The roots (or zeroes) are sqrt(2), 3i, and, the conjugate, -3i. Your polynomial must satisfy:
(x-sqrt(2))(x-3i)(x+3i) = 0
Note that the highest order term is x3 , so the leading coefficient is 1 (You could make it any number by multiplying every term in the polynomial by a constant)
Also the last two terms combine to x2+9
Distribute to get the polynomial.
Mark M.
Do not radical zeros come in pairs? E.g., sqrt 2 and - sqrt 2
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01/06/22
JACQUES D.
tutor
Good point and it may have been intended in the problem. You are right if the polynomial has rational coefficients and it has a root like a+sqrt(b). Otherwise, even with rational coefficients, uneven numbers of rational roots are possible. I assumed that irrational coefficients were ok.
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01/06/22
Umer S.
thanks a lot01/05/22