Douglas B. answered 06/05/20
Algebra tutor with masters degree in applied math
For 1), if we know some theory about polynomials, we can immediately see that the sqrt(6) cannot be a root. This is because (due to Galois), the roots of a fifth or higher degree polynomial cannot be expressed in terms of roots. We also know immediately that 2i cannot be a root. Complex roots of a polynomials with real coefficients always come in pairs. So, if 2i were a root, we would also have -2i being a root. But, because the polynomial is not an odd function, 2i and -2i cannot both be roots.
I used wolfram alpha to list out the roots, and they all appear to be irrational. So I don't know how you would be expected to calculate the roots yourself. Perhaps you made a typo in writing the polynomial.
For 2), we have g(x) = (x-2)^3-8+4 = (x-2)^3-4. Here, we have transformed the graph of f by shifting right by 2 units, and down by 4 units.
Hope this helps.