# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 5, 2+3i?

Aug 5, 2018

$p \left(x\right) = {x}^{3} - 9 {x}^{2} + 33 x - 65$

#### Explanation:

$\text{by the factor theorem}$

$\text{if "x=a" is a zero then }$

$\left(x - a\right) \text{ is a factor of the polynomial}$

$\text{note that complex zeros occur in conjugate pairs}$

$x = 2 + 3 i \text{ is a zero then "2color(red)(-)3i" is also a zero}$

$p \left(x\right) = \left(x - 5\right) \left(x - \left(2 + 3 i\right)\right) \left(x - \left(2 - 3 i\right)\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left(x - 5\right) \left(x - 2 - 3 i\right) \left(x - 2 + 3 i\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left(x - 5\right) \left({\left(x - 2\right)}^{2} - {\left(3 i\right)}^{2}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left(x - 5\right) \left({x}^{2} - 4 x + 4 + 9\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left(x - 5\right) \left({x}^{2} - 4 x + 13\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} - 9 {x}^{2} + 33 x - 65$