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:
$$\left(x-2\sqrt{15 (4-\sqrt{15})}-8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)\cdot
 \left(x-2\sqrt{4-\sqrt{15}}+\sqrt{15}\right) \\
\cdot \left(x+2\sqrt{4-\sqrt{15}}+\sqrt{15}\right)
\cdot\left(x+2\sqrt{15 (4-\sqrt{15})}+8\sqrt{4-\sqrt{15}}-\sqrt{15}\right)$$
 A: 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.
A: $$\sqrt{6}+\sqrt{10}+\sqrt{15}=\sqrt{2\cdot 3}+\sqrt{2\cdot 5}+\sqrt{3\cdot 5} = \frac{1}{2}\left(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2-(2+3+5)\right)$$
where $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an algebraic number of degree $8$ over $\mathbb{Q}$, having conjugates $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$, whose minimal polynomial is an even function. It follows that $\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2$ is an algebraic number of degree $4$ over $\mathbb{Q}$, and the conjugates of $\sqrt{6}+\sqrt{10}+\sqrt{15}$ are given by:
$$ \frac{1}{2}\left(\left(\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}\right)^2-(2+3+5)\right).$$
A: The conjugates of an algebraic number are (by definition) the roots of its minimal polynomial. The number of (distinct) roots of an irreducible polynomial over the rationals is equal to its degree, that is four. 
Thus once you know the minimal polynomial "it is clear."
There is some wiggling room as one might or might not count the number itself among its conjugates. But the former is more common. 
(This answer leaves open the question "why" the minimal polynomial has degree four or how it could be found as this is covered in other answers.)
