Question

Find an nth degree polynomial function with real coefficients satisfying the given conditions.

n=3; -5 and i are ​zeros; f (-3 ) = 60

Answer 1

keeping in mind that complex roots never come alone, their sister is always with them, the conjugate, so if we know that "i" or namely "0 + i" is there, her sister the conjugate "0 +i" must also be there too, so let's take a peek

`\begin{cases} x=-5\implies &x+5=0\\ x=0+i\implies &x-i=0\\ x=0-i\implies &x+i=0 \end{cases}\implies \underline{a(x+5)(x-i)(x+i)=\stackrel{0}{y}} \\\\[-0.35em] ~\dotfill`

`\underset{\textit{difference of squares}}{(x-i)(x+i)}\implies [x^2-i^2]\implies [x^2-(-1)]\implies (x^2+1) \\\\[-0.35em] ~\dotfill\\\\ \underline{a(x+5)(x^2+1)=y}~\hfill \stackrel{\textit{we also know that}}{f(\stackrel{x}{-3})=60}~\hfill a[(-3)+5][(-3)^2+1]=60 \\\\\\ a(2)(10)=60\implies 20a=60\implies a=\cfrac{60}{20}\implies a=3 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \underline{3(x+5)(x^2+1)=y}~\hfill`

Check the picture below.