# How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are 8, -i, i?

Jul 14, 2016

${x}^{3} - 8 {x}^{2} + x - 6$

#### Explanation:

The polynomial with zeros ${x}_{1} , {x}_{2} \mathmr{and} {x}_{3}$ is

${x}^{3} - \left({x}_{1} + {x}_{2} + {x}_{3}\right) {x}^{2} + \left({x}_{2} {x}_{3} + {x}_{3} {x}_{1} + {x}_{1} {x}_{2}\right) x - {x}_{1} {x}_{2} {x}_{3}$

Here, it is

${x}^{3} - \left(8 + i - i\right) {x}^{2} + \left(- {i}^{2} - 8 i + 8 i\right) x - \left(9\right) \left(i\right) \left(- i\right)$

$= {x}^{3} - 8 {x}^{2} + x - 6$