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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.

 

Thank you in advance!

 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
avatar+30921 
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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

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