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Irrational Roots

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There is a unique polynomial P(x) of degree 4 with rational coefficients and leading coefficient 1 which has $$\sqrt{2}+\sqrt{5}$$ as a root. What is P(1)?

I know one part of it is $$x^2-2x-3$$ already.

May 26, 2019

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$$\text{The constant term of a polynomial is the product of it's roots}\\ \text{So if \sqrt{2}+\sqrt{5} is a root and c_0 is rational it must be that\\ \sqrt{2}-\sqrt{5} is also a root}\\ \text{This is where you probably got your idea that x^2-2x-3 is part of it\\ You made a bit of an error. The actual factor is x^2-2\sqrt{2}x-3}\\ \text{Now given that is a factor, and we've got rational coefficients, it must be that \\x^2 +2\sqrt{2}x-3 is also a factor}\\ \text{That gets us \\P(x) = (x^2+2\sqrt{2}x-3)(x^2-2\sqrt{2}x-3) = x^4-14 x^2+9}\\ P(1)=1-14+9=-4$$

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May 26, 2019
edited by Rom  May 26, 2019