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

Jul 14, 2016

${x}^{2} - 6 x + 10$

#### Explanation:

If the roots are ${x}_{1} \mathmr{and} {x}_{2}$, the quadratic is

x^2-((x_1+x_2)x+x_1x_2.

Here, it is

x^2-((3+i+(3-i))x+(3-i)(3+i)

$= {x}^{2} - 6 x + \left({3}^{2} - {\left(i\right)}^{2}\right)$

$= {x}^{2} - 6 x + 10$