Question

Write a polynomial function of least degree that has given zeros. 2i, -2i

Answer 1

Reverse solve for x for the zeroes.
x = -2
x + 2 = 0

x = -1
x + 1 = 0

x = 1
x - 1 = 0

f(x) = (x + 2)(x + 1)(x - 1)
Can you multiply it out?

Answer 2

f(x) = (x+2)(x+1)(x-1)
Multiply it out
See that putting -1,-1 or 1 in the abuv function gives 0

Answer 3

hop these help

Answer 4

I don't get it

Answer 5

\(\large {
2i, -2i\quad
\begin{cases}
x=2i\to x-2i=0\to &(x-2i)=0\\
x=-2i\to x+2i=0\to &(x+2i)=0
\end{cases}
\\ \quad \\
(x-2i)(x+2i)=\textit{original polynomial}
}\)