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Find a monic quartic polynomial f(x) with rational coefficients whose roots include $$x=3-i\sqrt[4]2$$

Jun 13, 2020

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Coefficients are rational, which means they are real. This means the conjugate of x is also a root.

$$x = 3 - i\sqrt[4]{2}\text{ and }x = 3 + i \sqrt[4]2\text{ are roots.}$$

Now consider the product of roots. The product of roots is just the constant term, which means it must be rational as well.

By Fundamental Theorem of Algebra, the quartic polynomial must have 4 complex roots.

Let a and b be the other roots.

$$ab(3 - i\sqrt[4]2)(3 + i\sqrt[4]2) = ab(9 + \sqrt 2)$$

This means ab is a rational multiple of the conjugate of $$9 + \sqrt 2$$.

Notice that the polynomial is not unique, so we can let ab = $$9 - \sqrt 2$$ without loss of generality, where k is a rational constant.

Consider the sum of roots. The sum of roots is -1 * (coefficient of $$x^3$$), therefore the sum of roots must be rational as well.

This means that a and b are the conjugates of each other.

It is not hard to see that $$(a, b) = ((3 + \sqrt[4]2), (3 - \sqrt[4]2))$$ is a possible combination, where k1 and k2 are rational.

One of the possible polynomials are $$(x - (3 - i\sqrt[4]2))(x - (3 + i\sqrt[4]2))(x - (3 - \sqrt[4]2))(x - (3 + \sqrt[4]2))$$, which is $$f(x) = x^4 - 12x^3 + 54x^2 - 108x + 79$$

Jun 13, 2020