# How do you write a polynomial in standard form given the zeros x=2, -6, and 2+ 4i?

Sep 19, 2017

${x}^{4} - 8 {x}^{2} + 128 x - 240$

#### Explanation:

$\text{given the zeros } x = a , x = b , x = c , x = d$

$\text{then the factors are } \left(x - a\right) , \left(x - b\right) , \left(x - c\right) , \left(x - d\right)$

$\text{the polynomial is then the product of the factors}$

$p \left(x\right) = \left(x - a\right) \left(x - b\right) \left(x - c\right) \left(x - d\right)$

$\text{here one of the given zeros is complex } 2 + 4 i$

$\text{complex zeros always occur in "color(blue)"conjugate pairs}$

$\Rightarrow 2 - 4 i \text{ is also a zero of the polynomial}$

$\text{the four zeros are } x = 2 , x = - 6 , x = 2 \pm 4 i$

$\text{4 zeros indicate a polynomial of degree 4}$

$\Rightarrow p \left(x\right) = \left(x - 2\right) \left(x + 6\right) \left(x - 2 - 4 i\right) \left(x - 2 + 4 i\right)$

$\textcolor{w h i t e}{\Rightarrow p \left(x\right)} = \left({x}^{2} + 4 x - 12\right) \left({x}^{2} - 4 x + 20\right)$

$\textcolor{w h i t e}{\Rightarrow p \left(x\right)} = {x}^{4} - 8 {x}^{2} + 128 x - 240 \text{ possible polynomial}$