Is irrational roots occuring in conjugate pairs valid only for Binomial surds Its well know that if a polynomial has Rational coefficients then irrational roots occur in conjugate pairs. 
Will it be true for trinomial surds like for example if a polynomial has a root 
$\sqrt{2}+\sqrt{3}+1$ Then we get a polynomial as 
$$x=\sqrt{2}+\sqrt{3}+1$$
$$x-1=\sqrt{2}+\sqrt{3}$$ Squaring both sides
$$x^2-2x+1=5+2\sqrt{6}$$
$$x^2-2x-4=2\sqrt{6}$$ again squaring we get
$$x^4-4x^3-4x^2+16x-8=0$$ will the other root be its rationalizing factor
 A: If one finds the least degree integer polynomial having $c=\sqrt{2}+\sqrt{3}+\sqrt{6}$ as one of its roots [when set to $0$], then that polynomial $p(x)$ turns out to have degree $4.$ Since it has only degree $4$ one would expect that not all eight possible variations on $c$ obtained by changing $0$ or more signs on the three radicals will also be roots of $p(x)=0,$ and indeed one can check that the four roots are $c$ itself (no surprise there) along with the three other roots obtained from $c$ by changing exactly two signs before the three radicals. 
So in general, if one defines a conjugate of a surd as anything obtained by changing one or more signs before the radicals, the results may or may not be zeros of the starting polynomial generated from the initial surd.
I also looked at the case of $c'=\sqrt{2}+\sqrt{3}+\sqrt{5}$ in the same way; this time the least degree polynomial $q(x)$ for $c'$ turns out to have degree $8,$ and a check reveals that here all eight "conjugates" obtained from $c'$ by changing zero or more signs are roots of $q(x)=0.$
Though I'm not really up on the topic, I think the relevant thing to look at for this question is Galois Theory.
