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

Jul 11, 2016

${x}^{3} - {x}^{2} i + 9 x - 9 i = 0$

Explanation:

When we talk about the zeros of a polynomial, we can express them as $x = \text{the zero}$. So with the terms listed above, we can say:

$x = i , - 3 i , 3 i$ and we can therefore say:

$x - i = 0 , x + 3 i = 0 , x - 3 i = 0$

$\left(x - i\right) \left(x + 3 i\right) \left(x - 3 i\right) = 0$

Expanding and collecting like terms for coefficients in the cubic

${x}^{3} - {x}^{2} i + 9 x - 9 i = 0$.

Also, we can use the form

${x}^{3}$-

(sum of the roots)x^2+

sum of the products of the roots, taken two at a time)x-

product of the roots

= 0.

Here, it is

${x}^{3} - \left(i + 3 i - 3 i\right) {x}^{2} + \left(3 - 3 + 9\right) x - \left(9 i\right) = 0$. Simplifying,

${x}^{3} - {x}^{2} i + 9 x - 9 i = 0$