2006/07 AHSMC Part 1 Problem 13

Since $\dfrac{\sqrt{2}}{3}$ and $\dfrac{\sqrt{3}}{2}$ are roots, we must also have $-\dfrac{\sqrt{2}}{3}$ and $-\dfrac{\sqrt{3}}{2}$ as roots in order to have rational coefficients. Thus the degree of the polynomial cannot be less than $4$, but $4$ is sufficient. Of these polynomials, the one with $1$ as the leading coefficient is $$\left(x - \dfrac{\sqrt{2}}{3}\right)\left(x + \dfrac{\sqrt{2}}{3}\right)\left(x - \dfrac{\sqrt{3}}{2}\right)\left(x + \dfrac{\sqrt{3}}{2}\right) = \left(x^2 - \dfrac{2}{9}\right)\left(x^2 - \dfrac{3}{4}\right) = x^4 - \dfrac{35}{36}x^2 + \dfrac{1}{6}.$$ Since we want integral coefficients, we clear the denominators to obtain $36x^4 - 35x^2 + 6$. Since the coefficients are relatively prime (though not pairwise relatively prime) and the smallest absolute value among the coefficients is $6$, the smallest positive coefficient is also $6$. The answer is (b).