# How do you write a polynomial with zeros -2, -2, 3, -4i and leading coefficient 1?

Mar 31, 2017

${\left(x + 2\right)}^{2} \left(x - 3\right) \left({x}^{2} + 16\right)$

#### Explanation:

We want the following roots:

$- 2 , - 2 , 3 , - 4 i$

Complex roots appear in conjugate pairs so the complex root $- 4 i$ must also be paired with a complex root $+ 4 i$, so we require the following roots:

$- 2 , - 2 , 3 , 4 i , - 4 i$

By the factor theorem if $x = \alpha$ is a root of $f \left(x\right) = 0$, then $\left(x - \alpha\right)$ is a factor of $f \left(x\right)$.

For the roots $- 2 , - 2$ (double root) we require two factors:

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

For the root $3$ we require the factor:

$\left(x - 3\right)$

For the roots $\pm 4 i$ we require the (complex) factors;

$\left(x - 4 i\right) \left(x + 4 i\right) = {x}^{2} + 16$

Combining these results we get:

$f \left(x\right) = {\left(x + 2\right)}^{2} \left(x - 3\right) \left({x}^{2} + 16\right)$