Question

How do you determine the remaining zeroes for `h(x)=3x^4+5x^3+25x^2+45x-18` if 3i is a zero?

Answer 1

See below

Explanation:

If `3i` is a zero, then `-3i` must be a zero, so using those zeroes, I can write these factors:
`(x-3i)(x+3i) = (x^2+9)`

Now we can use long division to divide `3x^4+5x^3+25x^2+45x-18` by `(x^2+9)` to get `3x^2+5x-2`

`color(white)(mmmmmmm)3x^2+5x-2`
`x^2+0x+9|3x^4+5x^3+25x^2+45x-18`
`color(white)(mmmmm)-(3x^4+0x^3+27x^2)`
`color(white)(mmmmmmmmmm)5x^3-2x^2+45x`
`color(white)(mmmmmmmm)-(5x^3+0x^2+45x)`
`color(white)(mmmmmmmmmmmm) -2x^2-18`
`color(white)(mmmmmmmmmm)-(-2x^2-18)`
`color(white)(mmmmmmmmmmmmmmmmm)0`

Now factor:
`3x^2+5x-2`
`3x^2+6x-x-2`
`3x(x+2)-1(x+2)`
`(3x-1)(x+2)`

`x=-2, 1/3, +-3i`

graph{3x^4+5x^3+25x^2+45x-18 [-10, 10, -5, 5]}