3i)^2] = y\implies (x-5)~~~[(x)^2~~-~~(3^2i^2)] = y \\\\\\ (x-5)~~~[(x)^2~~-~~(9\cdot -1)] = y\implies (x-5)(x^2+9)=y \\\\\\ x^3+9x-5x^2-45=y\implies x^3-5x^2+9x-45=y](/tpl/images/0171/5398/e77df.png)
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