# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -2, -2, 3, -4i?

Jun 2, 2018

Construct it from linear terms including the given roots

#### Explanation:

$x$ minus each of the zeroes is equal to zero, so $f \left(x\right) = {\left(x + 2\right)}^{2} \left(x - 3\right) \left(x + 4 i\right)$ does the job and has no more degrees than needed - assuming that the question-setter requires the repeated root at $x = - 2$. If they don't, then $f \left(x\right) = \left(x + 2\right) \left(x - 3\right) \left(x + 4 i\right)$ is better.

This multiplies out to be
$f \left(x\right) = {x}^{3} - \left(1 - 4 i\right) {x}^{2} - \left(6 + 4 i\right) x - 24 i$
in the non-repeated root case and
$f \left(x\right) = {x}^{4} + \left(1 + r i\right) {x}^{3} - \left(8 - 4 i\right) {x}^{2} - \left(12 + 32 i\right) x - 48 i$
if we include the root at $x = - 2$ twice.