A fourth-degree polynomial with integer coefficients has zeros at 1 and 3 +[tex] \sqrt{5} [/tex] Which number cannot also be a zero of this polynomial?A. 1B. -3C. 3-[tex] \sqrt{5 [/tex]D. 3+[tex] \sqrt{2}[/tex]

If a radical be a zero of a polynomial then it's conjugate will also be the zero. Like for the given problem has a zero 3+ √5. So, it's conjugate 3- √5 must be the other zero.Hence, the zeroes of the polynomial are 1, 3+ √5 and 3-√5.According to the problem, the degree of the polynomial is 4. We already have three zeroes, so only one zero is unknown. Because degree of polynomial = number of zeroes.Other zeroes can be 1 and -3. We have already found that 3 - √5 is one of the zero.So, A, B, C cannot be the correct choice.But if 3+ √2 will be a zero then 3- √2 will also be the zero. Then the polynomial will have 5 zeroes which is not possible as the polynomial has degree four.So, D: 3+ √2 cannot also be a zero of this polynomial.Hope this helps you!