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)?

Guest Jun 7, 2021

#1**0 **

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,

$p(1)=1^4-5-2\cdot 1^2= 1-5-2=-6$.

Bginner Jun 7, 2021