Question

Find the polynomial function that would have the following solutions. Write it in standard form. x=3 and x= -2i

Answer 1

keeping in mind that complex roots never come all by their lonesome, their sister is always with them, namely the conjugate, so if we know that there's a complex root of -2i or namely 0 - 2i, its conjugate is also there too, namely 0 + 2i, or just 2i, so then

`\begin{cases} x=3\implies &x-3=0\\ x=-2i\implies &x+2i=0\\ x=2i\implies &x-2i=0 \end{cases}\qquad \implies (x-3)(x+2i)(x-2i)=\stackrel{0}{y} \\\\[-0.35em] ~\dotfill\\\\ \underset{\textit{difference of squares}}{(x+2i)(x-2i)}\implies [(x)^2-(2i)^2] \\\\\\\ [x^2-(4i^2)]\implies [x^2-[4(-1)]] \implies x^2+4 \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x^2+4)=y\implies x^3-3x^2+4x-12=y`