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

Oct 3, 2016

y = (x -5)(x - 3i)(x - -3i) = x³ - 5x² + 9x - 45

#### Explanation:

With any polynomial the with roots ${r}_{1} , {r}_{2} , {r}_{3} , \ldots {r}_{n}$, the factors are:

$y = k \left(x - {r}_{1}\right) \left(x - {r}_{2}\right) \left(x - {r}_{3}\right) \ldots \left(x - {r}_{n}\right)$

In this case, we are given that k = 1, and the roots are $5 , 3 i , \mathmr{and} - 3 i$. This makes the factors:

$y = \left(x - 5\right) \left(x - 3 i\right) \left(x - - 3 i\right)$

I will multiply the factors one-pair-at-a-time:

y = (x - 5)(x² + 9)

y = x³ - 5x² + 9x + 45