# How do you find a third degree polynomial given roots -4 and 4i?

Dec 27, 2017

${x}^{3} + 4 {x}^{2} + 16 x + 64 = 0$

#### Explanation:

Since the question ask for a third degree polynomial, I am going to assume that you want a polynomial with real coefficients, with $- 4 i$ as the third root, being the complex conjugate of $4 i$.

Each zero $a$ corresponds to a factor $\left(x - a\right)$, so we can write:

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

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

$\textcolor{w h i t e}{f \left(x\right)} = \left(x + 4\right) \left({x}^{2} + 16\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} + 4 {x}^{2} + 16 x + 64$

So we can write a cubic equation:

${x}^{3} + 4 {x}^{2} + 16 x + 64 = 0$ with the desired roots.