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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
 #1
avatar+6042 
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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

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