Find the conjuages of a given algebraic element over QQ

Find the minimal polynomial and conjugates of each of the following elements over \mathbb{Q}: \sqrt{3}, 4, \sqrt{-3}/5, and \sqrt[3]{5}.


Note that p(x) = x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3}) has \sqrt{3} as a root. p(x) is Eisenstein at 3, hence irreducible over \mathbb{Q}. Thus p(x) is the minimal polynomial of \sqrt{3} over \mathbb{Q}. The conjugates of \sqrt{3} are \pm \sqrt{3}.

Note that q(x) = x - 4 has 4 as a root. Since q(x) is linear, it is irreducible, and so is the minimal polynomial of 4 over \mathbb{Q}. The only conjugate of 4 is 4 itself.

Note that t(x) = 25x^2 + 3 = (5x + \sqrt{3})(5x - \sqrt{3}) has \sqrt{-3}/5 as a root. t(x) is Eisenstein at 3, hence irreducible over \mathbb{Q}. Thus t(x) is the minimal polynomial of \sqrt{-3}/5. The conjugates of \sqrt{-3}/5 are \pm \sqrt{-3}/5.

Note that s(x) = x^3-5 = (x - \sqrt[3]{5})(x - \sqrt[3]{5}(\frac{-1}{2} + \frac{\sqrt{3}}{2}i))(x - \sqrt[3]{5}(\frac{-1}{2} - \frac{\sqrt{3}}{2}i)) has \sqrt[3]{5} as a root.s(x) is Eisenstein at 5, hence irreducible over \mathbb{Q}. The conjugates of \sqrt[3]{5} are \sqrt[3]{5}, \sqrt[3]{5}(\frac{-1}{2} + \frac{\sqrt{3}}{2}i), and \sqrt[3]{5}(\frac{-1}{2} - \frac{\sqrt{3}}{2}i).

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