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1. Find a monic quartic polynomial $$f(x)$$ with rational coefficients whose roots include $$x=1-\sqrt 2$$ and $$x=2+\sqrt 5$$. Give your answer in expanded form.

So I know I'm supposed to do something with their conjugates, but after that idk.

2.

(a) Find a cubic polynomial with integer coefficients that has $$\sqrt[3]{2} + \sqrt[3]{4}$$ as a root.

(b) Prove that $$\sqrt[3]{2} + \sqrt[3]{4}$$ is irrational.

I know for (a) I should simplify it, but after that idk.

For part (b) I don't know.

Aug 11, 2020

#1
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2.

Since are looking for a cubic polynomial, let it be ax^3 + bx^2 + cx + d = 0.  So

$$a(\sqrt[3]{2} + \sqrt[3]{4})^3 + b(\sqrt[3]{2} + \sqrt[3]{4})^2 + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0$$

This expands to

$$a (2 + 3 \sqrt[3]{2^2} \sqrt[3]{4} + 3 \sqrt[3]{2} \sqrt[3]{4^2} + 4) + b(\sqrt[3]{2^2} + 2 \sqrt[3]{2} \sqrt[3]{4} + \sqrt[3]{4^2}) + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0$$

Expanding everything out, and comparing the coefficients, we get

a + 3b + 3c + d = 16,
-6b + 3c + 3d = -24,
c - 3d = 22,
d = -6.

The solution to this system is a = 1, b = 3, c = 4, d = -6, so the cubic is x^3 + 3x^2 + 4x - 6.

Aug 11, 2020
#2
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1.   Try  $$f(x) = (x-\sqrt2)(x+\sqrt2)(x+\sqrt5)(x-\sqrt5)$$

I'll leave you to expand it.

Aug 11, 2020