Use linear algebra to find an $n$th-degree polynomial with a zero at $x = \sqrt{3} + \sqrt{5} + \sqrt{3 + 4\sqrt{5}}$.

Question

It was easy enough (though tedious) to remove the radicals in the given zero algebraically. I got (and double-checked with software): $$x^8 - 44x^6 - 160x^5 + 254x^4 + 1600x^3 + 4740x^2 + 8800x - 5975$$But, I'm wondering how I might solve it with matrix methods. I can see a system of equations. Letting $a=\sqrt{3}, b=\sqrt{5}$, and $c =\sqrt{3 + 4\sqrt{5}}$, we get $$x-a-b-c = 0$$ $$ a^2 - 3 = 0$$ $$b^2 -5 = 0$$ $$c^2-4b-3 = 0$$But I'm stuck here. How can I set this up (and solve it) as a linear algebra problem?

Answer 1

Here's a way to solve it using matrix methods, but it's very tedious.

The field $K$ generated over the rationals by $\sqrt3$, $\sqrt5$, and $\sqrt{3+4\sqrt5}$ is an 8-dimensional vector space over the rationals, with basis $$ \{1,\sqrt3,\sqrt5,\sqrt{15},\sqrt{3+4\sqrt5},\sqrt{9+12\sqrt5},\sqrt{15+20\sqrt5},\sqrt{45+60\sqrt5}\} $$ Express each of the nine numbers $1,x,x^2,\dots,x^8$ as a linear combination of these basis elements. Then $a_0+a_1x+a_2x^2+\cdots+a_7x^7+x^8=0$ leads to eight linear equations in the eight unknowns $a_0,\dots,a_7$.