Answer:
The remaining zeros of f is (2 - i) and (-3 + i).
Explanation:
We are given a degree six polynomial f and four of its zeros:

And we want to find the remaining zeros of f.
By the Fundamental Theorem of Algebra, the number of zeros of any polynomial is equal to its degree.
Hence, a sixth degree polynomial must have six zeros.
Because we are given four zeros, f has two more zeros.
To find the remaining two zeros, recall the Complex Conjugate Root Theorem:

Our two complex zeros are (2 + i) and (-3 - i).
Then by the above theorem, (2 - i) and (-3 + i) is the two remaining zeros of f.