Question

Find all other zeros of P(x)=x^3-3x^2+54.

Given that 3+3i is a zero.

Answer 1

-3, 3-3i Is that correct?

Answer 2

If 3+3i is a zero, then 3-3i is also.
Then there must be one real zero.
Let's try -3: P(-3)=(-3)³-3(-3)²+54=-27-27+54=-54+54=0. Bingo!

Answer 3

Are those the only ones @ZeHanz ? My roots of 54 are +/-{1,2,27,54,9,6,18,3. My roots of 3 are +/- 1,3.

Answer 4

A 3rd degree polynomial equation has exactly three solutions.
You are given one solution: 3+3i.
Then 3-3i is also a (nonreal) solution.
There is therefore only one solution left. Is must be real, because a 3rd degree polynomial has at least one real solution.
Think about it: the graph of a 3d deg polynomial looks like this:
So -3 works and thus there are no more.

Your DIVIDERS of 54 are the numbers you mentioned,
Your DIVIDERS of 3 are the numbers you mentioned.
According to the Rational Root Theorem the POSSIBLE rational roots are the dividers of 54 divided by those of 3.
The number -3 we've identified as root fits this: e.g. 9/-3=-3.