# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 4i, -4i?

Feb 25, 2018

$f \left(x\right) = {x}^{2} + 16$

#### Explanation:

If $x = a$ is a zero of a polynomial, then $\left(x - a\right)$ is a factor.

Hence to get the required zeros we can write:

$f \left(x\right) = \left(x - 4 i\right) \left(x + 4 i\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{2} - {\left(4 i\right)}^{2}$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{2} + 16$

Any polynomial in $x$ with these zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$.