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Given that $\left(\sqrt3+\sqrt5\right)$ is one zero of a fourth-degree polynomial with integer coefficients and leading coefficient 1, how can the constant term of this polynomial be found?

I know that $\left(\sqrt3-\sqrt5\right)$ must also be a root because it is the conjugate. How can I determine the other two roots (and ultimately, the constant term) beyond what I have right now?

$(x^2-2\sqrt3x-2)(x-r_1)(x-r_2)$

Thanks!

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  • $\begingroup$ The other two roots are $-\sqrt3+\sqrt5$ and $-\sqrt3-\sqrt5$ $\endgroup$ – Brian Cheung Jan 8 '17 at 6:27
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We know that $\sqrt 3+\sqrt 5$ is a root of the polynomial $\sqrt 3+\sqrt 5=x$, so we begin from there and eliminate radicals to obtain a polynomial over $\Bbb Z$:

\begin{align} \sqrt 3+\sqrt 5&=x\\ \sqrt 3&=x-\sqrt 5\\ 3&=x^2-2\sqrt 5 x+5\\ 2\sqrt 5x&=x^2+2\\ 20x^2&=x^4+4x^2+4\\ 0&=x^4-16x^2+4. \end{align}

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