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

Feb 18, 2017

$P \left(x\right) = {x}^{4} - 12 {x}^{3} + 53 {x}^{2} - 104 x + 80$

#### Explanation:

$4 , 4 , 2 + i$
Root $4$ is multiplicity $2$
$2 + i$ Complex Conjugate Zero is $2 - i$, we have:
$P \left(x\right) = {\left(x - 4\right)}^{2} \left(x - 2 - i\right) \left(x - 2 + i\right)$
$P \left(x\right) = \left({x}^{2} - 8 x + 16\right) \left[{\left(x - 2\right)}^{2} - {i}^{2}\right]$
$P \left(x\right) = \left({x}^{2} - 8 x + 16\right) \left({x}^{2} - 4 x + 4 - \left(- 1\right)\right]$
$P \left(x\right) = \left({x}^{2} - 8 x + 16\right) \left({x}^{2} - 4 x + 5\right)$
$P \left(x\right) = {x}^{4} - 4 {x}^{3} + 5 {x}^{2} - 8 {x}^{3} + 32 {x}^{2} - 40 x + 16 {x}^{2} - 64 x + 80$
$P \left(x\right) = {x}^{4} - 12 {x}^{3} + 53 {x}^{2} - 104 x + 80$