Why does $\sqrt{6} + \sqrt{10} + \sqrt{15}$ have four conjugates? I am having trouble understanding how algebraic number $\sqrt{6} + \sqrt{10} + \sqrt{15}$ has four conjugates.
Minimal polynomial is $x^4-62 x^2-240 x-239$ according to Wolfram Alpha.
Factorized:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)\cdot
 \left(x-2\sqrt{4-\sqrt{15}}+\sqrt{15}\right) \\
\cdot \left(x+2\sqrt{4-\sqrt{15}}+\sqrt{15}\right)
\cdot\left(x+2\sqrt{15 (4-\sqrt{15})}+8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)$$

In general we would guess that $\sqrt a+\sqrt b+\sqrt c$ has eight conjugates, obtainable by toggling signs individually for the surds. However, in this special case we see that $\sqrt a\sqrt b\sqrt c=30$, which cannot change its sign. Hence once we picked the sign of two of the surds, the sign of the third is determined.
