A polynomial of degree 4 will have 4 zeros. The zero of 3 with a multiplicity of 2 counts as two of these zeros. You've also been given a zero of -i. So it seems that we only have 3 of the 4 zeros. We need all 4 to be able to form the desired polynomial.
You didn't mention that the polynomial should have real coefficients. But I'm going to assume it because otherwise we can't solve the problem. If a polynomial with real coefficients has a complex zero, then the complex conjugate of that complex zero will also be a zero. This is how we will find the 4th zero.
A complex number in standard form is a + bi. Its complex conjugate is a + (-bi) or a - bi. The complex zero we were given is -i. In standard form this would be 0 + (-i). The complex conjugate of 0 + (-i) would be 0 - (-i) or 0 + i or just i.
So the four zeros of the desired polynomial are 3 (twice), -i and i. If a number "z" is a zero of a polynomial then (x - z) is a factor of that polynomial. Using this fact and the four zeros we can write a factored form of our desired polynomial:
f%28x%29+=+a%28x-3%29%28x-3%29%28x-%28-i%29%29%28x-i%29
Simplifying the 3rd factor we get:
f%28x%29+=+a%28x-3%29%28x-3%29%28x%2Bi%29%28x-i%29
where the "a" is any non-zero number. Since there are an infinite number of non-zero numbers, there will be an infinite number of possible correct solutions to this problem. To make things easy we will choose to make "a" be 1. This makes our factored polynomial:
f%28x%29+=+%28x-3%29%28x-3%29%28x%2Bi%29%28x-i%29
All we have to do now is multiply this out. We can simplify this process if we take advantage of some factoring patterns. The first two factors can be multiplied using:
%28a-b%29%5E2+=+a%5E2-2ab%2Bb%5E2
And the last two factors can be multiplied using:
%28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2
With these patterns we get:
f%28x%29+=+%28%28x%29%5E2-2%28x%29%283%29%2B%283%29%5E2%29%28%28x%29%5E2-%28i%29%5E2%29
Simplifying inside each factor we get:
f%28x%29+=+%28x%5E2-6x%2B9%29%28x%5E2-i%5E2%29
And since i%5E2+=+-1 this becomes:
f%28x%29+=+%28x%5E2-6x%2B9%29%28x%5E2-%28-1%29%29
or
f%28x%29+=+%28x%5E2-6x%2B9%29%28x%5E2%2B1%29
We have one more multiplication. There are no shortcuts for this one. We have to multiply each term of one factor times each term of the other and then add like terms, if any:
f%28x%29+=+x%5E2%2Ax%5E2%2Bx%5E2%2A1%2B%28-6x%29%2Ax%5E2%2B%28-6x%29%2A1%2B9%2Ax%5E2%2B9%2A1
Simplifying...
f%28x%29+=+x%5E4%2Bx%5E2%2B%28-6x%5E3%29%2B%28-6x%29%2B9x%5E2%2B9
f%28x%29+=+x%5E4%2B10x%5E2%2B%28-6x%5E3%29%2B%28-6x%29%2B9
Re-ordering into standard form:
f%28x%29+=+x%5E4%2B%28-6x%5E3%29%2B10x%5E2%2B%28-6x%29%2B9
This is a polynomial function of degree 4 with the given zeros.
P.S. As mentioned earlier, this is not the only possible correct answer. This is the answer when you make "a" equal to 1. Other values for "a" give us different polynomial functions which also have the same zeros. For example, if we had chosen a = 2 we would have gotten f%28x%29+=+2x%5E4%2B%28-12x%5E3%29%2B20x%5E2%2B%28-12x%29%2B18.