# Tagged Questions

267 views

### Polynomials with Integer Coefficients and irrational roots

Is there a polynomial with integer coefficients which has √2 +√7 ￼ as a root?
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### Understanding non-solvable algebraic numbers

Background We know from Galois theory that the zeros of a polynomial with rational coefficients whose Galois group is solvable can be expressed in a formula that involves rational powers of the ...
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### Polynomial in $\mathbb{Q}[x]$ with root $\sqrt[3]{2}+\sqrt[3]{3}$

What is a polynomial $P(x)\in \mathbb{Q}[x]$ with root $\sqrt[3]{2}+\sqrt[3]{3}$? I write $x=\sqrt[3]{2}+\sqrt[3]{3}$, so $(x-\sqrt[3]{2})^3=3$, but the expansion of the left side contains two cube ...
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### The Conjugate Roots Theorem for Irrational Roots

The Conjugate Roots Theorem for Irrational Roots states that for a polynomial $f(x)$ with integer coefficients, if a root of the equation $f(x) = 0$ is expressed as $a+\alpha$, where $a\in\mathbb{Q}$ ...
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### Third degree polynomial with integer coefficient and three irrational roots

There are some polynomial with the above characteristic, and real roots of such polynomials cannot be found using rational number theorem and irrational conjugate theorem. The example of such function ...
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### Proof $\sqrt{1 + \sqrt[3]{2}}$ is irrational using the theorem about rational roots of a polynomial

I'm having trouble with this specific problem at the moment. The theorem states that if $n/m$ is a rational root of a polynomial with integer coefficients, the leading coefficient is divisible by m ...
The title pretty much asks my question: Does $f\in\mathbb{Q}[x]$ such that $$f(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3),$$ where $\alpha_1, \alpha_2, \alpha_3\in\mathbb{R}\setminus\mathbb{Q}\$ ...