# How do you find a fourth degree polynomial given roots 2+i and 1-sqrt5?

##### 1 Answer
Feb 15, 2017

${x}^{4} - 6 {x}^{3} + 10 {x}^{2} + 2 x - 15$

#### Explanation:

If 2+i is a root, its conjugate 2-i would also be a root. Likewise, if $1 - \sqrt{5}$ is a root, its conjugate $1 + \sqrt{5}$ would also be root. The polynomial would thus be

$\left(x - 2 - i\right) \left(x - 2 + i\right) \left(x - 1 + \sqrt{5}\right) \left(x - 1 - \sqrt{5}\right)$

$= \left({\left(x - 2\right)}^{2} + 1\right) \left({\left(x - 1\right)}^{2} - 5\right)$

$= \left({x}^{2} - 4 x + 5\right) \left({x}^{2} - 2 x - 3\right)$

$= {x}^{4} - 6 {x}^{3} + 10 {x}^{2} + 2 x - 15$