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There is a unique polynomial P(x) of degree 4 with rational coefficients and leading coefficient 1 which has sqrt(1 + sqrt(6)) as a root. What is P(1)?

 Jun 7, 2021
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Let $x=\sqrt{1+\sqrt6}$.  Then $x^4=7+2\sqrt6$, that is, $x^4-7=2\sqrt6$,
so that $(x^4-7)^2 =24$, so that $(x^4-7)^2-24=0$.
This means $\sqrt{1+\sqrt6}$ is a zero of the polynomial
$q(x)=(x^4-7)^2-24$.
Now $q(x)=(x^4-5-2x^2)(x^4-5+2x^2)$.   Let $p(x) = x^4-5-2x^2$.
We then find that $\sqrt{1+\sqrt6}$ is a zero of $p(x)$, and $p(x)$ is a
degree-4 polynomial with integer coefficients and leading coefficient 1.
It's given that such a polynomial $p$ is unique. Thus,
$p(1)=1^4-5-2\cdot 1^2= 1-5-2=-6$.

 Jun 7, 2021

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