You can put this solution on YOUR website! If a root is 3+i, there is a root 3-i, because complex roots are conjugate pairs.
Two factors are (x-2)(x+4)
To get roots of 3+/- i, we need b=-6 and sqrt(b^2-4ac) to equal sqrt(-4) since we are dividing by 2 and that will give +/- i
b^2=36, a is 1, so c must be 10 so that 4c=40
(x^2-6x+10) is the quadratic
The polynomial is (x^2+2x-8)(x^2-6x+10)
or x^4-4x^3-10x^2+68x-80.
